এ অধ্যায়ে আমরা যে বিষয় গুলি আলোচনা করব।
- সার সংক্ষেপ
- ত্রিকোণমিতিক সমীকরণের সাধারণ সমাধান
- \(\sin{\theta}=0\) এর সমাধান \(\theta=n\pi\)
- \(\cos{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
- \(\tan{\theta}=0\) এর সমাধান \(\theta=n\pi\)
- \(\cot{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
- \(\sin{\theta}=\sin{\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
- \(cosec \ {\theta}=cosec \ {\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
- \(\cos{\theta}=\cos{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
- \(\sec{\theta}=\sec{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
- \(\tan{\theta}=\tan{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)
- \(\cot{\theta}=\cot{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)
- \(\sin{\theta}=1\) এর সমাধান \(\theta=(4n+1)\frac{\pi}{2}\)
- \(\cos{\theta}=1\) এর সমাধান \(\theta=2n\pi\)
- \(\sin{\theta}=-1\) এর সমাধান \(\theta=(4n-1)\frac{\pi}{2}\)
- \(\cos{\theta}=-1\) এর সমাধান \(\theta=(2n+1)\pi\)
- ত্রিকোণমিতিক সমীকরণের অবান্তর মূল
- নির্দিষ্ট ব্যবধিতে ত্রিকোণমিতিক সমীকরণের সমাধান
- অধ্যায় \(7H\)-এর উদাহরণসমুহ
- অধ্যায় \(7H\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
- অধ্যায় \(7H\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(7H\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(7H\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(7H\) / \(Q.5\)-এর সৃজনশীল প্রশ্নসমূহ
- অধ্যায় \(7H\) / \(Q.6\)-এর ভর্তি পরীক্ষায় আশা প্রশ্নসমূহ

সার সংক্ষেপ
ত্রিকোণমিতিক সমীকরণের সাধারণ সমাধান
\(\sin{\theta}=0\) এর সমাধান \(\theta=n\pi\) \(\cos{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\) \(\tan{\theta}=0\) এর সমাধান \(\theta=n\pi\) \(\cot{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\) \(\sin{\theta}=\sin{\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\) \(cosec \ {\theta}=cosec \ {\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\) \(\cos{\theta}=\cos{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\) \(\sec{\theta}=\sec{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\) \(\tan{\theta}=\tan{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\) \(\cot{\theta}=\cot{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\) \(\sin{\theta}=1\) এর সমাধান \(\theta=(4n+1)\frac{\pi}{2}\) \(\cos{\theta}=1\) এর সমাধান \(\theta=2n\pi\) \(\sin{\theta}=-1\) এর সমাধান \(\theta=(4n-1)\frac{\pi}{2}\)\(\cos{\theta}=-1\) এর সমাধান \(\theta=(2n+1)\pi\) ত্রিকোণমিতিক সমীকরণের অবান্তর মূলনির্দিষ্ট ব্যবধিতে ত্রিকোণমিতিক সমীকরণের সমাধান
ত্রিকোণমিতিক সমীকরণের সাধারণ সমাধান
General solution of trigonometric equations
এক বা একাধিক ত্রিকোণমিতিক অনুপাত সম্বলিত সমীকরণকে ত্রিকোণমিতিক সমীকরণ বলা হয়। সমীকরণটির সংশ্লিষ্ট কোণকেই চলরাশি বলা হয়। চলরাশির যে সকল মানের জন্য প্রদত্ত সমীকরণটি সিদ্ধ হয়, সেগুলিকে সমীকরণটির সমাধান বলা হয়। সকল সমাধানকে অভিন্ন রাশিমালাতে প্রকাশ করা হলে সেটিকে সাধারণ সমাধান বলে।
ত্রিকোণমিতিক সমীকরণকে উৎপাদকে প্রকাশ করে যে সকল সরল সমীকরণ পাওয়া যায় সেগুলিকে মূল সমীকরণের প্রতীক সমীকরণ বলা হয়।
যেমনঃ \(2\tan^2{\theta}-3\tan{\theta}+1=0\) একটি ত্রিকোণমিতিক সমীকরণ।
এই সমীকরণটিকে উৎপাদকে বিশ্লেষণ করলে পাওয়া যায়,
\((2\tan{\theta}-1)(\tan{\theta}-1)=0\)
এখানে, প্রতীক সমীকরণ \(2\tan{\theta}-1=0\) এবং \(\tan{\theta}-1=0\)
ত্রিকোণমিতিক সমীকরণের এক বা একাধিক প্রতীক সমীকরণ থাকতে পারে। সে ক্ষেত্রে একটি প্রতীক সমীকরণের সাধারণ সমাধান মূল সমীকরণের সাধারণ সমাধান নয়। সকল প্রতীক সমীকরণের সাধারণ সমাধানই একত্রে মূল সমীকরণের সাধারণ সমাধান। এ অধ্যায়ে আমরা ত্রিকোণমিতিক সমীকরণের সাধারণ সমাধান নির্ণয় অধ্যয়ন করব।
ত্রিকোণমিতিক সমীকরণকে উৎপাদকে প্রকাশ করে যে সকল সরল সমীকরণ পাওয়া যায় সেগুলিকে মূল সমীকরণের প্রতীক সমীকরণ বলা হয়।
যেমনঃ \(2\tan^2{\theta}-3\tan{\theta}+1=0\) একটি ত্রিকোণমিতিক সমীকরণ।
এই সমীকরণটিকে উৎপাদকে বিশ্লেষণ করলে পাওয়া যায়,
\((2\tan{\theta}-1)(\tan{\theta}-1)=0\)
এখানে, প্রতীক সমীকরণ \(2\tan{\theta}-1=0\) এবং \(\tan{\theta}-1=0\)
ত্রিকোণমিতিক সমীকরণের এক বা একাধিক প্রতীক সমীকরণ থাকতে পারে। সে ক্ষেত্রে একটি প্রতীক সমীকরণের সাধারণ সমাধান মূল সমীকরণের সাধারণ সমাধান নয়। সকল প্রতীক সমীকরণের সাধারণ সমাধানই একত্রে মূল সমীকরণের সাধারণ সমাধান। এ অধ্যায়ে আমরা ত্রিকোণমিতিক সমীকরণের সাধারণ সমাধান নির্ণয় অধ্যয়ন করব।
প্রয়োজনীয় ও স্বরণীয় সূত্রসমূহ
Necessary and memorable formulas
\(\sin{\theta}=0\) এর সমাধান \(\theta=n\pi\)\(\cos{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
\(\tan{\theta}=0\) এর সমাধান \(\theta=n\pi\)\(\cot{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
\(\sin{\theta}=\sin{\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)\(cosec \ {\theta}=cosec \ {\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
\(\cos{\theta}=\cos{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)\(\sec{\theta}=\sec{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
\(\tan{\theta}=\tan{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)\(\cot{\theta}=\cot{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)
\(\sin{\theta}=1\) এর সমাধান \(\theta=(4n+1)\frac{\pi}{2}\)\(\cos{\theta}=1\) এর সমাধান \(\theta=2n\pi\)
\(\sin{\theta}=-1\) এর সমাধান \(\theta=(4n-1)\frac{\pi}{2}\)\(\cos{\theta}=-1\) এর সমাধান \(\theta=(2n+1)\pi\)
ত্রিকোণমিতিক সমীকরণের অবান্তর মূল
Extraneous roots of trigonometric equations
ত্রিকোণমিতিক সমীকরণ বিভিন্ন উপায়ে সমাধান করা হয় এবং বিভিন্ন প্রক্রিয়ায় সমাধান করে প্রাপ্ত মূলগুলি দৃশ্যত ভিন্ন আকারের হলেও সেগুলি সমতুল্য। কিছু কিছু সমীকরণকে বর্গ করে সমাধান নির্ণয় করে হয়। এ প্রক্রিয়া কিছুটা ত্রুটিপূর্ণ বলে প্রাপ্ত মূলগুলির কোনো কোনোটি প্রদত্ত সমীকরণকে সিদ্ধ করে না। এরূপ মূলকে অবান্তর মূল বলে। সুতরাং প্রকৃত মূল নির্ণয় করার জন্য প্রাপ্ত মূলগুলি দিয়ে প্রদত্ত সমীকরণ সিদ্ধ হয় কিনা তা পরীক্ষা করার প্রয়োজন হয়। যেমনঃ \(\sqrt{3}\sin{\theta}+\cos{\theta}=\sqrt{2}\)
\(\Rightarrow \sqrt{3}\sin{\theta}=\sqrt{2}-\cos{\theta}\)
\(\Rightarrow 3\sin^2{\theta}=2-2\sqrt{2}\cos{\theta}+\cos^2{\theta}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow 3(1-\cos^2{\theta})-2+2\sqrt{2}\cos{\theta}-\cos^2{\theta}=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 3-3\cos^2{\theta}-2+2\sqrt{2}\cos{\theta}-\cos^2{\theta}=0\)
\(\Rightarrow -4\cos^2{\theta}+2\sqrt{2}\cos{\theta}+1=0\)
\(\Rightarrow -(4\cos^2{\theta}-2\sqrt{2}\cos{\theta}-1)=0\)
\(\Rightarrow 4\cos^2{\theta}-2\sqrt{2}\cos{\theta}-1=0\)
\(\Rightarrow \cos{\theta}=\frac{2\sqrt{2}\pm{\sqrt{(-2\sqrt{2})^2-4\times4\times{-1}}}}{2\times4}\) ➜ \(\because ax^2+bx+c=0\)
\(\therefore x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}\)
\(\Rightarrow \cos{\theta}=\frac{2\sqrt{2}\pm{\sqrt{8+16}}}{8}\)
\(\Rightarrow \cos{\theta}=\frac{2\sqrt{2}\pm{\sqrt{24}}}{8}\)
\(\Rightarrow \cos{\theta}=\frac{2\sqrt{2}\pm{2\sqrt{6}}}{8}\)
\(\Rightarrow \cos{\theta}=\frac{2(\sqrt{2}\pm{\sqrt{6}})}{8}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{2}\pm{\sqrt{6}}}{4}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{2}\pm{\sqrt{2}\sqrt{3}}}{4}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{2}(1\pm{\sqrt{3}})}{2\times\sqrt{2}\times\sqrt{2}}\)
\(\Rightarrow \cos{\theta}=\frac{1\pm{\sqrt{3}}}{2\sqrt{2}}\)
\(\Rightarrow \cos{\theta}=\frac{1+\sqrt{3}}{2\sqrt{2}}, \ \cos{\theta}=\frac{1-\sqrt{3}}{2\sqrt{2}}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{3}+1}{2\sqrt{2}}, \ \cos{\theta}=-\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\Rightarrow \cos{\theta}=\cos{15^{o}}, \ \cos{\theta}=-\sin{15^{o}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{12}}, \ \cos{\theta}=\cos{(90^{o}+15^{o})}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{12}}, \ \cos{\theta}=\cos{(105^{o})}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{12}}, \ \cos{\theta}=\cos{\frac{7\pi}{12}}\)
\(\therefore \theta=2n\pi\pm{\frac{\pi}{12}}, \ \theta=2n\pi\pm{\frac{7\pi}{12}}\)
\(\theta=2n\pi-\frac{\pi}{12}, \ \theta=2n\pi-\frac{7\pi}{12}\) বসিয়ে দেখা যায় যে, সমীকরণটি সিদ্ধ করে না। সমীকরটি বর্গ করা হয়েছে বলে এই ভুল সমাধান বের হয়েছে।
নির্দিষ্ট ব্যবধিতে ত্রিকোণমিতিক সমীকরণের সমাধান
Solution of trigonometric equation in a finite interval
ত্রিকোণমিতিক ফাংশন পর্যায়বৃত্ত হওয়ায় এর মানগুলো পর্যায়ক্রমে আবর্তিত হয়। নির্দিষ্ট ব্যবধিতেও সমাধানের পুনরাবৃত্তি হয়। ত্রিকোণমিতিক সমীকরণের সাধারণ সমাধান থেকে নির্দিষ্ট ব্যবধিতে অবস্থিত মানগুলি নির্ণয় করা যায়। আবার লেখচিত্রের সাহায্যে নির্দিষ্ট ব্যবধিতে সমীকরণের সমাধানগুলি নির্ণয় করা যায়। নিম্নে লেখের সাহায্যে সমাধান নির্ণয়ের কৌশল আলোচনা করা হলো। \((1)\) প্রদত্ত ব্যবধিতে ফাংশনের লেখচিত্র অঙ্কন করতে হবে।
\((2)\) ফাংশনের নির্দেশিত স্থানে \(x\) অক্ষের সমান্তরাল রেখা আঁকতে হবে।
\((3)\) সমান্তরাল রেখাটি ফাংশনকে যতবার ছেদ করবে ঠিক ততটি সমাধান বিদ্যমান থাকবে।
\((4)\) ছেদবিন্দুগুলির স্থানাংক, প্রতিসমতা বা ফাংশনের পর্যায়কাল ব্যবহার করে নির্ণয় করতে হবে। অতঃপর ছেদবিন্দুগুলির \(x\) স্থানাংকই ফাংশনের সমাধান হবে।
\(\sin{\theta}=0\) এর সমাধান \(\theta=n\pi\)\(\cos{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
প্রমাণঃ
প্রথম প্রমাণঃ
দেওয়া আছে,\(\sin{\theta}=0\)
\(\Rightarrow \sin{\theta}=\sin{0^{o}}\)
\(\Rightarrow \sin{\theta}=\sin{\left(\pm{\pi}\right)}\)
\(\Rightarrow \sin{\theta}=\sin{\left(\pm{2\pi}\right)}\)
\(\Rightarrow \sin{\theta}=\sin{\left(\pm{3\pi}\right)}\)
\(....................\)
\(....................\)
\(\Rightarrow \sin{\theta}=\sin{\left(n\pi\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{\theta}=0\) এর সমাধান \(\theta=n\pi\)
দ্বিতীয় প্রমাণঃ
দেওয়া আছে,\(\cos{\theta}=0\)
\(\Rightarrow \cos{\theta}=\cos{\left(\frac{\pi}{2}\right)}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pm\frac{3\pi}{2}\right)}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pm\frac{5\pi}{2}\right)}\)
\(....................\)
\(....................\)
\(\Rightarrow \cos{\theta}=\cos{\left\{\frac{(2n+1)\pi}{2}\right\}}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
\(\tan{\theta}=0\) এর সমাধান \(\theta=n\pi\)\(\cot{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
প্রমাণঃ
প্রথম প্রমাণঃ
দেওয়া আছে,\(\tan{\theta}=0\)
\(\Rightarrow \tan{\theta}=\tan{0^{o}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm{\pi}\right)}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm{2\pi}\right)}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm{3\pi}\right)}\)
\(....................\)
\(....................\)
\(\Rightarrow \tan{\theta}=\tan{\left(n\pi\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{\theta}=0\) এর সমাধান \(\theta=n\pi\)
দ্বিতীয় প্রমাণঃ
দেওয়া আছে,\(\cot{\theta}=0\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}=0\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \cos{\theta}=0\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pm\frac{\pi}{2}\right)}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pm\frac{3\pi}{2}\right)}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pm\frac{5\pi}{2}\right)}\)
\(....................\)
\(....................\)
\(\Rightarrow \cos{\theta}=\cos{\left\{\frac{(2n+1)\pi}{2}\right\}}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cot{\theta}=0\) এর সমাধান \(\theta=(2n+1)\frac{\pi}{2}\)
\(\sin{\theta}=\sin{\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)\(cosec \ {\theta}=cosec \ {\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
প্রমাণঃ
প্রথম প্রমাণঃ
দেওয়া আছে,\(\sin{\theta}=\sin{\alpha}\)
\(\Rightarrow \sin{\theta}-\sin{\alpha}=0\)
\(\Rightarrow 2\cos{\frac{\theta+\alpha}{2}}\sin{\frac{\theta-\alpha}{2}}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\therefore \cos{\frac{\theta+\alpha}{2}}=0, \ \sin{\frac{\theta-\alpha}{2}}=0\)
এখন,
\(\cos{\frac{\theta+\alpha}{2}}=0\)
\(\Rightarrow \frac{\theta+\alpha}{2}=(2m+1)\frac{\pi}{2}\) যেখানে, \(m\in{\mathbb{Z}}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2m+1)\frac{\pi}{2}\)
\(\Rightarrow \theta+\alpha=(2m+1)\pi\)
\(\Rightarrow \theta=(2m+1)\pi-\alpha\)
\(\therefore \theta=(2m+1)\pi+(-1)^{2m+1}\alpha\)
\(n\) যে কোনো বিজোড় পূর্ণসংখ্যা হলে,
অর্থাৎ \(n=2m+1\) হলে, \(\theta=n\pi+(-1)^{n}\alpha\)
আবার,
\(\sin{\frac{\theta-\alpha}{2}}=0\)
\(\Rightarrow \frac{\theta-\alpha}{2}=m\pi\) যেখানে, \(m\in{\mathbb{Z}}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=m\pi\)
\(\Rightarrow \theta-\alpha=2m\pi\)
\(\Rightarrow \theta=2m\pi+\alpha\)
\(\therefore \theta=2m\pi+(-1)^{2m}\alpha\)
\(n\) যে কোনো জোড় পূর্ণসংখ্যা হলে,
অর্থাৎ \(n=2m\) হলে, \(\theta=n\pi+(-1)^{n}\alpha\)
\(n\) জোড় বা বিজোড় যে কোনো পূর্ণসংখ্যা অর্থাৎ \(n\in{\mathbb{Z}}\) হলে,
\(\theta=n\pi+(-1)^{n}\alpha\)
\(\therefore \sin{\theta}=\sin{\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
\(\sin{\theta}=\sin{\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
দ্বিতীয় প্রমাণঃ
দেওয়া আছে,\(cosec \ {\theta}=cosec \ {\alpha}\)
\(\Rightarrow \frac{1}{\sin{\theta}}=\frac{1}{\sin{\alpha}}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \sin{\theta}=\sin{\alpha}\)
\(\therefore \theta=n\pi+(-1)^{n}\alpha\)
\(cosec \ {\theta}=cosec \ {\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
\(cosec \ {\theta}=cosec \ {\alpha}\) এর সমাধান \(\theta=n\pi+(-1)^{n}\alpha\)
\(\cos{\theta}=\cos{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)\(\sec{\theta}=\sec{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
প্রমাণঃ
প্রথম প্রমাণঃ
দেওয়া আছে,\(\cos{\theta}=\cos{\alpha}\)
\(\Rightarrow \cos{\theta}-\cos{\alpha}=0\)
\(\Rightarrow \cos{\alpha}-\cos{\theta}=0\)
\(\Rightarrow 2\sin{\frac{\alpha+\theta}{2}}\sin{\frac{\theta-\alpha}{2}}=0\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\therefore \sin{\frac{\alpha+\theta}{2}}=0, \ \sin{\frac{\theta-\alpha}{2}}=0\)
এখন,
\(\sin{\frac{\alpha+\theta}{2}}=0\)
\(\Rightarrow \frac{\alpha+\theta}{2}=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\)
\(\Rightarrow \alpha+\theta=2n\pi\)
\(\therefore \theta=2n\pi-\alpha .........(1)\)
আবার,
\(\sin{\frac{\theta-\alpha}{2}}=0\)
\(\Rightarrow \frac{\theta-\alpha}{2}=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\)
\(\Rightarrow \theta-\alpha=2n\pi\)
\(\therefore \theta=2n\pi+\alpha .........(2)\)
\((1)\) ও \((2)\) কে সমন্বয় করে,
\(\theta=2n\pi\pm{\alpha}\)
\(\therefore \cos{\theta}=\cos{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
\(\cos{\theta}=\cos{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
দ্বিতীয় প্রমাণঃ
দেওয়া আছে,\(\sec{\theta}=\sec{\alpha}\)
\(\Rightarrow \frac{1}{\cos{\theta}}=\frac{1}{\cos{\alpha}}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \cos{\theta}=\cos{\alpha}\)
\(\Rightarrow \theta=2n\pi\pm{\alpha}\)
\(\therefore \sec{\theta}=\sec{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
\(\sec{\theta}=\sec{\alpha}\) এর সমাধান \(\theta=2n\pi\pm{\alpha}\)
\(\tan{\theta}=\tan{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)\(\cot{\theta}=\cot{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)
প্রমাণঃ
প্রথম প্রমাণঃ
দেওয়া আছে,\(\tan{\theta}=\tan{\alpha}\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}=\frac{\sin{\alpha}}{\cos{\alpha}}\)
\(\Rightarrow \sin{\theta}\cos{\alpha}=\cos{\theta}\sin{\alpha}\)
\(\Rightarrow \sin{\theta}\cos{\alpha}-\cos{\theta}\sin{\alpha}=0\)
\(\Rightarrow \sin{(\theta-\alpha)}=0\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \theta-\alpha=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\)
\(\therefore \theta=n\pi+\alpha\)
\(\therefore \tan{\theta}=\tan{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)
\(\tan{\theta}=\tan{\alpha}\) এর সমাধান \(\theta=n\pi+\alpha\)
দ্বিতীয় প্রমাণঃ
দেওয়া আছে,\(\cot{\theta}=\cot{\alpha}\)
\(\Rightarrow \frac{1}{\tan{\theta}}=\frac{1}{\tan{\alpha}}\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(\Rightarrow \tan{\theta}=\tan{\alpha}\)
\(\Rightarrow \theta=n\pi+{\alpha}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\)
\(\therefore \cot{\theta}=\cot{\alpha}\) এর সমাধান \(\theta=n\pi+{\alpha}\)
\(\cot{\theta}=\cot{\alpha}\) এর সমাধান \(\theta=n\pi+{\alpha}\)
\(\sin{\theta}=1\) এর সমাধান \(\theta=(4n+1)\frac{\pi}{2}\)\(\cos{\theta}=1\) এর সমাধান \(\theta=2n\pi\)
প্রমাণঃ
প্রথম প্রমাণঃ
দেওয়া আছে,\(\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{2}}\)
\(\therefore \theta=m\pi+(-1)^{m}\frac{\pi}{2}, \ m\in{\mathbb{Z}}\) ➜ \(\because \sin{\theta}=\sin{\alpha}\)
\(\Rightarrow \theta=n\pi+(-1)^{n}\alpha\)
\(m\) জোড় সংখ্যা হলে,
অর্থাৎ \(m=2n\) হলে,
\(\Rightarrow \theta=2n\pi+(-1)^{2n}\frac{\pi}{2}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{2}\)
\(\therefore \theta=(4n+1)\frac{\pi}{2}\)
আবার,
\(m\) বিজোড় সংখ্যা হলে,
অর্থাৎ \(m=2n+1\) হলে,
\(\Rightarrow \theta=(2n+1)\pi+(-1)^{2n+1}\frac{\pi}{2}\)
\(\Rightarrow \theta=(2n+1)\pi-\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n+2-1)\frac{\pi}{2}\)
\(\therefore \theta=(4n+1)\frac{\pi}{2}\)
\(\therefore\) উভয় ক্ষেত্রেই \(\theta=(4n+1)\frac{\pi}{2}\)
\(\therefore \sin{\theta}=1\) এর সমাধান \(\theta=(4n+1)\frac{\pi}{2}\)
\(\sin{\theta}=1\) এর সমাধান \(\theta=(4n+1)\frac{\pi}{2}\)
দ্বিতীয় প্রমাণঃ
দেওয়া আছে,\(\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\cos{0^{o}}\)
\(\Rightarrow \theta=2n\pi\pm{0^{o}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm{\alpha}\)
\(\Rightarrow \theta=2n\pi\)
\(\therefore \cos{\theta}=1\) এর সমাধান \(\theta=2n\pi\)
\(\cos{\theta}=1\) এর সমাধান \(\theta=2n\pi\)
\(\sin{\theta}=-1\) এর সমাধান \(\theta=(4n-1)\frac{\pi}{2}\)\(\cos{\theta}=-1\) এর সমাধান \(\theta=(2n+1)\pi\)
প্রমাণঃ
প্রথম প্রমাণঃ
দেওয়া আছে,\(\sin{\theta}=-1\)
\(\Rightarrow \sin{\theta}=-\sin{\frac{\pi}{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\left(-\frac{\pi}{2}\right)}\)
\(\Rightarrow \theta=m\pi+(-1)^{m}\left(-\frac{\pi}{2}\right), \ m\in{\mathbb{Z}}\) ➜ \(\because \sin{\theta}=\sin{\alpha}\)
\(\Rightarrow \theta=n\pi+(-1)^{n}\alpha\)
\(\Rightarrow \theta=m\pi-(-1)^{m}\frac{\pi}{2}\)
\(m\) জোড় সংখ্যা হলে,
অর্থাৎ \(m=2n\) হলে,
\(\Rightarrow \theta=2n\pi-(-1)^{2n}\frac{\pi}{2}\)
\(\Rightarrow \theta=2n\pi-\frac{\pi}{2}\)
\(\therefore \theta=(4n-1)\frac{\pi}{2}\)
আবার,
\(m\) বিজোড় সংখ্যা হলে,
অর্থাৎ \(m=2n-1\) হলে,
\(\Rightarrow \theta=(2n-1)\pi-(-1)^{2n-1}\frac{\pi}{2}\)
\(\Rightarrow \theta=(2n-1)\pi+\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n-2+1)\frac{\pi}{2}\)
\(\therefore \theta=(4n-1)\frac{\pi}{2}\)
\(\therefore\) উভয় ক্ষেত্রেই \(\theta=(4n-1)\frac{\pi}{2}\)
\(\therefore \sin{\theta}=-1\) এর সমাধান \(\theta=(4n-1)\frac{\pi}{2}\)
\(\sin{\theta}=-1\) এর সমাধান \(\theta=(4n-1)\frac{\pi}{2}\)
দ্বিতীয় প্রমাণঃ
দেওয়া আছে,\(\cos{\theta}=-1\)
\(\Rightarrow \cos{\theta}=\cos{(\pi)}\)
\(\Rightarrow \theta=2n\pi\pm{\pi}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm{\alpha}\)
\(\Rightarrow \theta=(2n\pm{1})\pi\)
\(\Rightarrow \theta=(2n+1)\pi, \ \theta=(2n-1)\pi\)
এখানে, \((2n+1), \ (2n-1)\) উভয়েই বিজোড় সংখ্যা তাই \(\theta\) এর মানের পুনরাাবৃত্তি ঘটে।
সুতরাং ঋণাত্মক চিহ্ন বর্জন করে,
\(\theta=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \cos{\theta}=-1\) এর সমাধান \(\theta=(2n+1)\pi\)
\(\cos{\theta}=-1\) এর সমাধান \(\theta=(2n+1)\pi\)
অধ্যায় \(7H\)-এর উদাহরণসমুহ
উদাহরণ \(1.\) সাধারণ সমাধান নির্ণয় করঃ
\((a)\) \(\sin{\theta}=\frac{1}{2}\)
\((b)\) \(\sin{\theta}=-\frac{1}{2}\)
\((c)\) \(\cos{\theta}=\frac{1}{2}\)
\((d)\) \(\cos{\theta}=-\frac{1}{2}\)
\((e)\) \(\tan{\theta}=\sqrt{3}\)
\((f)\) \(\tan{\theta}=-\sqrt{3}\)
উত্তরঃ \((a) \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((d) \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((e) \ n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((f) \ n\pi-\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(2.\) সমাধান করঃ \(3\tan{x}+\cot{x}=cosec \ {x}\)
উত্তরঃ \(\text{সমাধান অবান্তর বা অপ্রাসঙ্গিক।}\)
উদাহরণ \(3.\) \(\cos{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\) এর সমাধান লেখের সাহায্যে নির্ণয় কর।
উত্তরঃ \(60^{o}; \ 300^{o}\)
উদাহরণ \(4.\) সাধারণ সমাধান নির্ণয় করঃ
\((a)\) \(4\sin{\theta}\cos{\theta}=\sqrt{3}\)
\((b)\) \(2(\cos^2{\theta}-\sin^2{\theta})=1\)
\((c)\) \(\tan^2{\theta}+\cot^2{\theta}=2\)
উত্তরঃ \((a) \ \frac{n\pi}{2}+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(5.\) সমাধান করঃ \(\tan{2\theta}\tan{\theta}=1\)
উত্তরঃ \(n\pi\pm\frac{\pi}{6} \ \text{অথবা } (2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(6.\) সমাধান করঃ \(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\le{\theta}\le{2\pi}\)
উত্তরঃ \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
উদাহরণ \(7.\) সমাধান করঃ \(\sin{5x}-\sin{3x}-\sin{x}=0, \ 0\le{x}\le{360^{o}}\)
উত্তরঃ \(0, \ 15^{o}, \ 75^{o}, \ 105^{o}, \ 165^{o}, \ 180^{o}, \ 195^{o}, \ 255^{o}, \ 285^{o}, \ 345^{o}\) এবং \(360^{o}\)
উদাহরণ \(8.\) সমাধান করঃ \(a\cos{\theta}+b\sin{\theta}=c, \ |c|\le{\sqrt{a^2+b^2}}\)
উত্তরঃ \(\theta=2n\pi+\alpha\pm\beta\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}, \ n\in{\mathbb{Z}}\)
উদাহরণ \(9.\) সমাধান করঃ \(\cos{x}+\sqrt{3}\sin{x}=\sqrt{2}\)
উত্তরঃ \(2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(10.\) সমাধান করঃ \(\sqrt{3}\cos{x}+\sin{x}=1, \ -2\pi\le{x}\le{2\pi}\)
উত্তরঃ \(-\frac{3\pi}{2}, \ -\frac{\pi}{6}, \ \frac{\pi}{2}\) এবং \(\frac{11\pi}{6}\)
উদাহরণ \(11.\) সমাধান করঃ \(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{180^{o}}\)
উত্তরঃ \((2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(12.\) সমাধান করঃ \(\cot{\theta}+\tan{\theta}=2\sec{\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
উত্তরঃ \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
উদাহরণ \(13.\) \(f(x)=\sin{x}\) এবং \(g(x)=\cos{x}\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\frac{1-x}{1+x}}\)
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, দেখাও যে, \(\theta=\pm{\frac{\pi}{4}}+\cos{\frac{1}{2\sqrt{2}}}\)
\((c)\) সমাধান করঃ \(g(x)+f(x)=g(2x)+f(2x)\)
উত্তরঃ \((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(14.\) \(\sin{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\) এর সমাধান লেখের সাহায্যে নির্ণয় কর।
উত্তরঃ \(30^{o}; \ 150^{o}\)
\((a)\) \(\sin{\theta}=\frac{1}{2}\)
\((b)\) \(\sin{\theta}=-\frac{1}{2}\)
\((c)\) \(\cos{\theta}=\frac{1}{2}\)
\((d)\) \(\cos{\theta}=-\frac{1}{2}\)
\((e)\) \(\tan{\theta}=\sqrt{3}\)
\((f)\) \(\tan{\theta}=-\sqrt{3}\)
উত্তরঃ \((a) \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((d) \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((e) \ n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((f) \ n\pi-\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(2.\) সমাধান করঃ \(3\tan{x}+\cot{x}=cosec \ {x}\)
উত্তরঃ \(\text{সমাধান অবান্তর বা অপ্রাসঙ্গিক।}\)
উদাহরণ \(3.\) \(\cos{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\) এর সমাধান লেখের সাহায্যে নির্ণয় কর।
উত্তরঃ \(60^{o}; \ 300^{o}\)
উদাহরণ \(4.\) সাধারণ সমাধান নির্ণয় করঃ
\((a)\) \(4\sin{\theta}\cos{\theta}=\sqrt{3}\)
\((b)\) \(2(\cos^2{\theta}-\sin^2{\theta})=1\)
\((c)\) \(\tan^2{\theta}+\cot^2{\theta}=2\)
উত্তরঃ \((a) \ \frac{n\pi}{2}+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(5.\) সমাধান করঃ \(\tan{2\theta}\tan{\theta}=1\)
উত্তরঃ \(n\pi\pm\frac{\pi}{6} \ \text{অথবা } (2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ ২০১৫; বঃ ২০১২,২০১০,২০০৫; ঢাঃ২০১১,২০০৮; কুঃ২০১০,২০০৭,২০০০; যঃ২০১০; রাঃ২০০৯; সিঃ,মাঃ২০০৫; রুয়েটঃ,চুয়েটঃ ২০১৩-২০১৪; বুয়েটঃ২০০৬-২০০৭ ।
উদাহরণ \(6.\) সমাধান করঃ \(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\le{\theta}\le{2\pi}\)
উত্তরঃ \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
দিঃ২০১২; চঃ২০১১,২০০১; সিঃ২০১০,২০০৭; রাঃ,বঃ২০১০; যঃ২০০৪; কুঃ,চঃ ২০০১ ।
উদাহরণ \(7.\) সমাধান করঃ \(\sin{5x}-\sin{3x}-\sin{x}=0, \ 0\le{x}\le{360^{o}}\)
উত্তরঃ \(0, \ 15^{o}, \ 75^{o}, \ 105^{o}, \ 165^{o}, \ 180^{o}, \ 195^{o}, \ 255^{o}, \ 285^{o}, \ 345^{o}\) এবং \(360^{o}\)
উদাহরণ \(8.\) সমাধান করঃ \(a\cos{\theta}+b\sin{\theta}=c, \ |c|\le{\sqrt{a^2+b^2}}\)
উত্তরঃ \(\theta=2n\pi+\alpha\pm\beta\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}, \ n\in{\mathbb{Z}}\)
উদাহরণ \(9.\) সমাধান করঃ \(\cos{x}+\sqrt{3}\sin{x}=\sqrt{2}\)
উত্তরঃ \(2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
দিঃ২০১৭; ঢাঃ২০১৫,২০০৭; চঃ২০১৪,২০০৭; সিঃ২০১৪,২০০৬; যঃ২০১২,২০০৯; বঃ২০০৮,২০০২; কুঃ২০০৬,২০০২; রাঃ২০০৫,২০০১; মাঃ২০১০,২০০৬ ।
উদাহরণ \(10.\) সমাধান করঃ \(\sqrt{3}\cos{x}+\sin{x}=1, \ -2\pi\le{x}\le{2\pi}\)
উত্তরঃ \(-\frac{3\pi}{2}, \ -\frac{\pi}{6}, \ \frac{\pi}{2}\) এবং \(\frac{11\pi}{6}\)
দিঃ২০১৫; ঢাঃ২০১১,২০০৩; সিঃ২০০৮,২০০২; যঃ২০১৪,২০১০,২০০৭; বঃ২০১৪,২০০৪; কুঃ২০০৮; রাঃ২০১৩,২০০৮,২০০৬; মাঃ২০০৮,২০০৫ ।
উদাহরণ \(11.\) সমাধান করঃ \(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{180^{o}}\)
উত্তরঃ \((2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
ঢাঃ২০০৮; কুঃ২০১১; চুয়েটঃ২০০৩-২০০৪ ।
উদাহরণ \(12.\) সমাধান করঃ \(\cot{\theta}+\tan{\theta}=2\sec{\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
উত্তরঃ \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
সিঃ২০১৬,২০১৫,২০১২; ঢাঃ২০১৫,২০১০; বঃ২০১৫,২০০৮; কুঃ২০১৪,২০০৭; চঃ২০১৩; যঃ২০১২; দিঃ২০১১; রাঃ২০০৭; মাঃ২০১২ ।
উদাহরণ \(13.\) \(f(x)=\sin{x}\) এবং \(g(x)=\cos{x}\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\frac{1-x}{1+x}}\)
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, দেখাও যে, \(\theta=\pm{\frac{\pi}{4}}+\cos{\frac{1}{2\sqrt{2}}}\)
\((c)\) সমাধান করঃ \(g(x)+f(x)=g(2x)+f(2x)\)
উত্তরঃ \((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১৫,২০০৫; ঢাঃ২০১৩,২০০৬,২০০২,২০০০; বঃ২০০৪; কুঃ২০১৪,২০০৫,২০০১; চঃ২০১২,২০০৬; যঃ২০১৪,২০১১,২০০৭; দিঃ২০১৪; রাঃ২০১২,২০০৮,২০০৪; মাঃ২০১৩,২০০৯ ।
উদাহরণ \(14.\) \(\sin{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\) এর সমাধান লেখের সাহায্যে নির্ণয় কর।
উত্তরঃ \(30^{o}; \ 150^{o}\)
উদাহরণ \(15.\) লেখের সাহায্যে \(\tan{x}=\sqrt{3};\) \(-180^{o}\le{x}\le{180^{o}}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(-150^{o}, \ -60^{o}, \ 30^{o}, \ 120^{o}\)
উদাহরণ \(16.\) সমাধান করঃ \(4(\sin^2{x}+\cos{x})=5\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\)
উদাহরণ \(17.\) লেখের সাহায্যে \(2\cos^2{x}+\sin{x}=1;\) \(0\le{x}\le{2\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(\frac{7\pi}{6}, \ \frac{11\pi}{6}\)
উদাহরণ \(18.\) লেখের সাহায্যে \(\cos{x}=-\frac{1}{2};\) \(0\le{x}\le{360^{o}}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(120^{o}, \ 240^{o}\)
উদাহরণ \(19.\) লেখের সাহায্যে \(\tan{x}=-1;\) \(-\pi\le{x}\le{\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(-\frac{\pi}{4}, \ \frac{3\pi}{4}\)
উদাহরণ \(20.\) লেখের সাহায্যে \(\cos{2x}=\frac{\sqrt{3}}{2};\) \(0\le{x}\le{2\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{12}, \ \frac{11\pi}{12}, \ \frac{13\pi}{12}, \ \frac{23\pi}{12}\)
উদাহরণ \(21.\) \(f(x)=\cos{x}\) এবং \(g(x)=\sin^{-1}{x}\)
\((a)\) \(f(x)\) এর পর্যায় কাল ও \(g(x)\) এর মূখ্যমান কত?
\((b)\) প্রমাণ কর যে, \(g\left\{\sqrt{2} f\left(\frac{\pi}{2}-\theta\right)\right\}+g\left\{\sqrt{f(2\theta)}\right\}=\frac{\pi}{2}\)
\((c)\) \(4f(x)f(2x)f(3x)=1\) হলে, \(0\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয় কর।
উত্তরঃ \((a) \ \left[-\frac{\pi}{2}, \ \frac{\pi}{2}\right]\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}\)
উদাহরণ \(22.\) দুইটি আতসবাজি বিস্ফোরণের ফলে \(y_{1}=a\cos{x_{1}}\) এবং \(y_{2}=b\cos{x_{2}}\) পরিবর্তনশীল সরণ বিশিষ্ট দুইটি শব্দ তরঙ্গ উৎপন্ন হয়। উপরিপাতনের ফলে তরঙ্গদ্বয়ের মিলিত তরঙ্গের সরণ হলো \(y=y_{1}+y_{2}\) । এখানে \(y\) হলো তরঙ্গের সরণ এবং \(x\) হলো সময়।
\((a)\) \(x_{1}=cosec^{-1}{\frac{a}{\sqrt{a^2-1}}}\) হলে, \(y_{1}\) নির্ণয় কর।
\((b)\) \(a=b=1\) এবং \(x_{1}+x_{2}=\frac{\pi}{2}\) হলে, কত সময় পরে মিলিত তরঙ্গের সরণ \(\sqrt{2}\) একক হবে?
\((c)\) \(x_{1}+x_{2}=\theta\) হলে প্রমাণ কর যে, \(\frac{y_{1}^2}{a^2}-\frac{2y_{1}y_{2}\cos{\theta}}{ab}+\frac{y_{2}^2}{b^2}=\sin^2{\theta}.\)
উত্তরঃ \((a) \ 1\)
\((b) \ \frac{\pi}{4}\)
উদাহরণ \(23.\) সমাধান করঃ \(2\cos^2{\theta}+2\sqrt{2}\sin{\theta}=3;\) \(0\le{\theta}\le{\frac{\pi}{2}}\)
উত্তরঃ \(\frac{\pi}{4}\)
উদাহরণ \(24.\) \(f(x)=\tan^{-1}{x}\)
\((a)\) দেখাও যে, \(2f(x)=\sin^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\((b)\) প্রমাণ কর যে, \(\tan{\left\{2f(x)\right\}}=2\tan{\left\{f(x)+f(x^3)\right\}}\)
\((c)\) \(f(x)=\tan{x}\) হলে, সমাধান করঃ \(f(x)f(2x)=1; \ 0\le{x}\le{\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
উদাহরণ \(25.\) \(\sin{(\pi\cos{\theta})}=\cos{(\pi\sin{\theta})} ........(1)\)
এবং \(\sqrt{3}\sin{\theta}-\cos{\theta}=\sqrt{2} ........(2)\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\left(\frac{1-x}{1+x}\right)}\)
\((b)\) উদ্দীপকের \((1)\) নং হতে প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) \(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে উদ্দীপকের \((2)\) নং সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{19\pi}{12}, \ -\frac{13\pi}{12}, \ \frac{5\pi}{12}, \ \frac{11\pi}{12}\)
উদাহরণ \(26.\) \(f(x)=\cos{x}, \ g(x)=\sin{x}\)
\((a)\) \(\frac{g(2\theta)g(\theta)}{f(2\theta)f(\theta)}=1\) এর সমাধান কর।
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) সমাধান করঃ \(f(x)-f(2x)=g(2x)-g(x)\)
উত্তরঃ \((a) \ n\pi\pm\frac{\pi}{6}\)
\((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
উদাহরণ \(27.\) সমাধান করঃ \(\cos{6\theta}+\cos{4\theta}+\cos{2\theta}+1=0\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\)
উদাহরণ \(28.\) সমাধান করঃ \(\tan{\theta}+\cot{\theta}=2cosec \ {\theta}, \ 0\lt{\theta}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
উদাহরণ \(29.\) \(a\tan{\theta}+b\sec{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\) হলে, প্রমাণ কর যে, \(\tan{(\alpha+\beta)}=\frac{2ac}{a^2-c^2}\)
উদাহরণ \(30.\) সমাধান করঃ \(\cot{x}-\cot{2x}=2\)
উত্তরঃ \(\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
উত্তরঃ \(-150^{o}, \ -60^{o}, \ 30^{o}, \ 120^{o}\)
উদাহরণ \(16.\) সমাধান করঃ \(4(\sin^2{x}+\cos{x})=5\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\)
উদাহরণ \(17.\) লেখের সাহায্যে \(2\cos^2{x}+\sin{x}=1;\) \(0\le{x}\le{2\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(\frac{7\pi}{6}, \ \frac{11\pi}{6}\)
উদাহরণ \(18.\) লেখের সাহায্যে \(\cos{x}=-\frac{1}{2};\) \(0\le{x}\le{360^{o}}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(120^{o}, \ 240^{o}\)
উদাহরণ \(19.\) লেখের সাহায্যে \(\tan{x}=-1;\) \(-\pi\le{x}\le{\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(-\frac{\pi}{4}, \ \frac{3\pi}{4}\)
উদাহরণ \(20.\) লেখের সাহায্যে \(\cos{2x}=\frac{\sqrt{3}}{2};\) \(0\le{x}\le{2\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{12}, \ \frac{11\pi}{12}, \ \frac{13\pi}{12}, \ \frac{23\pi}{12}\)
উদাহরণ \(21.\) \(f(x)=\cos{x}\) এবং \(g(x)=\sin^{-1}{x}\)
\((a)\) \(f(x)\) এর পর্যায় কাল ও \(g(x)\) এর মূখ্যমান কত?
\((b)\) প্রমাণ কর যে, \(g\left\{\sqrt{2} f\left(\frac{\pi}{2}-\theta\right)\right\}+g\left\{\sqrt{f(2\theta)}\right\}=\frac{\pi}{2}\)
\((c)\) \(4f(x)f(2x)f(3x)=1\) হলে, \(0\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয় কর।
উত্তরঃ \((a) \ \left[-\frac{\pi}{2}, \ \frac{\pi}{2}\right]\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}\)
উদাহরণ \(22.\) দুইটি আতসবাজি বিস্ফোরণের ফলে \(y_{1}=a\cos{x_{1}}\) এবং \(y_{2}=b\cos{x_{2}}\) পরিবর্তনশীল সরণ বিশিষ্ট দুইটি শব্দ তরঙ্গ উৎপন্ন হয়। উপরিপাতনের ফলে তরঙ্গদ্বয়ের মিলিত তরঙ্গের সরণ হলো \(y=y_{1}+y_{2}\) । এখানে \(y\) হলো তরঙ্গের সরণ এবং \(x\) হলো সময়।
\((a)\) \(x_{1}=cosec^{-1}{\frac{a}{\sqrt{a^2-1}}}\) হলে, \(y_{1}\) নির্ণয় কর।
\((b)\) \(a=b=1\) এবং \(x_{1}+x_{2}=\frac{\pi}{2}\) হলে, কত সময় পরে মিলিত তরঙ্গের সরণ \(\sqrt{2}\) একক হবে?
\((c)\) \(x_{1}+x_{2}=\theta\) হলে প্রমাণ কর যে, \(\frac{y_{1}^2}{a^2}-\frac{2y_{1}y_{2}\cos{\theta}}{ab}+\frac{y_{2}^2}{b^2}=\sin^2{\theta}.\)
উত্তরঃ \((a) \ 1\)
\((b) \ \frac{\pi}{4}\)
উদাহরণ \(23.\) সমাধান করঃ \(2\cos^2{\theta}+2\sqrt{2}\sin{\theta}=3;\) \(0\le{\theta}\le{\frac{\pi}{2}}\)
উত্তরঃ \(\frac{\pi}{4}\)
উদাহরণ \(24.\) \(f(x)=\tan^{-1}{x}\)
\((a)\) দেখাও যে, \(2f(x)=\sin^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\((b)\) প্রমাণ কর যে, \(\tan{\left\{2f(x)\right\}}=2\tan{\left\{f(x)+f(x^3)\right\}}\)
\((c)\) \(f(x)=\tan{x}\) হলে, সমাধান করঃ \(f(x)f(2x)=1; \ 0\le{x}\le{\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
উদাহরণ \(25.\) \(\sin{(\pi\cos{\theta})}=\cos{(\pi\sin{\theta})} ........(1)\)
এবং \(\sqrt{3}\sin{\theta}-\cos{\theta}=\sqrt{2} ........(2)\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\left(\frac{1-x}{1+x}\right)}\)
\((b)\) উদ্দীপকের \((1)\) নং হতে প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) \(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে উদ্দীপকের \((2)\) নং সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{19\pi}{12}, \ -\frac{13\pi}{12}, \ \frac{5\pi}{12}, \ \frac{11\pi}{12}\)
উদাহরণ \(26.\) \(f(x)=\cos{x}, \ g(x)=\sin{x}\)
\((a)\) \(\frac{g(2\theta)g(\theta)}{f(2\theta)f(\theta)}=1\) এর সমাধান কর।
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) সমাধান করঃ \(f(x)-f(2x)=g(2x)-g(x)\)
উত্তরঃ \((a) \ n\pi\pm\frac{\pi}{6}\)
\((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
উদাহরণ \(27.\) সমাধান করঃ \(\cos{6\theta}+\cos{4\theta}+\cos{2\theta}+1=0\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\)
উদাহরণ \(28.\) সমাধান করঃ \(\tan{\theta}+\cot{\theta}=2cosec \ {\theta}, \ 0\lt{\theta}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
উদাহরণ \(29.\) \(a\tan{\theta}+b\sec{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\) হলে, প্রমাণ কর যে, \(\tan{(\alpha+\beta)}=\frac{2ac}{a^2-c^2}\)
উদাহরণ \(30.\) সমাধান করঃ \(\cot{x}-\cot{2x}=2\)
উত্তরঃ \(\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
উদাহরণ \(1.\) সাধারণ সমাধান নির্ণয় করঃ
\((a)\) \(\sin{\theta}=\frac{1}{2}\)
\((b)\) \(\sin{\theta}=-\frac{1}{2}\)
\((c)\) \(\cos{\theta}=\frac{1}{2}\)
\((d)\) \(\cos{\theta}=-\frac{1}{2}\)
\((e)\) \(\tan{\theta}=\sqrt{3}\)
\((f)\) \(\tan{\theta}=-\sqrt{3}\)
উত্তরঃ \((a) \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((d) \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((e) \ n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((f) \ n\pi-\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((a)\) \(\sin{\theta}=\frac{1}{2}\)
\((b)\) \(\sin{\theta}=-\frac{1}{2}\)
\((c)\) \(\cos{\theta}=\frac{1}{2}\)
\((d)\) \(\cos{\theta}=-\frac{1}{2}\)
\((e)\) \(\tan{\theta}=\sqrt{3}\)
\((f)\) \(\tan{\theta}=-\sqrt{3}\)
উত্তরঃ \((a) \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((d) \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((e) \ n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((f) \ n\pi-\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
\((a)\)
প্রদত্ত সমীকরণ, \(\sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{x}=\sin{\alpha}\)
\(\Rightarrow x=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) নির্ণেয় সমাধান, \( n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b)\)
প্রদত্ত সমীকরণ, \(\sin{\theta}=-\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \sin{\theta}=\sin{\left(-\frac{\pi}{6}\right)}\) ➜ \(\because \sin{(-\theta)}=-\sin{\theta}\)
\(\Rightarrow \theta=n\pi+(-1)^n\left(-\frac{\pi}{6}\right)\) ➜ \(\because \sin{x}=\sin{\alpha}\)
\(\Rightarrow x=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi-(-1)^n\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \( n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
প্রদত্ত সমীকরণ, \(\cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((d)\)
প্রদত্ত সমীকরণ, \(\cos{\theta}=-\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cos{x}=\cos{(\pi-x)}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{3\pi-\pi}{3}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((e)\)
প্রদত্ত সমীকরণ, \(\tan{\theta}=\sqrt{3}\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{3}\) ➜ \(\because \tan{x}=\tan{\alpha}\)
\(\Rightarrow x=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{3}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((f)\)
প্রদত্ত সমীকরণ, \(\tan{\theta}=-\sqrt{3}\)
\(\Rightarrow \tan{\theta}=-\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(-\frac{\pi}{3}\right)}\) ➜ \(\because \tan{(-\theta)}=-\tan{\theta}\)
\(\Rightarrow \theta=n\pi+\left(-\frac{\pi}{3}\right)\) ➜ \(\because \tan{x}=\tan{\alpha}\)
\(\Rightarrow x=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi-\frac{\pi}{3}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=n\pi-\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(2.\) সমাধান করঃ \(3\tan{x}+\cot{x}=cosec \ {x}\)
উত্তরঃ \(\text{সমাধান অবান্তর বা অপ্রাসঙ্গিক।}\)
উত্তরঃ \(\text{সমাধান অবান্তর বা অপ্রাসঙ্গিক।}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(3\tan{x}+\cot{x}=cosec \ {x}\)
\(\Rightarrow 3\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}=\frac{1}{\sin{x}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{3\sin^2{x}+\cos^2{x}}{\sin{x}\cos{x}}=\frac{1}{\sin{x}}\)
\(\Rightarrow \frac{3\sin^2{x}+\cos^2{x}}{\cos{x}}=1\)
\(\Rightarrow 3\sin^2{x}+\cos^2{x}=\cos{x}\)
\(\Rightarrow 3(1-\cos^2{x})+\cos^2{x}=\cos{x}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 3-3\cos^2{x}+\cos^2{x}-\cos{x}=0\)
\(\Rightarrow -2\cos^2{x}-\cos{x}+3=0\)
\(\Rightarrow 2\cos^2{x}+\cos{x}-3=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow 2\cos^2{x}+3\cos{x}-2\cos{x}-3=0\)
\(\Rightarrow \cos{x}(2\cos{x}+3)-1(2\cos{x}+3)=0\)
\(\Rightarrow (2\cos{x}+3)(\cos{x}-1)=0\)
\(\Rightarrow 2\cos{x}+3=0, \ \cos{x}-1=0\)
\(\Rightarrow 2\cos{x}=-3, \ \cos{x}=1\)
\(\Rightarrow \cos{x}=-\frac{3}{2}, \ \cos{x}=1\)
\(\because \cos{x}=-\frac{3}{2}\) গ্রহণযোগ্য নয় কারণ, \(-1\le{\cos{x}}\le{1}\)
\(\Rightarrow \cos{x}=1\)
\(\Rightarrow \cos{x}=\cos{0^{o}}\)
\(\Rightarrow x=2n\pi\pm{0^{o}}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\)
\(x=2n\pi\) হলে, \(\sin{x}=0\) হবে।
সে ক্ষেত্রে সমীকরণটি অর্থপূর্ণ হবে না।
অতএব প্রাপ্ত সমাধান অবান্তর বা অপ্রাসঙ্গিক।
\(3\tan{x}+\cot{x}=cosec \ {x}\)
\(\Rightarrow 3\frac{\sin{x}}{\cos{x}}+\frac{\cos{x}}{\sin{x}}=\frac{1}{\sin{x}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{3\sin^2{x}+\cos^2{x}}{\sin{x}\cos{x}}=\frac{1}{\sin{x}}\)
\(\Rightarrow \frac{3\sin^2{x}+\cos^2{x}}{\cos{x}}=1\)
\(\Rightarrow 3\sin^2{x}+\cos^2{x}=\cos{x}\)
\(\Rightarrow 3(1-\cos^2{x})+\cos^2{x}=\cos{x}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 3-3\cos^2{x}+\cos^2{x}-\cos{x}=0\)
\(\Rightarrow -2\cos^2{x}-\cos{x}+3=0\)
\(\Rightarrow 2\cos^2{x}+\cos{x}-3=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow 2\cos^2{x}+3\cos{x}-2\cos{x}-3=0\)
\(\Rightarrow \cos{x}(2\cos{x}+3)-1(2\cos{x}+3)=0\)
\(\Rightarrow (2\cos{x}+3)(\cos{x}-1)=0\)
\(\Rightarrow 2\cos{x}+3=0, \ \cos{x}-1=0\)
\(\Rightarrow 2\cos{x}=-3, \ \cos{x}=1\)
\(\Rightarrow \cos{x}=-\frac{3}{2}, \ \cos{x}=1\)
\(\because \cos{x}=-\frac{3}{2}\) গ্রহণযোগ্য নয় কারণ, \(-1\le{\cos{x}}\le{1}\)
\(\Rightarrow \cos{x}=1\)
\(\Rightarrow \cos{x}=\cos{0^{o}}\)
\(\Rightarrow x=2n\pi\pm{0^{o}}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\)
\(x=2n\pi\) হলে, \(\sin{x}=0\) হবে।
সে ক্ষেত্রে সমীকরণটি অর্থপূর্ণ হবে না।
অতএব প্রাপ্ত সমাধান অবান্তর বা অপ্রাসঙ্গিক।
উদাহরণ \(3.\) \(\cos{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\) এর সমাধান লেখের সাহায্যে নির্ণয় কর।
উত্তরঃ \(60^{o}; \ 300^{o}\)
উত্তরঃ \(60^{o}; \ 300^{o}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\cos{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\)
\((a) \ [0, 360^{o}]\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \cos{x}=\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(60^{o}, \frac{1}{2}\right)\) ও \(\left(300^{o}, \frac{1}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\cos{x}=\frac{1}{2}\) ফাংশনের সমাধান, \(x=60^{o}, \ 300^{o}\)
\(\cos{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\)
\((a) \ [0, 360^{o}]\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \cos{x}=\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(60^{o}, \frac{1}{2}\right)\) ও \(\left(300^{o}, \frac{1}{2}\right)\) নির্ণয় করি।

উদাহরণ \(4.\) সাধারণ সমাধান নির্ণয় করঃ
\((a)\) \(4\sin{\theta}\cos{\theta}=\sqrt{3}\)
\((b)\) \(2(\cos^2{\theta}-\sin^2{\theta})=1\)
\((c)\) \(\tan^2{\theta}+\cot^2{\theta}=2\)
উত্তরঃ \((a) \ \frac{n\pi}{2}+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((a)\) \(4\sin{\theta}\cos{\theta}=\sqrt{3}\)
\((b)\) \(2(\cos^2{\theta}-\sin^2{\theta})=1\)
\((c)\) \(\tan^2{\theta}+\cot^2{\theta}=2\)
উত্তরঃ \((a) \ \frac{n\pi}{2}+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
\((a)\)
প্রদত্ত সমীকরণ, \(4\sin{\theta}\cos{\theta}=\sqrt{3}\)
\(\Rightarrow 2\times2\sin{\theta}\cos{\theta}=\sqrt{3}\)
\(\Rightarrow 2\sin{2\theta}=\sqrt{3}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{2\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{2\theta}=\sin{\frac{\pi}{3}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \sin{x}=\sin{\alpha}\)
\(\Rightarrow x=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{2}+(-1)^n\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=\frac{n\pi}{2}+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b)\)
প্রদত্ত সমীকরণ, \(2(\cos^2{\theta}-\sin^2{\theta})=1\)
\(\Rightarrow 2\cos{2\theta}=1\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
প্রদত্ত সমীকরণ, \(\tan^2{\theta}+\cot^2{\theta}=2\)
\(\Rightarrow \tan^2{\theta}+\frac{1}{\tan^2{\theta}}=2\) ➜ \(\because \cot^2{A}=\frac{1}{\tan^2{A}}\)
\(\Rightarrow \tan^2{\theta}+\frac{1}{\tan^2{\theta}}-2=0\)
\(\Rightarrow \tan^2{\theta}+\frac{1}{\tan^2{\theta}}-2\tan{\theta}\times\frac{1}{\tan{\theta}}=0\)
\(\Rightarrow \left(\tan{\theta}-\frac{1}{\tan{\theta}}\right)^2=0\) ➜ \(\because a^2+b^2-2ab=(a-b)^2\)
\(\Rightarrow \tan{\theta}-\frac{1}{\tan{\theta}}=0\)
\(\Rightarrow \tan{\theta}=\frac{1}{\tan{\theta}}\)
\(\Rightarrow \tan^2{\theta}=1\)
\(\Rightarrow \tan^2{\theta}=\pm\sqrt{1}\)
\(\Rightarrow \tan{\theta}=\pm1\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{x}=\tan{\alpha}\)
\(\Rightarrow x=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi\pm\frac{\pi}{4}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(5.\) সমাধান করঃ \(\tan{2\theta}\tan{\theta}=1\)
উত্তরঃ \(n\pi\pm\frac{\pi}{6} \ \text{অথবা } (2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
উত্তরঃ \(n\pi\pm\frac{\pi}{6} \ \text{অথবা } (2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ ২০১৫; বঃ ২০১২,২০১০,২০০৫; ঢাঃ২০১১,২০০৮; কুঃ২০১০,২০০৭,২০০০; যঃ২০১০; রাঃ২০০৯; সিঃ,মাঃ২০০৫; রুয়েটঃ,চুয়েটঃ ২০১৩-২০১৪; বুয়েটঃ২০০৬-২০০৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{2\theta}\tan{\theta}=1\)
\(\Rightarrow \frac{\sin{2\theta}\sin{\theta}}{\cos{2\theta}\cos{\theta}}=1\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \cos{2\theta}\cos{\theta}=\sin{2\theta}\sin{\theta}\)
\(\Rightarrow \cos{2\theta}\cos{\theta}-\sin{2\theta}\sin{\theta}=0\)
\(\Rightarrow \cos{(2\theta+\theta)}=0\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \cos{3\theta}=0\)
\(\Rightarrow 3\theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{x}=0\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=(2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{2\theta}\tan{\theta}=1\)
\(\Rightarrow \frac{2\tan{\theta}}{1-tan^2{\theta}}\times\tan{\theta}=1\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\Rightarrow \frac{2\tan^2{\theta}}{1-tan^2{\theta}}=1\)
\(\Rightarrow 2\tan^2{\theta}=1-tan^2{\theta}\)
\(\Rightarrow 2\tan^2{\theta}+tan^2{\theta}=1\)
\(\Rightarrow 3tan^2{\theta}=1\)
\(\Rightarrow tan^2{\theta}=\frac{1}{3}\)
\(\Rightarrow tan{\theta}=\pm\sqrt{\frac{1}{3}}\)
\(\Rightarrow tan{\theta}=\pm\frac{1}{\sqrt{3}}\)
\(\Rightarrow tan{\theta}=\pm\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow tan{\theta}=\tan{\left(\pm\frac{\pi}{6}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{6}\right)\) ➜ \(\because \tan{x}=\tan{\alpha}\)
\(\Rightarrow x=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{2\theta}\tan{\theta}=1\)
\(\Rightarrow \frac{\sin{2\theta}\sin{\theta}}{\cos{2\theta}\cos{\theta}}=1\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \cos{2\theta}\cos{\theta}=\sin{2\theta}\sin{\theta}\)
\(\Rightarrow \cos{2\theta}\cos{\theta}-\sin{2\theta}\sin{\theta}=0\)
\(\Rightarrow \cos{(2\theta+\theta)}=0\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \cos{3\theta}=0\)
\(\Rightarrow 3\theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{x}=0\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=(2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
বিকল্প পদ্ধতিঃ
প্রদত্ত সমীকরণ, \(\tan{2\theta}\tan{\theta}=1\)
\(\Rightarrow \frac{2\tan{\theta}}{1-tan^2{\theta}}\times\tan{\theta}=1\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\Rightarrow \frac{2\tan^2{\theta}}{1-tan^2{\theta}}=1\)
\(\Rightarrow 2\tan^2{\theta}=1-tan^2{\theta}\)
\(\Rightarrow 2\tan^2{\theta}+tan^2{\theta}=1\)
\(\Rightarrow 3tan^2{\theta}=1\)
\(\Rightarrow tan^2{\theta}=\frac{1}{3}\)
\(\Rightarrow tan{\theta}=\pm\sqrt{\frac{1}{3}}\)
\(\Rightarrow tan{\theta}=\pm\frac{1}{\sqrt{3}}\)
\(\Rightarrow tan{\theta}=\pm\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow tan{\theta}=\tan{\left(\pm\frac{\pi}{6}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{6}\right)\) ➜ \(\because \tan{x}=\tan{\alpha}\)
\(\Rightarrow x=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(6.\) সমাধান করঃ \(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\le{\theta}\le{2\pi}\)
উত্তরঃ \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
উত্তরঃ \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
দিঃ২০১২; চঃ২০১১,২০০১; সিঃ২০১০,২০০৭; রাঃ,বঃ২০১০; যঃ২০০৪; কুঃ,চঃ ২০০১ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow 4(1-\cos^2{\theta}+\cos{\theta})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-4\cos^2{\theta}+4\cos{\theta}-5=0\)
\(\Rightarrow -4\cos^2{\theta}+4\cos{\theta}-1=0\)
\(\Rightarrow 4\cos^2{\theta}-4\cos{\theta}+1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow (2\cos{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\cos{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\)
\(\therefore \theta=\frac{\pi}{3}, \ -\frac{\pi}{3}\) উভয়ে গ্রহনযোগ্য।
যখন, \(n=1,\) \(\theta=2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=-\frac{5\pi}{3}\)
যখন, \(n=2,\) \(\theta=4\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{13\pi}{3}, \ \frac{11\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
\(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow 4(1-\cos^2{\theta}+\cos{\theta})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-4\cos^2{\theta}+4\cos{\theta}-5=0\)
\(\Rightarrow -4\cos^2{\theta}+4\cos{\theta}-1=0\)
\(\Rightarrow 4\cos^2{\theta}-4\cos{\theta}+1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow (2\cos{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\cos{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\)
\(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\pm\frac{\pi}{3}\)\(\therefore \theta=\frac{\pi}{3}, \ -\frac{\pi}{3}\) উভয়ে গ্রহনযোগ্য।
যখন, \(n=1,\) \(\theta=2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=-\frac{5\pi}{3}\)
যখন, \(n=2,\) \(\theta=4\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{13\pi}{3}, \ \frac{11\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
উদাহরণ \(7.\) সমাধান করঃ \(\sin{5x}-\sin{3x}-\sin{x}=0, \ 0\le{x}\le{360^{o}}\)
উত্তরঃ \(0, \ 15^{o}, \ 75^{o}, \ 105^{o}, \ 165^{o}, \ 180^{o}, \ 195^{o}, \ 255^{o}, \ 285^{o}, \ 345^{o}, \ 360^{o}\)
উত্তরঃ \(0, \ 15^{o}, \ 75^{o}, \ 105^{o}, \ 165^{o}, \ 180^{o}, \ 195^{o}, \ 255^{o}, \ 285^{o}, \ 345^{o}, \ 360^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{5x}-\sin{3x}-\sin{x}=0, \ 0\le{x}\le{360^{o}}\)
\(\Rightarrow 2\cos{\frac{5x+3x}{2}}\sin{\frac{5x-3x}{2}}-\sin{x}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{8x}{2}}\sin{\frac{2x}{2}}-\sin{x}=0\)
\(\Rightarrow 2\cos{4x}\sin{x}-\sin{x}=0\)
\(\Rightarrow \sin{x}(2\cos{4x}-1)=0\)
\(\Rightarrow \sin{x}=0, \ 2\cos{4x}-1=0\)
\(\Rightarrow \sin{x}=0, \ 2\cos{4x}=1\)
\(\Rightarrow \sin{x}=0, \ \cos{4x}=\frac{1}{2}\)
\(\Rightarrow \sin{x}=0, \ \cos{4x}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x=n\pi, \ 4x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=n\pi, \ x=\frac{n\pi}{2}\pm\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi, \ \frac{n\pi}{2}\pm\frac{\pi}{12}\)
\(\Rightarrow x=0, \ 15^{o}, \ -15^{o}\) প্রথম মানদ্বয় গ্রহনযোগ্য।
\(\therefore x=0, \ 15^{o}\)
যখন, \(n=1,\) \(x=\pi, \ \frac{\pi}{2}\pm\frac{\pi}{12}\)
\(\Rightarrow x=180^{o}, \ 90^{o}-15^{o}, \ 90^{o}+15^{o}\)
\(\Rightarrow x=180^{o}, \ 75^{o}, \ 105^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore x=75^{o}, \ 105^{o}, \ 180^{o}\)
যখন, \(n=-1,\) \(x=-\pi, \ -\frac{\pi}{2}\pm\frac{\pi}{12}\)
\(\Rightarrow x=-180^{o}, \ -90^{o}-15^{o}, \ -90^{o}+15^{o}\)
\(\Rightarrow x=-180^{o}, \ -105^{o}, \ -75^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi, \ \pi\pm\frac{\pi}{12}\)
\(\Rightarrow x=360^{o}, \ 180^{o}-15^{o}, \ 180^{o}+15^{o}\)
\(\Rightarrow x=360^{o}, \ 165^{o}, \ 195^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore x=165^{o}, \ 195^{o}, \ 360^{o}\)
যখন, \(n=3,\) \(x=3\pi, \ \frac{3\pi}{2}\pm\frac{\pi}{12}\)
\(\Rightarrow x=540^{o}, \ 270^{o}-15^{o}, \ 270^{o}+15^{o}\)
\(\Rightarrow x=540^{o}, \ 255^{o}, \ 285^{o}\) শেষ মানদ্বয় গ্রহনযোগ্য।
\(\therefore x=255^{o}, \ 285^{o}\)
যখন, \(n=4,\) \(x=4\pi, \ 2\pi\pm\frac{\pi}{12}\)
\(\Rightarrow x=720^{o}, \ 360^{o}-15^{o}, \ 360^{o}+15^{o}\)
\(\Rightarrow x=720^{o}, \ 345^{o}, \ 375^{o}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore x=345^{o}\)
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(0, \ 15^{o}, \ 75^{o}, \ 105^{o}, \ 165^{o}, \ 180^{o}, \ 195^{o}, \ 255^{o}, \ 285^{o}, \ 345^{o}, \ 360^{o}\)
\(\sin{5x}-\sin{3x}-\sin{x}=0, \ 0\le{x}\le{360^{o}}\)
\(\Rightarrow 2\cos{\frac{5x+3x}{2}}\sin{\frac{5x-3x}{2}}-\sin{x}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{8x}{2}}\sin{\frac{2x}{2}}-\sin{x}=0\)
\(\Rightarrow 2\cos{4x}\sin{x}-\sin{x}=0\)
\(\Rightarrow \sin{x}(2\cos{4x}-1)=0\)
\(\Rightarrow \sin{x}=0, \ 2\cos{4x}-1=0\)
\(\Rightarrow \sin{x}=0, \ 2\cos{4x}=1\)
\(\Rightarrow \sin{x}=0, \ \cos{4x}=\frac{1}{2}\)
\(\Rightarrow \sin{x}=0, \ \cos{4x}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x=n\pi, \ 4x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=n\pi, \ x=\frac{n\pi}{2}\pm\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi, \ \frac{n\pi}{2}\pm\frac{\pi}{12}\)
\(0\le{x}\le{360^{o}}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=0, \ \pm\frac{\pi}{12}\)\(\Rightarrow x=0, \ 15^{o}, \ -15^{o}\) প্রথম মানদ্বয় গ্রহনযোগ্য।
\(\therefore x=0, \ 15^{o}\)
যখন, \(n=1,\) \(x=\pi, \ \frac{\pi}{2}\pm\frac{\pi}{12}\)
\(\Rightarrow x=180^{o}, \ 90^{o}-15^{o}, \ 90^{o}+15^{o}\)
\(\Rightarrow x=180^{o}, \ 75^{o}, \ 105^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore x=75^{o}, \ 105^{o}, \ 180^{o}\)
যখন, \(n=-1,\) \(x=-\pi, \ -\frac{\pi}{2}\pm\frac{\pi}{12}\)
\(\Rightarrow x=-180^{o}, \ -90^{o}-15^{o}, \ -90^{o}+15^{o}\)
\(\Rightarrow x=-180^{o}, \ -105^{o}, \ -75^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi, \ \pi\pm\frac{\pi}{12}\)
\(\Rightarrow x=360^{o}, \ 180^{o}-15^{o}, \ 180^{o}+15^{o}\)
\(\Rightarrow x=360^{o}, \ 165^{o}, \ 195^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore x=165^{o}, \ 195^{o}, \ 360^{o}\)
যখন, \(n=3,\) \(x=3\pi, \ \frac{3\pi}{2}\pm\frac{\pi}{12}\)
\(\Rightarrow x=540^{o}, \ 270^{o}-15^{o}, \ 270^{o}+15^{o}\)
\(\Rightarrow x=540^{o}, \ 255^{o}, \ 285^{o}\) শেষ মানদ্বয় গ্রহনযোগ্য।
\(\therefore x=255^{o}, \ 285^{o}\)
যখন, \(n=4,\) \(x=4\pi, \ 2\pi\pm\frac{\pi}{12}\)
\(\Rightarrow x=720^{o}, \ 360^{o}-15^{o}, \ 360^{o}+15^{o}\)
\(\Rightarrow x=720^{o}, \ 345^{o}, \ 375^{o}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore x=345^{o}\)
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(0, \ 15^{o}, \ 75^{o}, \ 105^{o}, \ 165^{o}, \ 180^{o}, \ 195^{o}, \ 255^{o}, \ 285^{o}, \ 345^{o}, \ 360^{o}\)
উদাহরণ \(8.\) সমাধান করঃ \(a\cos{\theta}+b\sin{\theta}=c, \ |c|\le{\sqrt{a^2+b^2}}\)
উত্তরঃ \(\theta=2n\pi+\alpha\pm\beta\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}, \ n\in{\mathbb{Z}}\)
উত্তরঃ \(\theta=2n\pi+\alpha\pm\beta\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
ধরি,
\(r\cos{\alpha}=a, \ r\sin{\alpha}=b\) যেখানে, \(\alpha\) একটি ক্ষুদ্রতম ধনাত্মক কোণ এবং \(r\) ধনাত্মক।
\(\Rightarrow r^2\cos^2{\alpha}+r^2\sin^2{\alpha}=a^2+b^2\)
\(\Rightarrow r^2(\sin^2{\alpha}+\cos^2{\alpha})=a^2+b^2\)
\(\Rightarrow r^2=a^2+b^2\) ➜ \(\because \sin^2{\alpha}+\cos^2{\alpha}=1\)
\(\therefore r=\sqrt{a^2+b^2}\)
আবার,
\(\Rightarrow \cos{\alpha}=\frac{a}{r}, \ \sin{\alpha}=\frac{b}{r}\)
\(\therefore \cos{\alpha}=\frac{a}{\sqrt{a^2+b^2}}, \ \sin{\alpha}=\frac{b}{\sqrt{a^2+b^2}}\)
প্রদত্ত সমীকরণ,
\(a\cos{\theta}+b\sin{\theta}=c\)
\(\Rightarrow r\cos{\alpha}\cos{\theta}+r\sin{\alpha}\sin{\theta}=c\) ➜ \(\because r\cos{\alpha}=a, \ r\sin{\alpha}=b\)
\(\Rightarrow r(\cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha})=c\)
\(\Rightarrow r\cos{(\theta-\alpha)}=c\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \cos{(\theta-\alpha)}=\frac{c}{r}\)
\(\Rightarrow \cos{(\theta-\alpha)}=\frac{c}{\sqrt{a^2+b^2}}\) ➜ \(\because r=\sqrt{a^2+b^2}\)
\(\Rightarrow \cos{(\theta-\alpha)}=\cos{\beta}\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}\)
\(\Rightarrow \theta-\alpha=2n\pi\pm{\beta}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\beta+\alpha\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=2n\pi+\alpha\pm\beta\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}, \ n\in{\mathbb{Z}}\)
\(r\cos{\alpha}=a, \ r\sin{\alpha}=b\) যেখানে, \(\alpha\) একটি ক্ষুদ্রতম ধনাত্মক কোণ এবং \(r\) ধনাত্মক।
\(\Rightarrow r^2\cos^2{\alpha}+r^2\sin^2{\alpha}=a^2+b^2\)
\(\Rightarrow r^2(\sin^2{\alpha}+\cos^2{\alpha})=a^2+b^2\)
\(\Rightarrow r^2=a^2+b^2\) ➜ \(\because \sin^2{\alpha}+\cos^2{\alpha}=1\)
\(\therefore r=\sqrt{a^2+b^2}\)
আবার,
\(\Rightarrow \cos{\alpha}=\frac{a}{r}, \ \sin{\alpha}=\frac{b}{r}\)
\(\therefore \cos{\alpha}=\frac{a}{\sqrt{a^2+b^2}}, \ \sin{\alpha}=\frac{b}{\sqrt{a^2+b^2}}\)
প্রদত্ত সমীকরণ,
\(a\cos{\theta}+b\sin{\theta}=c\)
\(\Rightarrow r\cos{\alpha}\cos{\theta}+r\sin{\alpha}\sin{\theta}=c\) ➜ \(\because r\cos{\alpha}=a, \ r\sin{\alpha}=b\)
\(\Rightarrow r(\cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha})=c\)
\(\Rightarrow r\cos{(\theta-\alpha)}=c\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \cos{(\theta-\alpha)}=\frac{c}{r}\)
\(\Rightarrow \cos{(\theta-\alpha)}=\frac{c}{\sqrt{a^2+b^2}}\) ➜ \(\because r=\sqrt{a^2+b^2}\)
\(\Rightarrow \cos{(\theta-\alpha)}=\cos{\beta}\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}\)
\(\Rightarrow \theta-\alpha=2n\pi\pm{\beta}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\beta+\alpha\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=2n\pi+\alpha\pm\beta\) যেখানে, \(\cos{\beta}=\frac{c}{\sqrt{a^2+b^2}}, \ n\in{\mathbb{Z}}\)
উদাহরণ \(9.\) সমাধান করঃ \(\cos{x}+\sqrt{3}\sin{x}=\sqrt{2}\)
উত্তরঃ \(2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
উত্তরঃ \(2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
দিঃ২০১৭; ঢাঃ২০১৫,২০০৭; চঃ২০১৪,২০০৭; সিঃ২০১৪,২০০৬; যঃ২০১২,২০০৯; বঃ২০০৮,২০০২; কুঃ২০০৬,২০০২; রাঃ২০০৫,২০০১; মাঃ২০১০,২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{x}+\sqrt{3}\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{1}{2}+\sin{x}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\frac{1}{2}+\sin{x}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\)
\(\Rightarrow \cos{x}\frac{1}{2}+\sin{x}\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{x}\cos{\frac{\pi}{3}}+\sin{x}\sin{\frac{\pi}{3}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}, \ \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{3}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow x-\frac{\pi}{3}=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{4}+\frac{\pi}{3}, \ 2n\pi-\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow x=2n\pi+\frac{3\pi+4\pi}{12}, \ 2n\pi+\frac{4\pi-3\pi}{12}\)
\(\therefore x=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\)
\(\therefore\) নির্ণেয় সমাধান, \(x=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{x}+\sqrt{3}\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{1}{2}+\sin{x}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\frac{1}{2}+\sin{x}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\)
\(\Rightarrow \cos{x}\frac{1}{2}+\sin{x}\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{x}\cos{\frac{\pi}{3}}+\sin{x}\sin{\frac{\pi}{3}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}, \ \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{3}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow x-\frac{\pi}{3}=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{4}+\frac{\pi}{3}, \ 2n\pi-\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow x=2n\pi+\frac{3\pi+4\pi}{12}, \ 2n\pi+\frac{4\pi-3\pi}{12}\)
\(\therefore x=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\)
\(\therefore\) নির্ণেয় সমাধান, \(x=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(10.\) সমাধান করঃ \(\sqrt{3}\cos{x}+\sin{x}=1, \ -2\pi\le{x}\le{2\pi}\)
উত্তরঃ \(-\frac{3\pi}{2}, \ -\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{11\pi}{6}\)
উত্তরঃ \(-\frac{3\pi}{2}, \ -\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{11\pi}{6}\)
দিঃ২০১৫; ঢাঃ২০১১,২০০৩; সিঃ২০০৮,২০০২; যঃ২০১৪,২০১০,২০০৭; বঃ২০১৪,২০০৪; কুঃ২০০৮; রাঃ২০১৩,২০০৮,২০০৬; মাঃ২০০৮,২০০৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{3}\cos{x}+\sin{x}=1\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+1^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{x}\sin{\frac{\pi}{6}}=\frac{1}{2}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}, \ \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}+\frac{\pi}{6}, \ 2n\pi-\frac{\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi+\pi}{6}, \ 2n\pi-\frac{2\pi-\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{3\pi}{6}, \ 2n\pi-\frac{\pi}{6}\)
\(\therefore x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{2}, \ -\frac{\pi}{6}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{2}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{5\pi}{2}, \ \frac{11\pi}{6}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore x=\frac{11\pi}{6}\)
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{2}, \ -2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{3\pi}{2}, \ -\frac{13\pi}{6}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore x=-\frac{3\pi}{2}\)
যখন, \(n=2,\) \(x=4\pi+\frac{\pi}{2}, \ 4\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(-\frac{3\pi}{2}, \ -\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{11\pi}{6}\)
\(\sqrt{3}\cos{x}+\sin{x}=1\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+1^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{x}\sin{\frac{\pi}{6}}=\frac{1}{2}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}, \ \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}+\frac{\pi}{6}, \ 2n\pi-\frac{\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi+\pi}{6}, \ 2n\pi-\frac{2\pi-\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{3\pi}{6}, \ 2n\pi-\frac{\pi}{6}\)
\(\therefore x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\le{x}\le{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{2}, \ -\frac{\pi}{6}\) সবগুলি মান গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{2}, \ -\frac{\pi}{6}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{2}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{5\pi}{2}, \ \frac{11\pi}{6}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore x=\frac{11\pi}{6}\)
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{2}, \ -2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{3\pi}{2}, \ -\frac{13\pi}{6}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore x=-\frac{3\pi}{2}\)
যখন, \(n=2,\) \(x=4\pi+\frac{\pi}{2}, \ 4\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(-\frac{3\pi}{2}, \ -\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{11\pi}{6}\)
উদাহরণ \(11.\) সমাধান করঃ \(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{180^{o}}\)
উত্তরঃ \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 162^{o}\)
উত্তরঃ \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 162^{o}\)
ঢাঃ২০০৮; কুঃ২০১১; চুয়েটঃ২০০৩-২০০৪ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{180^{o}}\)
\(\Rightarrow \frac{1}{\cos{4\theta}}-\frac{1}{\cos{2\theta}}=2\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{2\theta}-\cos{4\theta}}{\cos{4\theta}\cos{2\theta}}=2\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=2\cos{4\theta}\cos{2\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{(4\theta-2\theta)}+\cos{(4\theta+2\theta)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{2\theta}+\cos{6\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}-\cos{2\theta}-\cos{6\theta}=0\)
\(\Rightarrow -\cos{4\theta}-\cos{6\theta}=0\)
\(\Rightarrow \cos{6\theta}+\cos{4\theta}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{\frac{6\theta+4\theta}{2}}\cos{\frac{6\theta-4\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{5\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{5\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 5\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=18^{o}, \ 90^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=18^{o}, \ 90^{o}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{10}, \ \frac{3\pi}{2}\)
\(\Rightarrow \theta=54^{o}, \ 270^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=54^{o}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{10}, \ -\frac{\pi}{2}\)
\(\Rightarrow \theta=-18^{o}, \ -90^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{10}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=\frac{\pi}{2}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=90^{o}, \ 450^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=90^{o}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{10}, \ \frac{7\pi}{2}\)
\(\Rightarrow \theta=126^{o}, \ 630^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=126^{o}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{10}, \ \frac{9\pi}{2}\)
\(\Rightarrow \theta=162^{o}, \ 810^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=5,\) \(\theta=\frac{11\pi}{10}, \ \frac{11\pi}{2}\)
\(\Rightarrow \theta=198^{o}, \ 990^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{x}\le{180^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 162^{o}\)
\(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{180^{o}}\)
\(\Rightarrow \frac{1}{\cos{4\theta}}-\frac{1}{\cos{2\theta}}=2\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{2\theta}-\cos{4\theta}}{\cos{4\theta}\cos{2\theta}}=2\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=2\cos{4\theta}\cos{2\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{(4\theta-2\theta)}+\cos{(4\theta+2\theta)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{2\theta}+\cos{6\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}-\cos{2\theta}-\cos{6\theta}=0\)
\(\Rightarrow -\cos{4\theta}-\cos{6\theta}=0\)
\(\Rightarrow \cos{6\theta}+\cos{4\theta}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{\frac{6\theta+4\theta}{2}}\cos{\frac{6\theta-4\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{5\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{5\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 5\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{180^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{10}, \ \frac{\pi}{2}\) \(\Rightarrow \theta=18^{o}, \ 90^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=18^{o}, \ 90^{o}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{10}, \ \frac{3\pi}{2}\)
\(\Rightarrow \theta=54^{o}, \ 270^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=54^{o}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{10}, \ -\frac{\pi}{2}\)
\(\Rightarrow \theta=-18^{o}, \ -90^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{10}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=\frac{\pi}{2}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=90^{o}, \ 450^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=90^{o}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{10}, \ \frac{7\pi}{2}\)
\(\Rightarrow \theta=126^{o}, \ 630^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=126^{o}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{10}, \ \frac{9\pi}{2}\)
\(\Rightarrow \theta=162^{o}, \ 810^{o}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=5,\) \(\theta=\frac{11\pi}{10}, \ \frac{11\pi}{2}\)
\(\Rightarrow \theta=198^{o}, \ 990^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{x}\le{180^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 162^{o}\)
উদাহরণ \(12.\) সমাধান করঃ \(\cot{\theta}+\tan{\theta}=2\sec{\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
উত্তরঃ \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
উত্তরঃ \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
সিঃ২০১৬,২০১৫,২০১২; ঢাঃ২০১৫,২০১০; বঃ২০১৫,২০০৮; কুঃ২০১৪,২০০৭; চঃ২০১৩; যঃ২০১২; দিঃ২০১১; রাঃ২০০৭; মাঃ২০১২ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{\theta}+\tan{\theta}=2\sec{\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}+\frac{\sin{\theta}}{\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
\(\Rightarrow \frac{1}{\sin{\theta}}=2, \ \because \cos{\theta}\ne{0}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{\theta}=\sin{\alpha}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{7\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{7\pi}{6}\)
যখন, \(n=2,\) \(\theta=2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{13\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(\theta=-2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{11\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{17\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-3,\) \(\theta=-3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{19\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
\(\cot{\theta}+\tan{\theta}=2\sec{\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}+\frac{\sin{\theta}}{\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
\(\Rightarrow \frac{1}{\sin{\theta}}=2, \ \because \cos{\theta}\ne{0}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{\theta}=\sin{\alpha}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন,\(n=0,\) \(\theta=\frac{\pi}{6}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{7\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{7\pi}{6}\)
যখন, \(n=2,\) \(\theta=2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{13\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(\theta=-2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{11\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{17\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-3,\) \(\theta=-3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{19\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
উদাহরণ \(13.\) \(f(x)=\sin{x}\) এবং \(g(x)=\cos{x}\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\frac{1-x}{1+x}}\)
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, দেখাও যে, \(\theta=\pm{\frac{\pi}{4}}+\cos{\frac{1}{2\sqrt{2}}}\)
\((c)\) সমাধান করঃ \(g(x)+f(x)=g(2x)+f(2x)\)
উত্তরঃ \((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\frac{1-x}{1+x}}\)
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, দেখাও যে, \(\theta=\pm{\frac{\pi}{4}}+\cos{\frac{1}{2\sqrt{2}}}\)
\((c)\) সমাধান করঃ \(g(x)+f(x)=g(2x)+f(2x)\)
উত্তরঃ \((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১৫,২০০৫; ঢাঃ২০১৩,২০০৬,২০০২,২০০০; বঃ২০০৪; কুঃ২০১৪,২০০৫,২০০১; চঃ২০১২,২০০৬; যঃ২০১৪,২০১১,২০০৭; দিঃ২০১৪; রাঃ২০১২,২০০৮,২০০৪; মাঃ২০১৩,২০০৯ ।
সমাধানঃ
\((a)\)
\(L.S=\tan^{-1}{\sqrt{x}}\)\(=\frac{1}{2}\times2\tan^{-1}{\sqrt{x}}\)
\(=\frac{1}{2}\cos^{-1}{\frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2}}\) ➜ \(\because 2\tan^{-1}{A}=\cos^{-1}{\frac{1-A^2}{1+A^2}}\)
\(=\frac{1}{2}\cos^{-1}{\frac{1-x}{1+x}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,\(f(x)=\sin{x}, \ g(x)=\cos{x}\) এবং \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\)
\(\Rightarrow f\{\pi\cos{\theta}\}=g\{\pi \sin{\theta}\}\) ➜ \(\because f(x)=\sin{x}, \ g(x)=\cos{x}\)
\(\Rightarrow \sin{\left(\pi\cos{\theta}\right)}=\cos{\left(\pi\sin{\theta}\right)}\) ➜ \(\because f(x)=\sin{x}, \ g(x)=\cos{x}\)
\(\Rightarrow \sin{\left(\pi\cos{\theta}\right)}=\sin{\left(\frac{\pi}{2}\pm\pi\sin{\theta}\right)}\) ➜ \(\because \cos{A}=\sin{\left(\frac{\pi}{2}\pm{A}\right)}\)
\(\Rightarrow \pi\cos{\theta}=\frac{\pi}{2}\pm\pi\sin{\theta}\)
\(\Rightarrow \pi\cos{\theta}=\pi\left(\frac{1}{2}\pm\sin{\theta}\right)\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\pm\sin{\theta}\) ➜ উভয় পার্শে \(\pi\) ভাগ করে,
\(\Rightarrow \cos{\theta}\pm\sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}\pm\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{\theta}\) ও \(\sin{\theta}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}\pm\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{2\sqrt{2}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta\pm\frac{\pi}{4}\right)}=\frac{1}{2\sqrt{2}}\) ➜ \(\because \cos{A}\cos{B}\pm\sin{A}\sin{B}=\cos{(A\pm{B})}\)
\(\Rightarrow \theta\pm\frac{\pi}{4}=\cos^{-1}{\left(\frac{1}{2\sqrt{2}}\right)}\)
\(\therefore\ \theta=\pm\frac{\pi}{4}+\cos^{-1}{\left(\frac{1}{2\sqrt{2}}\right)}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,\(f(x)=\sin{x}, \ g(x)=\cos{x}\) এবং \(g(x)+f(x)=g(2x)+f(2x)\)
\(\Rightarrow \cos{x}+\sin{x}=\cos{2x}+\sin{2x}\) ➜ \(\because f(x)=\sin{x}, \ g(x)=\cos{x}\)
\(\Rightarrow \cos{x}-\cos{2x}=\sin{2x}-\sin{x}\)
\(\Rightarrow 2\sin{\frac{x+2x}{2}}\sin{\frac{2x-x}{2}}=2\cos{\frac{2x+x}{2}}\sin{\frac{2x-x}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}=\cos{\frac{3x}{2}}\sin{\frac{x}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}-\cos{\frac{3x}{2}}\sin{\frac{x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}\left(\sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}=\cos{\frac{3x}{2}}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \frac{\sin{\frac{3x}{2}}}{\cos{\frac{3x}{2}}}=1\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{x}{2}=n\pi, \ \frac{3x}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
\(\therefore\) নির্ণেয় সাধারণ সমাধান, \(x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(14.\) \(\sin{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\) এর সমাধান লেখের সাহায্যে নির্ণয় কর।
উত্তরঃ \(30^{o}; \ 150^{o}\)
উত্তরঃ \(30^{o}; \ 150^{o}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\sin{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \sin{x}=\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(30^{o}, \frac{1}{2}\right)\) ও \(\left(150^{o}, \frac{1}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\sin{x}=\frac{1}{2}\) ফাংশনের সমাধান, \(x=30^{o}, \ 150^{o}\)
\(\sin{x}=\frac{1}{2}\) যখন, \(0\le{x}\le{360^{o}}\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \sin{x}=\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(30^{o}, \frac{1}{2}\right)\) ও \(\left(150^{o}, \frac{1}{2}\right)\) নির্ণয় করি।

উদাহরণ \(15.\) লেখের সাহায্যে \(\tan{x}=\sqrt{3};\) \(-180^{o}\le{x}\le{180^{o}}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(-150^{o}, \ -60^{o}, \ 30^{o}, \ 120^{o}\)
উত্তরঃ \(-150^{o}, \ -60^{o}, \ 30^{o}, \ 120^{o}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\tan{x}=\sqrt{3};\) \(-180^{o}\le{x}\le{180^{o}}\)
\((a) \ -180^{o}\le{x}\le{180^{o}}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \tan{x}=\sqrt{3}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে চারটি বিন্দুতে ছেদ করে। সুতরাং এর চারটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(-150^{o}, \sqrt{3}\right),\) \(\left(-60^{o}, \sqrt{3}\right),\) \(\left(30^{o}, \sqrt{3}\right)\) ও \(\left(120^{o}, \sqrt{3}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\tan{x}=\sqrt{3}\) ফাংশনের সমাধান, \(x=-150^{o}, \ -60^{o}, \ 30^{o}, \ 120^{o}\)
\(\tan{x}=\sqrt{3};\) \(-180^{o}\le{x}\le{180^{o}}\)
\((a) \ -180^{o}\le{x}\le{180^{o}}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \tan{x}=\sqrt{3}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে চারটি বিন্দুতে ছেদ করে। সুতরাং এর চারটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(-150^{o}, \sqrt{3}\right),\) \(\left(-60^{o}, \sqrt{3}\right),\) \(\left(30^{o}, \sqrt{3}\right)\) ও \(\left(120^{o}, \sqrt{3}\right)\) নির্ণয় করি।

উদাহরণ \(16.\) সমাধান করঃ \(4(\sin^2{x}+\cos{x})=5\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4(\sin^2{x}+\cos{x})=5\)
\(\Rightarrow 4(1-\cos^2{x}+\cos{x})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-4\cos^2{x}+4\cos{x}-5=0\)
\(\Rightarrow -4\cos^2{x}+4\cos{x}-1=0\)
\(\Rightarrow 4\cos^2{x}-4\cos{x}+1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow (2\cos{x}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\cos{x}-1=0\)
\(\Rightarrow 2\cos{x}=1\)
\(\Rightarrow \cos{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm\frac{\pi}{3}\)
\(4(\sin^2{x}+\cos{x})=5\)
\(\Rightarrow 4(1-\cos^2{x}+\cos{x})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-4\cos^2{x}+4\cos{x}-5=0\)
\(\Rightarrow -4\cos^2{x}+4\cos{x}-1=0\)
\(\Rightarrow 4\cos^2{x}-4\cos{x}+1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow (2\cos{x}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\cos{x}-1=0\)
\(\Rightarrow 2\cos{x}=1\)
\(\Rightarrow \cos{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm\frac{\pi}{3}\)
উদাহরণ \(17.\) লেখের সাহায্যে \(2\cos^2{x}+\sin{x}=1;\) \(0\le{x}\le{2\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(\frac{7\pi}{6}, \ \frac{11\pi}{6}\)
উত্তরঃ \(\frac{7\pi}{6}, \ \frac{11\pi}{6}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\cos^2{x}+\sin{x}=1; \ 0\le{x}\le{2\pi}\)
\(\Rightarrow 2(1-\sin^2{x})+\sin{x}=1\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2-2\sin^2{x}+\sin{x}-1=0\)
\(\Rightarrow -2\sin^2{x}+\sin{x}+1=0\)
\(\Rightarrow 2\sin^2{x}-\sin{x}-1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow 2\sin^2{x}-2\sin{x}+\sin{x}-1=0\)
\(\Rightarrow 2\sin{x}(\sin{x}-1)+1(\sin{x}-1)=0\)
\(\Rightarrow (\sin{x}-1)(2\sin{x}+1)=0\)
\(\Rightarrow \sin{x}-1=0, \ 2\sin{x}+1=0\)
\(\Rightarrow \sin{x}=1, \ 2\sin{x}=-1\)
\(\Rightarrow \sin{x}=1, \ \sin{x}=-\frac{1}{2}\)
\(\therefore \sin{x}=1 ......(1)\)
\(\sin{x}=-\frac{1}{2} ......(2)\)
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \sin{x}=1\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে একটি বিন্দুতে ছেদ করে। সুতরাং এর একটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{2}, 1\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\sin{x}=1\) ফাংশনের সমাধান, \(x=\frac{\pi}{2}\)
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \sin{x}=-\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{7\pi}{6}, -\frac{1}{2}\right), \ \left(\frac{11\pi}{6}, -\frac{1}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\sin{x}=-\frac{1}{2}\) ফাংশনের সমাধান, \(x=\frac{7\pi}{6}, \ \frac{11\pi}{6}\)
\(2\cos^2{x}+\sin{x}=1; \ 0\le{x}\le{2\pi}\)
\(\Rightarrow 2(1-\sin^2{x})+\sin{x}=1\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2-2\sin^2{x}+\sin{x}-1=0\)
\(\Rightarrow -2\sin^2{x}+\sin{x}+1=0\)
\(\Rightarrow 2\sin^2{x}-\sin{x}-1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow 2\sin^2{x}-2\sin{x}+\sin{x}-1=0\)
\(\Rightarrow 2\sin{x}(\sin{x}-1)+1(\sin{x}-1)=0\)
\(\Rightarrow (\sin{x}-1)(2\sin{x}+1)=0\)
\(\Rightarrow \sin{x}-1=0, \ 2\sin{x}+1=0\)
\(\Rightarrow \sin{x}=1, \ 2\sin{x}=-1\)
\(\Rightarrow \sin{x}=1, \ \sin{x}=-\frac{1}{2}\)
\(\therefore \sin{x}=1 ......(1)\)
\(\sin{x}=-\frac{1}{2} ......(2)\)
\((1)\) নং সমীকরণের ক্ষেত্রে
\((a) \ 0\le{x}\le{2\pi}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \sin{x}=1\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে একটি বিন্দুতে ছেদ করে। সুতরাং এর একটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{2}, 1\right)\) নির্ণয় করি।

\((2)\) নং সমীকরণের ক্ষেত্রে
\((a) \ 0\le{x}\le{2\pi}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \sin{x}=-\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{7\pi}{6}, -\frac{1}{2}\right), \ \left(\frac{11\pi}{6}, -\frac{1}{2}\right)\) নির্ণয় করি।

উদাহরণ \(18.\) লেখের সাহায্যে \(\cos{x}=-\frac{1}{2};\) \(0\le{x}\le{360^{o}}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(120^{o}, \ 240^{o}\)
উত্তরঃ \(120^{o}, \ 240^{o}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\cos{x}=-\frac{1}{2};\) \(0\le{x}\le{360^{o}}\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \cos{x}=-\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(120^{o}, -\frac{1}{2}\right),\) \(\left(240^{o}, -\frac{1}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\cos{x}=-\frac{1}{2}\) ফাংশনের সমাধান, \(x=120^{o}, \ 240^{o}\)
\(\cos{x}=-\frac{1}{2};\) \(0\le{x}\le{360^{o}}\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \cos{x}=-\frac{1}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(120^{o}, -\frac{1}{2}\right),\) \(\left(240^{o}, -\frac{1}{2}\right)\) নির্ণয় করি।

উদাহরণ \(19.\) লেখের সাহায্যে \(\tan{x}=-1;\) \(-\pi\le{x}\le{\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(-\frac{\pi}{4}, \ \frac{3\pi}{4}\)
উত্তরঃ \(-\frac{\pi}{4}, \ \frac{3\pi}{4}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\tan{x}=-1;\) \(-\pi\le{x}\le{\pi}\)
\((a) \ -\pi\le{x}\le{\pi}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \tan{x}=-1\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(-\frac{\pi}{4}, -1\right),\) \(\left(\frac{3\pi}{4}, -1\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\tan{x}=-1\) ফাংশনের সমাধান, \(x=-\frac{\pi}{4}, \ \frac{3\pi}{4}\)
\(\tan{x}=-1;\) \(-\pi\le{x}\le{\pi}\)
\((a) \ -\pi\le{x}\le{\pi}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \tan{x}=-1\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(-\frac{\pi}{4}, -1\right),\) \(\left(\frac{3\pi}{4}, -1\right)\) নির্ণয় করি।

উদাহরণ \(20.\) লেখের সাহায্যে \(\cos{2x}=\frac{\sqrt{3}}{2};\) \(0\le{x}\le{2\pi}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{12}, \ \frac{11\pi}{12}, \ \frac{13\pi}{12}, \ \frac{23\pi}{12}\)
উত্তরঃ \(\frac{\pi}{12}, \ \frac{11\pi}{12}, \ \frac{13\pi}{12}, \ \frac{23\pi}{12}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\cos{2x}=\frac{\sqrt{3}}{2};\) \(0\le{x}\le{2\pi}\)
\((a) \ 0\le{x}\le{2\pi}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \cos{2x}=\frac{\sqrt{3}}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে চারটি বিন্দুতে ছেদ করে। সুতরাং এর চারটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{12}, \frac{\sqrt{3}}{2}\right),\) \(\left(\frac{11\pi}{12}, \frac{\sqrt{3}}{2}\right),\) \(\left(\frac{13\pi}{12}, \frac{\sqrt{3}}{2}\right)\) এবং \(\left(\frac{23\pi}{12}, \frac{\sqrt{3}}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে \(\cos{2x}=\frac{\sqrt{3}}{2}\) ফাংশনের সমাধান, \(x=\frac{\pi}{12}, \ \frac{11\pi}{12}, \ \frac{13\pi}{12}, \ \frac{23\pi}{12}\)
\(\cos{2x}=\frac{\sqrt{3}}{2};\) \(0\le{x}\le{2\pi}\)
\((a) \ 0\le{x}\le{2\pi}\) ব্যবধিতে ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=\frac{\pi}{18}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ \cos{2x}=\frac{\sqrt{3}}{2}\) নির্দেশিত স্থানে, \(x\) অক্ষের সমান্তরাল সরলরেখা ফাংশনের লেখকে চারটি বিন্দুতে ছেদ করে। সুতরাং এর চারটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{12}, \frac{\sqrt{3}}{2}\right),\) \(\left(\frac{11\pi}{12}, \frac{\sqrt{3}}{2}\right),\) \(\left(\frac{13\pi}{12}, \frac{\sqrt{3}}{2}\right)\) এবং \(\left(\frac{23\pi}{12}, \frac{\sqrt{3}}{2}\right)\) নির্ণয় করি।

উদাহরণ \(21.\) \(f(x)=\cos{x}\) এবং \(g(x)=\sin^{-1}{x}\)
\((a)\) \(f(x)\) এর পর্যায় কাল ও \(g(x)\) এর মূখ্যমান কত?
\((b)\) প্রমাণ কর যে, \(g\left\{\sqrt{2} f\left(\frac{\pi}{2}-\theta\right)\right\}+g\left\{\sqrt{f(2\theta)}\right\}=\frac{\pi}{2}\)
\((c)\) \(4f(x)f(2x)f(3x)=1\) হলে, \(0\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয় কর।
উত্তরঃ \((a) \ \left[-\frac{\pi}{2}, \ \frac{\pi}{2}\right]\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}\)
\((a)\) \(f(x)\) এর পর্যায় কাল ও \(g(x)\) এর মূখ্যমান কত?
\((b)\) প্রমাণ কর যে, \(g\left\{\sqrt{2} f\left(\frac{\pi}{2}-\theta\right)\right\}+g\left\{\sqrt{f(2\theta)}\right\}=\frac{\pi}{2}\)
\((c)\) \(4f(x)f(2x)f(3x)=1\) হলে, \(0\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয় কর।
উত্তরঃ \((a) \ \left[-\frac{\pi}{2}, \ \frac{\pi}{2}\right]\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}\)
সমাধানঃ
\((a)\)
দেওয়া আছে,\(f(x)=\cos{x}\) এবং \(g(x)=\sin^{-1}{x}\)
আমরা জানি,
\(\cos{x}\) ফাংশন পর্যায়ী এবং এর পর্যায়কাল \(2\pi\) ।
\(\therefore f(x)\) এর পর্যায়কাল \(2\pi\) ।
আবার,
\(g(x)=\sin^{-1}{x}\)
\(\sin^{-1}{x}\) ফাংশন \(\left[-\frac{\pi}{2}, \ -\frac{\pi}{2}\right]\) ব্যবধিতে এক-এক।
\(\therefore\) বিপরীত \(\sin^{-1}{x}\) এর মুখ্যমানের ব্যবধি \(\left[-\frac{\pi}{2}, \ -\frac{\pi}{2}\right]\) ।
\(\therefore g(x)\) এর মুখ্যমানের ব্যবধি \(\left[-\frac{\pi}{2}, \ -\frac{\pi}{2}\right]\) ।
\((b)\)
দেওয়া আছে,\(f(x)=\cos{x}\) এবং \(g(x)=\sin^{-1}{x}\)
\(L.S=g\left\{\sqrt{2} f\left(\frac{\pi}{2}-\theta\right)\right\}+g\left\{\sqrt{f(2\theta)}\right\}\)
\(=g\left\{\sqrt{2} \cos{\left(\frac{\pi}{2}-\theta\right)}\right\}+g\left\{\sqrt{\cos{2\theta}}\right\}\) ➜ \(\because f(x)=\cos{x}\)
\(\Rightarrow f\left(\frac{\pi}{2}-\theta\right)=\cos{\left(\frac{\pi}{2}-\theta\right)}\)
এবং \(f(2\theta)=\cos{2\theta}\)
\(=\sin^{-1}{\left\{\sqrt{2}\cos{\left(\frac{\pi}{2}-\theta\right)}\right\}}+\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\) ➜ \(\because g(x)=\sin^{-1}{x}\)
\(\Rightarrow g\left\{\sqrt{2} \cos{\left(\frac{\pi}{2}-\theta\right)}\right\}=\sin^{-1}{\left\{\sqrt{2}\cos{\left(\frac{\pi}{2}-\theta\right)}\right\}}\)
এবং \(g\left\{\sqrt{\cos{2\theta}}\right\}=\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\)
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}+\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\) ➜

প্রথম পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\sqrt{1-\cos{2\theta}}+\sqrt{\cos{2\theta}}\sqrt{1-2\sin^2{\theta}}\right)}\) ➜ \(\because \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)}\)
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\sqrt{2\sin^2{\theta}}+\sqrt{\cos{2\theta}}\times\sqrt{\cos{2\theta}}\right)}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1-2\sin^2{A}=\cos{2A}\)
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\times\sqrt{2}\sin{\theta}+\cos{2\theta}\right)}\)
\(=\sin^{-1}{\left(2\sin^2{\theta}+1-2\sin^2{\theta}\right)}\) ➜ \(\because \cos{2A}=1-2\sin^2{A}\)
\(=\sin^{-1}{\left(1\right)}\)
\(=\sin^{-1}{\left(\sin{\frac{\pi}{2}}\right)}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
\(=R.S\)
\(\therefore g\left\{\sqrt{2} f\left(\frac{\pi}{2}-\theta\right)\right\}+g\left\{\sqrt{f(2\theta)}\right\}=\frac{\pi}{2}\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,\(f(x)=\cos{x}\)
এবং \(4f(x)f(2x)f(3x)=1, \ 0\lt{x}\lt{\pi}\)
\(\Rightarrow 4\cos{x}\cos{2x}\cos{3x}=1\) ➜ \(\because f(x)=\cos{x}\)
\(\Rightarrow f(2x)=\cos{2x}\)
এবং \(f(3x)=\cos{3x}\)
\(\Rightarrow 2\cos{2x}\{\cos{(3x-x)}+\cos{(3x+x)}\}=1\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow 2\cos{2x}\{\cos{2x}+\cos{4x}\}-1=0\)
\(\Rightarrow 2\cos^2{2x}+2\cos{4x}\cos{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+2\cos^2{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+\cos{4x}=0\) ➜ \(\because 2\cos^2{A}-1=\cos{2A}\)
\(\Rightarrow \cos{4x}(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 4x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{8}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ -\frac{\pi}{3}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{8}, \ \frac{\pi}{3}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{8}, \ \pi+\frac{\pi}{3}, \ \pi-\frac{\pi}{3}\)
\(\Rightarrow x=\frac{3\pi}{8}, \ \frac{4\pi}{3}, \ \frac{2\pi}{3}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{8}, \ \frac{2\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{8}, \ -\pi+\frac{\pi}{3}, \ -\pi-\frac{\pi}{3}\)
\(\Rightarrow x=-\frac{\pi}{8}, \ -\frac{2\pi}{3}, \ -\frac{4\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{5\pi}{8}, \ 2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{8}\)
যখন, \(n=3,\) \(x=\frac{7\pi}{8}, \ 3\pi+\frac{\pi}{3}, \ 3\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{7\pi}{8}\)
যখন, \(n=4,\) \(x=\frac{9\pi}{8}, \ 4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
উদাহরণ \(22.\) দুইটি আতসবাজি বিস্ফোরণের ফলে \(y_{1}=a\cos{x_{1}}\) এবং \(y_{2}=b\cos{x_{2}}\) পরিবর্তনশীল সরণ বিশিষ্ট দুইটি শব্দ তরঙ্গ উৎপন্ন হয়। উপরিপাতনের ফলে তরঙ্গদ্বয়ের মিলিত তরঙ্গের সরণ হলো \(y=y_{1}+y_{2}\) । এখানে \(y\) হলো তরঙ্গের সরণ এবং \(x\) হলো সময়।
\((a)\) \(x_{1}=cosec^{-1}{\frac{a}{\sqrt{a^2-1}}}\) হলে, \(y_{1}\) নির্ণয় কর।
\((b)\) \(a=b=1\) এবং \(x_{1}+x_{2}=\frac{\pi}{2}\) হলে, কত সময় পরে মিলিত তরঙ্গের সরণ \(\sqrt{2}\) একক হবে?
\((c)\) \(x_{1}+x_{2}=\theta\) হলে প্রমাণ কর যে, \(\frac{y_{1}^2}{a^2}-\frac{2y_{1}y_{2}\cos{\theta}}{ab}+\frac{y_{2}^2}{b^2}=\sin^2{\theta}.\)
উত্তরঃ \((a) \ 1\)
\((b) \ \frac{\pi}{4}\)
\((a)\) \(x_{1}=cosec^{-1}{\frac{a}{\sqrt{a^2-1}}}\) হলে, \(y_{1}\) নির্ণয় কর।
\((b)\) \(a=b=1\) এবং \(x_{1}+x_{2}=\frac{\pi}{2}\) হলে, কত সময় পরে মিলিত তরঙ্গের সরণ \(\sqrt{2}\) একক হবে?
\((c)\) \(x_{1}+x_{2}=\theta\) হলে প্রমাণ কর যে, \(\frac{y_{1}^2}{a^2}-\frac{2y_{1}y_{2}\cos{\theta}}{ab}+\frac{y_{2}^2}{b^2}=\sin^2{\theta}.\)
উত্তরঃ \((a) \ 1\)
\((b) \ \frac{\pi}{4}\)
সমাধানঃ
\((a)\)
দেওয়া আছে,\(y_{1}=a\cos{x_{1}}\) এবং \(x_{1}=cosec^{-1}{\frac{a}{\sqrt{a^2-1}}},\)
\(\Rightarrow y_{1}=a\cos{\left(cosec^{-1}{\frac{a}{\sqrt{a^2-1}}}\right)}\) ➜ \(\because x_{1}=cosec^{-1}{\frac{a}{\sqrt{a^2-1}}}\)
\(=a\cos{\left(\cos^{-1}{\frac{1}{a}}\right)}\) ➜ এখানে,
\(\text{লম্ব}=\sqrt{a^2-1}, \ \text{অতিভুজ}=a\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{a^2-(\sqrt{a^2-1})^2}\)
\(=\sqrt{a^2-a^2+1}\)
\(=\sqrt{1}\)
\(=1\)
\(\therefore cosec^{-1}{\frac{a}{\sqrt{a^2-1}}}=\cos^{-1}{\frac{1}{a}}\)
\(=a\times\frac{1}{a}\)
\(=1\)
\(\therefore y_{1}=1\)
\((b)\)
দেওয়া আছে,দেওয়া আছে তরঙ্গদ্বয়ের মিলিত তরঙ্গের সরণ, \(y=y_{1}+y_{2},\) \(a=b=1,\) এবং \(x_{1}+x_{2}=\frac{\pi}{2}\)
শর্ত মতে, \(y=\sqrt{2}\)
\(\Rightarrow y_{1}+y_{2}=\sqrt{2}\) ➜ \(\because y=y_{1}+y_{2}\)
\(\Rightarrow a\cos{x_{1}}+b\cos{x_{2}}=\sqrt{2}\) ➜ \(\because y_{1}=a\cos{x_{1}}, \ y_{2}=b\cos{x_{2}}\)
\(\Rightarrow a\cos{x_{1}}+b\cos{\left(\frac{\pi}{2}-x_{1}\right)}=\sqrt{2}\) ➜ \(\because x_{1}+x_{2}=\frac{\pi}{2}\)
\(\Rightarrow x_{2}=\frac{\pi}{2}-x_{1}\)
\(\Rightarrow a\cos{x_{1}}+b\sin{x_{1}}=\sqrt{2}\) ➜

দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।
\(\Rightarrow 1.\cos{x_{1}}+1.\sin{x_{1}}=\sqrt{2}\) ➜ \(\because a=b=1\)
\(\Rightarrow \cos{x_{1}}+\sin{x_{1}}=\sqrt{2}\)
\(\Rightarrow \cos{x_{1}}\frac{1}{\sqrt{2}}+\sin{x_{1}}\frac{1}{\sqrt{2}}=1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{x_{1}}\cos{\frac{\pi}{4}}+\sin{x_{1}}\sin{\frac{\pi}{4}}=1\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(x_{1}-\frac{\pi}{4}\right)}=\cos{0^{o}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(1=\cos{0^{o}}\)
\(\Rightarrow x_{1}-\frac{\pi}{4}=0\)
\(\therefore x_{1}=\frac{\pi}{4}\)
\(\therefore\) প্রথম তরঙ্গ উৎপন্নের পর \(\frac{\pi}{4}\) সময় অতিক্রান্ত হলে তরঙ্গদ্বয় মিলিত হবে।
\((c)\)
দেওয়া আছে,\(y_{1}=a\cos{x_{1}}, \ y_{2}=b\cos{x_{2}}\) এবং \(x_{1}+x_{2}=\theta\)
\(\Rightarrow a\cos{x_{1}}=y_{1}, \ b\cos{x_{2}}=y_{2}\)
\(\Rightarrow \cos{x_{1}}=\frac{y_{1}}{a}, \ \cos{x_{2}}=\frac{y_{2}}{b}\)
\(\therefore x_{1}=\cos^{-1}{\frac{y_{1}}{a}}, \ x_{2}=\cos^{-1}{\frac{y_{2}}{b}}\)
আবার,
\(x_{1}+x_{2}=\theta\)
\(\Rightarrow \cos^{-1}{\frac{y_{1}}{a}}+\cos^{-1}{\frac{y_{2}}{b}}=\theta\) ➜ \(\because x_{1}=\cos^{-1}{\frac{y_{1}}{a}}\)
এবং \(x_{2}=\cos^{-1}{\frac{y_{2}}{b}}\)
\(\Rightarrow \cos^{-1}{\left\{\frac{y_{1}}{a}.\frac{y_{2}}{b}-\sqrt{\left(1-\frac{y_{1}^2}{a^2}\right)\left(1-\frac{y_{2}^2}{b^2}\right)}\right\}}=\theta\) ➜ \(\because \cos^{-1}{x}+\cos^{-1}{y}=\cos^{-1}{\left\{xy-\sqrt{(1-x^2)(1-y^2)}\right\}}\)
\(\Rightarrow \frac{y_{1}y_{2}}{ab}-\sqrt{\left(1-\frac{y_{1}^2}{a^2}\right)\left(1-\frac{y_{2}^2}{b^2}\right)}=\cos{\theta}\)
\(\Rightarrow \frac{y_{1}y_{2}}{ab}-\cos{\theta}=\sqrt{\left(1-\frac{y_{1}^2}{a^2}\right)\left(1-\frac{y_{2}^2}{b^2}\right)}\)
\(\Rightarrow \frac{y_{1}y_{2}}{ab}-\cos{\theta}=\sqrt{1-\frac{y_{1}^2}{a^2}-\frac{y_{2}^2}{b^2}+\frac{y_{1}^2y_{2}^2}{a^2b^2}}\)
\(\Rightarrow \left(\frac{y_{1}y_{2}}{ab}-\cos{\theta}\right)^2=1-\frac{y_{1}^2}{a^2}-\frac{y_{2}^2}{b^2}+\frac{y_{1}^2y_{2}^2}{a^2b^2}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{y_{1}^2y_{2}^2}{a^2b^2}-2\frac{y_{1}y_{2}}{ab}\cos{\theta}+\cos^2{\theta}=1-\frac{y_{1}^2}{a^2}-\frac{y_{2}^2}{b^2}+\frac{y_{1}^2y_{2}^2}{a^2b^2}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(\Rightarrow \frac{y_{1}^2y_{2}^2}{a^2b^2}-2\frac{y_{1}y_{2}}{ab}\cos{\theta}+\frac{y_{1}^2}{a^2}+\frac{y_{2}^2}{b^2}-\frac{y_{1}^2y_{2}^2}{a^2b^2}=1-\cos^2{\theta}\)
\(\Rightarrow \frac{y_{1}^2}{a^2}-2\frac{y_{1}y_{2}}{ab}\cos{\theta}+\frac{y_{2}^2}{b^2}=\sin^2{\theta}\) ➜ \(\because 1-\cos^2{A}=\sin^2{A}\)
\(\therefore \frac{y_{1}^2}{a^2}-\frac{2y_{1}y_{2}\cos{\theta}}{ab}+\frac{y_{2}^2}{b^2}=\sin^2{\theta}\)
(প্রমাণিত)
উদাহরণ \(23.\) সমাধান করঃ \(2\cos^2{\theta}+2\sqrt{2}\sin{\theta}=3;\) \(0\le{\theta}\le{\frac{\pi}{2}}\)
উত্তরঃ \(\frac{\pi}{4}\)
উত্তরঃ \(\frac{\pi}{4}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\cos^2{\theta}+2\sqrt{2}\sin{\theta}=3, \ 0\le{\theta}\le{\frac{\pi}{2}}\)
\(\Rightarrow 2(1-\sin^2{\theta})+2\sqrt{2}\sin{\theta}=3\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2-2\sin^2{\theta}+2\sqrt{2}\sin{\theta}-3=0\)
\(\Rightarrow -2\sin^2{\theta}+2\sqrt{2}\sin{\theta}-1=0\)
\(\Rightarrow 2\sin^2{\theta}-2\sqrt{2}\sin{\theta}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow (\sqrt{2}\sin{\theta})^2-2\sqrt{2}\sin{\theta}+1^2=0\)
\(\Rightarrow (\sqrt{2}\sin{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{2}\sin{\theta}-1=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{4}\)
যখন, \(n=1,\) \(\theta=\pi-\frac{\pi}{4}\)
\(\Rightarrow \theta=\frac{3\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{\frac{\pi}{2}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{4}\)
\(2\cos^2{\theta}+2\sqrt{2}\sin{\theta}=3, \ 0\le{\theta}\le{\frac{\pi}{2}}\)
\(\Rightarrow 2(1-\sin^2{\theta})+2\sqrt{2}\sin{\theta}=3\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2-2\sin^2{\theta}+2\sqrt{2}\sin{\theta}-3=0\)
\(\Rightarrow -2\sin^2{\theta}+2\sqrt{2}\sin{\theta}-1=0\)
\(\Rightarrow 2\sin^2{\theta}-2\sqrt{2}\sin{\theta}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow (\sqrt{2}\sin{\theta})^2-2\sqrt{2}\sin{\theta}+1^2=0\)
\(\Rightarrow (\sqrt{2}\sin{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{2}\sin{\theta}-1=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{\frac{\pi}{2}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{4}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{4}\)
যখন, \(n=1,\) \(\theta=\pi-\frac{\pi}{4}\)
\(\Rightarrow \theta=\frac{3\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{\frac{\pi}{2}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{4}\)
উদাহরণ \(24.\) \(f(x)=\tan^{-1}{x}\)
\((a)\) দেখাও যে, \(2f(x)=\sin^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\((b)\) প্রমাণ কর যে, \(\tan{\left\{2f(x)\right\}}=2\tan{\left\{f(x)+f(x^3)\right\}}\)
\((c)\) \(f(x)=\tan{x}\) হলে, সমাধান করঃ \(f(x)f(2x)=1; \ 0\le{x}\le{\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
\((a)\) দেখাও যে, \(2f(x)=\sin^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\((b)\) প্রমাণ কর যে, \(\tan{\left\{2f(x)\right\}}=2\tan{\left\{f(x)+f(x^3)\right\}}\)
\((c)\) \(f(x)=\tan{x}\) হলে, সমাধান করঃ \(f(x)f(2x)=1; \ 0\le{x}\le{\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
সমাধানঃ
\((a)\)
দেওয়া আছে,\(f(x)=\tan^{-1}{x}\)
\(L.S=2f(x)\)
\(=2\tan^{-1}{x}\) ➜ \(\because f(x)=\tan^{-1}{x}\)
ধরি,
\(\tan^{-1}{x}=\theta\)
\(\therefore x=\tan{\theta}\)
এখন
\(\sin{2\theta}=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\sin^{-1}{\left(\frac{2\tan{\theta}}{1+\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\sin^{-1}{\left(\frac{2x}{1+x^2}\right)}\) ➜ \(\because \tan^{-1}{x}=\theta\)
\(\Rightarrow x=\tan{\theta}\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,\(f(x)=\tan^{-1}{x}\)
\(\Rightarrow f(x^3)=\tan^{-1}{x^3}\)
\(L.S=\tan{\left\{2f(x)\right\}}\)
\(=\tan{\left\{2\tan^{-1}{x}\right\}}\) ➜ \(\because f(x)=\tan^{-1}{x}\)
\(=\tan{\left\{\tan^{-1}{\frac{2x}{1-x^2}}\right\}}\) ➜ \(\because 2\tan^{-1}{a}=\tan^{-1}{\frac{2a}{1-a^2}}\)
\(=\frac{2x}{1-x^2}\)
\(R.S=2\tan{\left\{f(x)+f(x^3)\right\}}\)
\(=2\tan{\left\{\tan^{-1}{x}+\tan^{-1}{x^3}\right\}}\) ➜ \(\because f(x)=\tan^{-1}{x}\)
\(\Rightarrow f(x^3)=\tan^{-1}{x^3}\)
\(=2\tan{\left\{\tan^{-1}{\frac{x+x^3}{1-x\times{x^3}}}\right\}}\) ➜ \(\because \tan^{-1}{a}+\tan^{-1}{b}=\tan^{-1}{\frac{a+b}{1-ab}}\)
\(=2\frac{x+x^3}{1-x^4}\)
\(=2\frac{x(1+x^2)}{(1-x^2)(1+x^2)}\)
\(=2\frac{x}{1-x^2}\)
\(=\frac{2x}{1-x^2}\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,\(f(x)=\tan{x}\)
\(\Rightarrow f(2x)=\tan{2x}\)
প্রদত্ত সমীকরণ,
\(f(x)f(2x)=1, \ 0\le{x}\le{\pi}\)
\(\Rightarrow \tan{x}\tan{2x}=1\)
\(\Rightarrow \tan{2x}\tan{x}=1\)
\(\Rightarrow \frac{\sin{2x}\sin{x}}{\cos{2x}\cos{x}}=1\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \cos{2x}\cos{x}=\sin{2x}\sin{x}\)
\(\Rightarrow \cos{2x}\cos{x}-\sin{2x}\sin{x}=0\)
\(\Rightarrow \cos{(2x+x)}=0\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \cos{3x}=0\)
\(\Rightarrow 3x=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{x}=0\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(x=(2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{x}\le{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{6}\) মানটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{6}\)
যখন, \(n=1,\) \(x=3\frac{\pi}{6}\)
\(\Rightarrow x=\frac{\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=5\frac{\pi}{6}\)
\(\Rightarrow x=\frac{5\pi}{6}\) মানটি গ্রহনযোগ্য ।
\(\therefore x=\frac{5\pi}{6}\)
যখন, \(n=-2,\) \(x=-3\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{6}, \ \frac{5\pi}{6}\)
উদাহরণ \(25.\) \(\sin{(\pi\cos{\theta})}=\cos{(\pi\sin{\theta})} ........(1)\)
এবং \(\sqrt{3}\sin{\theta}-\cos{\theta}=\sqrt{2} ........(2)\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\left(\frac{1-x}{1+x}\right)}\)
\((b)\) উদ্দীপকের \((1)\) নং হতে প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) \(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে উদ্দীপকের \((2)\) নং সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{19\pi}{12}, \ -\frac{13\pi}{12}, \ \frac{5\pi}{12}, \ \frac{11\pi}{12}\)
এবং \(\sqrt{3}\sin{\theta}-\cos{\theta}=\sqrt{2} ........(2)\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\left(\frac{1-x}{1+x}\right)}\)
\((b)\) উদ্দীপকের \((1)\) নং হতে প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) \(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে উদ্দীপকের \((2)\) নং সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{19\pi}{12}, \ -\frac{13\pi}{12}, \ \frac{5\pi}{12}, \ \frac{11\pi}{12}\)
সমাধানঃ
\((a)\)
\(L.S=\tan^{-1}{\sqrt{x}}\)\(=\frac{1}{2}\times2\tan^{-1}{\sqrt{x}}\)
\(=\frac{1}{2}\cos^{-1}{\left\{\frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2}\right\}}\) ➜ \(\because 2\tan^{-1}{a}=\cos^{-1}{\frac{1-a^2}{1+a^2}}\)
\(=\frac{1}{2}\cos^{-1}{\left(\frac{1-x}{1+x}\right)}\)
\(=R.S\)
\(\therefore \tan^{-1}{\sqrt{x}}=\frac{1}{2}\cos^{-1}{\left(\frac{1-x}{1+x}\right)}\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,\(\sin{(\pi\cos{\theta})}=\cos{(\pi\sin{\theta})}\)
\(\Rightarrow \sin{(\pi\cos{\theta})}=\sin{\left(\frac{\pi}{2}\pm\pi\sin{\theta}\right)}\)
\(\Rightarrow \pi\cos{\theta}=\frac{\pi}{2}\pm\pi\sin{\theta}\)
\(\Rightarrow \pi\cos{\theta}=\pi\left(\frac{1}{2}\pm\sin{\theta}\right)\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\pm\sin{\theta}\)
\(\Rightarrow \cos{\theta}\pm\sin{\theta}=\frac{1}{2}\)
\(\Rightarrow (\cos{\theta}\pm\sin{\theta})^2=\frac{1}{4}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \cos^2{\theta}\pm2\cos{\theta}\sin{\theta}+\sin^2{\theta}=\frac{1}{4}\) ➜ \(\because (a\pm{b})^2=a^2\pm2ab+b^2\)
\(\Rightarrow \sin^2{\theta}+\cos^2{\theta}\pm2\sin{\theta}\cos{\theta}=\frac{1}{4}\)
\(\Rightarrow 1\pm\sin{2\theta}=\frac{1}{4}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \pm\sin{2\theta}=\frac{1}{4}-1\)
\(\Rightarrow \pm\sin{2\theta}=\frac{1-4}{4}\)
\(\Rightarrow \pm\sin{2\theta}=-\frac{3}{4}\)
\(\Rightarrow \sin{2\theta}=\pm\frac{3}{4}\)
\(\Rightarrow 2\theta=\sin{\left(\pm\frac{3}{4}\right)}\)
\(\Rightarrow 2\theta=\pm\sin{\frac{3}{4}}\)
\(\therefore \theta=\pm\frac{1}{2}\sin{\frac{3}{4}}\)
(প্রমাণিত)
\((c)\)
প্রদত্ত সমীকরণ,\(\sqrt{3}\sin{\theta}-\cos{\theta}=\sqrt{2}, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow \cos{\theta}-\sqrt{3}\sin{\theta}=-\sqrt{2}\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{\theta}\frac{1}{2}-\sin{\theta}\frac{\sqrt{3}}{2}=-\frac{\sqrt{2}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}-\sin{\theta}\sin{\frac{\pi}{3}}=-\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}-\sin{\theta}\sin{\frac{\pi}{3}}=-\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{3}\right)}=-\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{3}\right)}=\cos{\left(\pi-\frac{\pi}{4}\right)}\) ➜ \(\because -\cos{A}=\cos{(\pi-A)}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{3}\right)}=\cos{\frac{3\pi}{4}}\)
\(\Rightarrow \theta+\frac{\pi}{3}=2n\pi\pm\frac{3\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{3\pi}{4}-\frac{\pi}{3}\)
\(\Rightarrow \theta=2n\pi+\frac{3\pi}{4}-\frac{\pi}{3}, \ 2n\pi-\frac{3\pi}{4}-\frac{\pi}{3}\)
\(\Rightarrow \theta=2n\pi+\frac{9\pi-4\pi}{12}, \ 2n\pi-\frac{9\pi+4\pi}{12}\)
\(\therefore \theta=2n\pi+\frac{5\pi}{12}, \ 2n\pi-\frac{13\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+\frac{5\pi}{12}, \ 2n\pi-\frac{13\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{5\pi}{12}, \ -\frac{13\pi}{12}\) মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=\frac{5\pi}{12}, \ -\frac{13\pi}{12}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{5\pi}{12}, \ 2\pi-\frac{13\pi}{12}\)
\(\Rightarrow \theta=\frac{29\pi}{12}, \ \frac{11\pi}{12}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{11\pi}{12}\)
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{5\pi}{12}, \ -2\pi-\frac{13\pi}{12}\)
\(\Rightarrow \theta=-\frac{19\pi}{12}, \ -\frac{37\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{19\pi}{12}\)
যখন, \(n=2,\) \(\theta=4\pi+\frac{5\pi}{12}, \ 4\pi-\frac{13\pi}{12}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(\theta=-4\pi+\frac{5\pi}{12}, \ -4\pi-\frac{13\pi}{12}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{19\pi}{12}, \ -\frac{13\pi}{12}, \ \frac{5\pi}{12}, \ \frac{11\pi}{12}\)
উদাহরণ \(26.\) \(f(x)=\cos{x}, \ g(x)=\sin{x}\)
\((a)\) \(\frac{g(2\theta)g(\theta)}{f(2\theta)f(\theta)}=1\) এর সমাধান কর।
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) সমাধান করঃ \(f(x)-f(2x)=g(2x)-g(x)\)
উত্তরঃ \((a) \ n\pi\pm\frac{\pi}{6}\)
\((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
\((a)\) \(\frac{g(2\theta)g(\theta)}{f(2\theta)f(\theta)}=1\) এর সমাধান কর।
\((b)\) \(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\) হলে, প্রমাণ কর যে, \(\theta=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) সমাধান করঃ \(f(x)-f(2x)=g(2x)-g(x)\)
উত্তরঃ \((a) \ n\pi\pm\frac{\pi}{6}\)
\((c) \ 2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
সমাধানঃ
\((a)\)
দেওয়া আছে,\(f(x)=\cos{x}, \ g(x)=\sin{x}\)
প্রদত্ত সমীকরণ,
\(\frac{g(2\theta)g(\theta)}{f(2\theta)f(\theta)}=1\)
\(\Rightarrow \frac{\sin{2\theta}\sin{\theta}}{\cos{2\theta}\cos{\theta}}=1\) ➜ \(\because f(x)=\cos{x}, \ g(x)=\sin{x}\)
\(\Rightarrow f(2\theta)=\cos{2\theta}, \ g(2\theta)=\sin{2\theta}\)
\(\Rightarrow f(\theta)=\cos{\theta}, \ g(\theta)=\sin{\theta}\)
\(\tan{2\theta}\tan{\theta}=1\)
\(\Rightarrow \frac{2\tan{\theta}}{1-\tan^2{\theta}}\times\tan{\theta}=1\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\Rightarrow \frac{2\tan^2{\theta}}{1-\tan^2{\theta}}=1\)
\(\Rightarrow 2\tan^2{\theta}=1-\tan^2{\theta}\)
\(\Rightarrow 2\tan^2{\theta}+\tan^2{\theta}=1\)
\(\Rightarrow 3\tan^2{\theta}=1\)
\(\Rightarrow \tan^2{\theta}=\frac{1}{3}\)
\(\Rightarrow \tan{\theta}=\pm\sqrt{\frac{1}{3}}\)
\(\Rightarrow \tan{\theta}=\pm\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{6}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{6}\right)\) ➜ \(\because \tan{x}=\tan{\alpha}\)
\(\Rightarrow x=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b)\)
দেওয়া আছে,\(f(x)=\cos{x}, \ g(x)=\sin{x}\)
এবং
\(f\{\pi g(\theta)\}=g\{\pi f(\theta)\}\)
\(\Rightarrow f(\pi\sin{\theta})=g(\pi\cos{\theta})\) ➜ \(\because f(x)=\cos{x}, \ g(x)=\sin{x}\)
\(\Rightarrow f(\theta)=\cos{\theta}, \ g(\theta)=\sin{\theta}\)
\(\Rightarrow \sin{(\pi\cos{\theta})}=\cos{(\pi\sin{\theta})}\) ➜ \(\because f(x)=\cos{x}, \ g(x)=\sin{x}\)
\(\Rightarrow f(\pi\sin{\theta})=\cos{(\pi\sin{\theta})}, \ g(\pi\cos{\theta})=\sin{(\pi\cos{\theta})}\)
\(\Rightarrow \sin{(\pi\cos{\theta})}=\sin{\left(\frac{\pi}{2}\pm\pi\sin{\theta}\right)}\)
\(\Rightarrow \pi\cos{\theta}=\frac{\pi}{2}\pm\pi\sin{\theta}\)
\(\Rightarrow \pi\cos{\theta}=\pi\left(\frac{1}{2}\pm\sin{\theta}\right)\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\pm\sin{\theta}\)
\(\Rightarrow \cos{\theta}\pm\sin{\theta}=\frac{1}{2}\)
\(\Rightarrow (\cos{\theta}\pm\sin{\theta})^2=\frac{1}{4}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \cos^2{\theta}\pm2\cos{\theta}\sin{\theta}+\sin^2{\theta}=\frac{1}{4}\) ➜ \(\because (a\pm{b})^2=a^2\pm2ab+b^2\)
\(\Rightarrow \sin^2{\theta}+\cos^2{\theta}\pm2\sin{\theta}\cos{\theta}=\frac{1}{4}\)
\(\Rightarrow 1\pm\sin{2\theta}=\frac{1}{4}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \pm\sin{2\theta}=\frac{1}{4}-1\)
\(\Rightarrow \pm\sin{2\theta}=\frac{1-4}{4}\)
\(\Rightarrow \pm\sin{2\theta}=-\frac{3}{4}\)
\(\Rightarrow \sin{2\theta}=\pm\frac{3}{4}\)
\(\Rightarrow 2\theta=\sin{\left(\pm\frac{3}{4}\right)}\)
\(\Rightarrow 2\theta=\pm\sin{\frac{3}{4}}\)
\(\therefore \theta=\pm\frac{1}{2}\sin{\frac{3}{4}}\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,\(f(x)=\cos{x}, \ g(x)=\sin{x}\)
এবং
\(f(x)-f(2x)=g(2x)-g(x)\)
\(\Rightarrow \cos{x}-\cos{2x}=\sin{2x}-\sin{x}\) ➜ \(\because f(x)=\cos{x}, \ g(x)=\sin{x}\)
\(\Rightarrow f(2x)=\cos{2x}, \ g(2x)=\sin{2x}\)
\(\Rightarrow 2\sin{\frac{x+2x}{2}}\sin{\frac{2x-x}{2}}=2\cos{\frac{2x+x}{2}}\sin{\frac{2x-x}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}=\cos{\frac{3x}{2}}\sin{\frac{x}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}-\cos{\frac{3x}{2}}\sin{\frac{x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}\left(\sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}=\cos{\frac{3x}{2}}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \frac{\sin{\frac{3x}{2}}}{\cos{\frac{3x}{2}}}=1\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{x}{2}=n\pi, \ \frac{3x}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
\(\therefore\) নির্ণেয় সাধারণ সমাধান, \(x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(27.\) সমাধান করঃ \(\cos{6\theta}+\cos{4\theta}+\cos{2\theta}+1=0\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{6\theta}+\cos{4\theta}+\cos{2\theta}+1=0\)
\(\Rightarrow \cos{6\theta}+\cos{2\theta}+1+\cos{4\theta}=0\)
\(\Rightarrow 2\cos{\frac{6\theta+2\theta}{2}}\cos{\frac{6\theta-2\theta}{2}}+2\cos^2{2\theta}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow 2\cos{\frac{8\theta}{2}}\cos{\frac{4\theta}{2}}+2\cos^2{2\theta}=0\)
\(\Rightarrow 2\cos{4\theta}\cos{2\theta}+2\cos^2{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}(\cos{4\theta}+\cos{2\theta})=0\)
\(\Rightarrow 2\cos{2\theta}=0, \ \cos{4\theta}+\cos{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}=0, \ 2\cos{\frac{4\theta+2\theta}{2}}\cos{\frac{4\theta-2\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{\frac{6\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 3\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{x}=0\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{6}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=(2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{6\theta}+\cos{4\theta}+\cos{2\theta}+1=0\)
\(\Rightarrow \cos{6\theta}+\cos{2\theta}+1+\cos{4\theta}=0\)
\(\Rightarrow 2\cos{\frac{6\theta+2\theta}{2}}\cos{\frac{6\theta-2\theta}{2}}+2\cos^2{2\theta}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow 2\cos{\frac{8\theta}{2}}\cos{\frac{4\theta}{2}}+2\cos^2{2\theta}=0\)
\(\Rightarrow 2\cos{4\theta}\cos{2\theta}+2\cos^2{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}(\cos{4\theta}+\cos{2\theta})=0\)
\(\Rightarrow 2\cos{2\theta}=0, \ \cos{4\theta}+\cos{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}=0, \ 2\cos{\frac{4\theta+2\theta}{2}}\cos{\frac{4\theta-2\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{\frac{6\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 3\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{x}=0\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{6}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=(2n+1)\frac{\pi}{2}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
উদাহরণ \(28.\) সমাধান করঃ \(\tan{\theta}+\cot{\theta}=2cosec \ {\theta}, \ 0\lt{\theta}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
উত্তরঃ \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{\theta}+\cot{\theta}=2cosec \ {\theta}, \ 0\lt{\theta}\lt{2\pi}\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\frac{2}{\sin{\theta}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}+\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{2}{\sin{\theta}}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=\frac{2}{\sin{\theta}}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow \frac{1}{\cos{\theta}}=2\) ➜ \(\because \sin{\theta}\ne{0}\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{3}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
\(\tan{\theta}+\cot{\theta}=2cosec \ {\theta}, \ 0\lt{\theta}\lt{2\pi}\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\frac{2}{\sin{\theta}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}+\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{2}{\sin{\theta}}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=\frac{2}{\sin{\theta}}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow \frac{1}{\cos{\theta}}=2\) ➜ \(\because \sin{\theta}\ne{0}\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{3}, \ -\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{3}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
উদাহরণ \(29.\) \(a\tan{\theta}+b\sec{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\) হলে, প্রমাণ কর যে, \(\tan{(\alpha+\beta)}=\frac{2ac}{a^2-c^2}\)
সমাধানঃ
দেওয়া আছে,
\(a\tan{\theta}+b\sec{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\)
\(\Rightarrow b\sec{\theta}=c-a\tan{\theta}\)
\(\Rightarrow c-a\tan{\theta}=b\sec{\theta}\)
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}=b^2\sec^2{\theta}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}=b^2(1+\tan^2{\theta})\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}=b^2+b^2\tan^2{\theta}\)
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}-b^2-b^2\tan^2{\theta}=0\)
\(\therefore (a^2-b^2)\tan^2{\theta}-2ca\tan{\theta}+(c^2-b^2)=0 .....(1)\)
\((1)\) নং \(\tan{\theta}\) এর দ্বিঘাত সমীকরণ।
শর্ত মতে সমীকরণটির মূলদুইটি হবে যথাক্রমে, \(\tan{\alpha}\) এবং \(\tan{\beta}\)
\(\therefore \tan{\alpha}+\tan{\beta}=-\frac{-2ca}{a^2-b^2}, \ \tan{\alpha}\tan{\beta}=\frac{c^2-b^2}{a^2-b^2}\) ➜ \(ax^2+bx+c=0\) এর মূলদ্বয় \(p\) ও \(q\)
\(\therefore p+q=-\frac{b}{a}, \ pq=\frac{c}{a}\)
\(\Rightarrow \tan{\alpha}+\tan{\beta}=\frac{2ca}{a^2-b^2}, \ \tan{\alpha}\tan{\beta}=\frac{c^2-b^2}{a^2-b^2}\)
এখন,
\(\tan{(\alpha+\beta)}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\)
\(=\frac{\frac{2ca}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}\) ➜ \(\because \tan{\alpha}+\tan{\beta}=\frac{2ca}{a^2-b^2}\)
এবং \(\tan{\alpha}\tan{\beta}=\frac{c^2-b^2}{a^2-b^2}\)
\(=\frac{2ca}{(a^2-b^2)-(c^2-b^2)}\) ➜ লব ও হরকে \((a^2-b^2)\) দ্বারা গুণ করে,
\(=\frac{2ca}{a^2-b^2-c^2+b^2}\)
\(=\frac{2ca}{a^2-c^2}\)
\(\therefore \tan{(\alpha+\beta)}=\frac{2ca}{a^2-c^2}\)
(প্রমাণিত)
\(a\tan{\theta}+b\sec{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\)
\(\Rightarrow b\sec{\theta}=c-a\tan{\theta}\)
\(\Rightarrow c-a\tan{\theta}=b\sec{\theta}\)
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}=b^2\sec^2{\theta}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}=b^2(1+\tan^2{\theta})\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}=b^2+b^2\tan^2{\theta}\)
\(\Rightarrow c^2-2ca\tan{\theta}+a^2\tan^2{\theta}-b^2-b^2\tan^2{\theta}=0\)
\(\therefore (a^2-b^2)\tan^2{\theta}-2ca\tan{\theta}+(c^2-b^2)=0 .....(1)\)
\((1)\) নং \(\tan{\theta}\) এর দ্বিঘাত সমীকরণ।
শর্ত মতে সমীকরণটির মূলদুইটি হবে যথাক্রমে, \(\tan{\alpha}\) এবং \(\tan{\beta}\)
\(\therefore \tan{\alpha}+\tan{\beta}=-\frac{-2ca}{a^2-b^2}, \ \tan{\alpha}\tan{\beta}=\frac{c^2-b^2}{a^2-b^2}\) ➜ \(ax^2+bx+c=0\) এর মূলদ্বয় \(p\) ও \(q\)
\(\therefore p+q=-\frac{b}{a}, \ pq=\frac{c}{a}\)
\(\Rightarrow \tan{\alpha}+\tan{\beta}=\frac{2ca}{a^2-b^2}, \ \tan{\alpha}\tan{\beta}=\frac{c^2-b^2}{a^2-b^2}\)
এখন,
\(\tan{(\alpha+\beta)}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\)
\(=\frac{\frac{2ca}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}\) ➜ \(\because \tan{\alpha}+\tan{\beta}=\frac{2ca}{a^2-b^2}\)
এবং \(\tan{\alpha}\tan{\beta}=\frac{c^2-b^2}{a^2-b^2}\)
\(=\frac{2ca}{(a^2-b^2)-(c^2-b^2)}\) ➜ লব ও হরকে \((a^2-b^2)\) দ্বারা গুণ করে,
\(=\frac{2ca}{a^2-b^2-c^2+b^2}\)
\(=\frac{2ca}{a^2-c^2}\)
\(\therefore \tan{(\alpha+\beta)}=\frac{2ca}{a^2-c^2}\)
(প্রমাণিত)
উদাহরণ \(30.\) সমাধান করঃ \(\cot{x}-\cot{2x}=2\)
উত্তরঃ \(\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
উত্তরঃ \(\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{x}-\cot{2x}=2\)
\(\Rightarrow \frac{\cos{x}}{\sin{x}}-\frac{\cos{2x}}{\sin{2x}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin{2x}\cos{x}-\cos{2x}\sin{x}}{\sin{2x}\sin{x}}=2\)
\(\Rightarrow \frac{\sin{(2x-x)}}{\sin{2x}\sin{x}}=2\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \frac{1}{\sin{2x}}=2, \ \because \sin{x}\ne{0}\)
\(\Rightarrow \sin{2x}=\frac{1}{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \sin{2x}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 2x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cot{x}-\cot{2x}=2\)
\(\Rightarrow \frac{\cos{x}}{\sin{x}}-\frac{\cos{2x}}{\sin{2x}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin{2x}\cos{x}-\cos{2x}\sin{x}}{\sin{2x}\sin{x}}=2\)
\(\Rightarrow \frac{\sin{(2x-x)}}{\sin{2x}\sin{x}}=2\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \frac{1}{\sin{2x}}=2, \ \because \sin{x}\ne{0}\)
\(\Rightarrow \sin{2x}=\frac{1}{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \sin{2x}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 2x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
অধ্যায় \(7H\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
সমাধান করঃ
\(Q.1.(i)\) \(\cot{\theta}-\tan{\theta}=2\)উত্তরঃ \((4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০০২।
\(Q.1.(ii)\) \(\tan{x}+\tan{3x}=0\)
উত্তরঃ \(\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০০৫।
\(Q.1.(iii)\) \(4\cos^2{x}+6\sin^2{x}=5\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{4}, \ 2n\pi\pm\frac{3\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(iv)\) \(\sin^2{2\theta}-3\cos^2{\theta}=0\)
উত্তরঃ \(n\pi\pm\frac{\pi}{3}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১১; কুঃ২০১০,২০০৯।
\(Q.1.(v)\) \(cosec \ {\theta}+\cot{\theta}=\sqrt{3}\)
উত্তরঃ \(2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
ডুয়েটঃ২০১৪-২০১৫;রুয়েটঃ২০০৮-২০০৯; মাঃ২০০৬ ।
\(Q.1.(vi)\) \(\frac{\cos{\theta}}{1+\sin{\theta}}+\tan{\theta}=2\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১৩ ।
\(Q.1.(vii)\) \(cosec \ {\theta}+\cot{\theta}=\frac{1}{\sqrt{2}}\)
উত্তরঃ\(\theta=2n\pi+2\alpha\) যেখানে, \(\alpha=\tan^{-1}{(\sqrt{2})}, \ n\in{\mathbb{Z}}\)
\(Q.1.(viii)\) \(\sec^2{\theta}+3 cosec^2{\theta}=8\)
উত্তরঃ \(n\pi\pm\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০০৭ ।
\(Q.1.(ix)\) \(3\tan{\theta}+\cot{\theta}=5 cosec \ {\theta}\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০১৩ ।
\(Q.1.(x)\) \(2\cos^2{\theta}-3\sin{\theta}=0\)
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xi)\) \(6\cos^2{x}+\sin{x}=5\)
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}, \ n\in{\mathbb{Z}}\)
\(Q.1.(xii)\) \(\sqrt{3}\cot^2{x}+4\cot{x}+\sqrt{3}=0\)
উত্তরঃ \(n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xiii)\) \(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)
উত্তরঃ \(2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
কুঃ২০০৫; রাঃ২০১৪; চুয়েটঃ২০১১-২০১২ ।
\(Q.1.(xiv)\) \(\cot{2\theta}-\cot{4\theta}=\sqrt{2}\)
উত্তরঃ \(\frac{n\pi}{4}+(-1)^n\frac{\pi}{16}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xv)\) \(4\sin^2{x}+\sqrt{3}=2(1+\sqrt{3})\sin{x}\)
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{3}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xvi)\) \(2\sin{x}\tan{x}+1=\tan{x}+2\sin{x}\)
উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
চুয়েটঃ২০০৫-২০০৬ ।
\(Q.1.(xvii)\) \(2\sin^2{\theta}+\sin^2{2\theta}=2\)
উত্তরঃ \((2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xviii)\) \(\cos^3{\theta}-\cos{\theta}\sin{\theta}-\sin^3{\theta}=1\)
উত্তরঃ \(2n\pi, \ 2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xix)\) \(2\sin{2x}+3\cos{x}=0\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
\(Q.1.(xx)\) \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})=-1\)
উত্তরঃ \(n\pi+(-1)^n\frac{7\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxi)\) \(2\cos^2{x}+\cos^2{2x}=2\)
উত্তরঃ \(n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxii)\) \(\tan^2{x}+\sec^2{x}=3\)উত্তরঃ \(n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxiii)\) \(\cot^2{\theta}+cosec^2{\theta}=3\)
উত্তরঃ \(n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxiv)\) \(\sqrt{2}\sec{x}+\tan{x}=1\)
উত্তরঃ \(2n\pi-\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxv)\) \(2\cos{x}+3\sin{x}=1\)
উত্তরঃ \(2n\pi\pm\alpha+\theta\) যেখানে, \(\cos{\theta}=\frac{2}{\sqrt{13}}, \ \cos{\alpha}=\frac{1}{\sqrt{13}}, \ n\in{\mathbb{Z}}\)
\(Q.1.(xxvi)\) \(5\tan^2{\theta}-\sec^2{\theta}=11\)
উত্তরঃ \(n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxvii)\) \(\cos{\theta}+\sec{\theta}=\frac{5}{2}\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxviii)\) \(8\sin^2{\theta}-2\cos{\theta}=5\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{3}, \ 2n\pi\pm\alpha\) যেখানে, \(\cos{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
\(Q.1.(xxix)\) \(\sin{\theta}+cosec \ {\theta}=\frac{3}{\sqrt{2}}\)
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxx)\) \(cosec \ {\theta}\cot{\theta}=2\sqrt{3}\)
উত্তরঃ \(2n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxxi)\) \(\sin{5\theta}+\sin{\theta}=\sin{3\theta}\)
উত্তরঃ \(\frac{n\pi}{3}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxxii)\) \(\sin{2\theta}+4\cos{\theta}=\sqrt{3}\sin{\theta}+2\sqrt{3}\)
উত্তরঃ \(n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxxiii)\) \(\tan{\left(\frac{\pi}{4}+\theta\right)}+\tan{\left(\frac{\pi}{4}-\theta\right)}=4\)
উত্তরঃ \(n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxxiv)\) \(\sin^2{2\theta}-\sin^2{\theta}=\frac{1}{3}\sin{3\theta}\)
উত্তরঃ \(n\pi, \ 2n\pi\pm\frac{\pi}{3}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=\frac{1}{3}, \ n\in{\mathbb{Z}}\)
\(Q.1.(xxxv)\) \(\tan^3{\theta}-\sec^2{\theta}=4\tan^2{\theta}-5\tan{\theta}\)
উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+\alpha, \ n\pi+\beta\) যেখানে, \(\tan{\alpha}=2+\sqrt{3}, \ \tan{\beta}=2-\sqrt{3}, \ n\in{\mathbb{Z}}\)
\(Q.1.(xxxvi)\) \(cosec \ {\theta}+\sec{\theta}=2\sqrt{2}\)
উত্তরঃ \((8n+3)\frac{\pi}{12}, \ (8n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxxvii)\) \(\cos{3\theta}=\cos{2\theta}\)
উত্তরঃ \(2n\pi, \ \frac{2}{5}n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxxviii)\) \(\sin{5\theta}\cos{\theta}=\sin{6\theta}\cos{2\theta}\)
উত্তরঃ \(n\pi, \ (2n+1)\frac{\pi}{14}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xxxix)\) \(4\sin{x}=\sec{x}\)
উত্তরঃ \(\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xL)\) \(\sqrt{2}(\cos^2{x}-\sin^2{x})=1\)
উত্তরঃ \(n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xLi)\) \(\cos{2x}=\cos{x}\sin{x}\)
উত্তরঃ \(\frac{1}{2}(n\pi+\alpha)\) যেখানে, \(\tan{\alpha}=2, \ n\in{\mathbb{Z}}\)
\(Q.1.(xLii)\) \(\sin{4\theta}\cos{\theta}=\frac{1}{4}+\sin{\frac{5\theta}{2}}\cos{\frac{5\theta}{2}}\)
উত্তরঃ \(\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xLiii)\) \(\tan{2\theta}-\tan{\theta}=\frac{1}{2}\sec{\theta}\sec{2\theta}\)
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.1.(xLiv)\) \(\sin{2\theta}=\cos{3\theta}\)
উত্তরঃ \((4n+1)\frac{\pi}{10}, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(i)\) \(\cot{\theta}-\tan{\theta}=2\)উত্তরঃ \((4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০০২।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{\theta}-\tan{\theta}=2\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=2\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=2\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{2\theta}\)
\(\Rightarrow \sin{2\theta}=\cos{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\cos{2A}\)
\(\Rightarrow \frac{\sin{2\theta}}{\cos{2\theta}}=1\)
\(\Rightarrow \tan{2\theta}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{2\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow 2\theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 2\theta=(4n+1)\frac{\pi}{4}\)
\(\therefore \theta=(4n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cot{\theta}-\tan{\theta}=2\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=2\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=2\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{2\theta}\)
\(\Rightarrow \sin{2\theta}=\cos{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\cos{2A}\)
\(\Rightarrow \frac{\sin{2\theta}}{\cos{2\theta}}=1\)
\(\Rightarrow \tan{2\theta}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{2\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow 2\theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 2\theta=(4n+1)\frac{\pi}{4}\)
\(\therefore \theta=(4n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(ii)\) \(\tan{x}+\tan{3x}=0\)উত্তরঃ \(\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০০৫।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{x}+\tan{3x}=0\)
\(\Rightarrow \frac{\sin{x}}{\cos{x}}+\frac{\sin{3x}}{\cos{3x}}=0\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\sin{x}\cos{3x}+\cos{x}\sin{3x}}{\cos{x}\cos{3x}}=0\)
\(\Rightarrow \sin{x}\cos{3x}+\cos{x}\sin{3x}=0\)
\(\Rightarrow \sin{(x+3x)}=0\)
\(\Rightarrow \sin{4x}=0\)
\(\Rightarrow 4x=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{x}+\tan{3x}=0\)
\(\Rightarrow \frac{\sin{x}}{\cos{x}}+\frac{\sin{3x}}{\cos{3x}}=0\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\sin{x}\cos{3x}+\cos{x}\sin{3x}}{\cos{x}\cos{3x}}=0\)
\(\Rightarrow \sin{x}\cos{3x}+\cos{x}\sin{3x}=0\)
\(\Rightarrow \sin{(x+3x)}=0\)
\(\Rightarrow \sin{4x}=0\)
\(\Rightarrow 4x=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(iii)\) \(4\cos^2{x}+6\sin^2{x}=5\)উত্তরঃ \(2n\pi\pm\frac{\pi}{4}, \ 2n\pi\pm\frac{3\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4\cos^2{x}+6\sin^2{x}=5\)
\(\Rightarrow 4\cos^2{x}+6(1-\cos^2{x})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4\cos^2{x}+6-6\cos^2{x}=5\)
\(\Rightarrow -2\cos^2{x}=5-6\)
\(\Rightarrow -2\cos^2{x}=-1\)
\(\Rightarrow 2\cos^2{x}=1\)
\(\Rightarrow \cos^2{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=\pm\sqrt{\frac{1}{2}}\)
\(\Rightarrow \cos{x}=\pm\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{x}=\frac{1}{\sqrt{2}}, \ \cos{x}=-\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{4}}, \ \cos{x}=-\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{4}}, \ \cos{x}=\cos{\left(\pi-\frac{\pi}{4}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{4}, \ x=2n\pi\pm\left(\pi-\frac{\pi}{4}\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{4}, \ x=2n\pi\pm\frac{4\pi-\pi}{4}\)
\(\therefore x=2n\pi\pm\frac{\pi}{4}, \ 2n\pi\pm\frac{3\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm\frac{\pi}{4}, \ 2n\pi\pm\frac{3\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(4\cos^2{x}+6\sin^2{x}=5\)
\(\Rightarrow 4\cos^2{x}+6(1-\cos^2{x})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4\cos^2{x}+6-6\cos^2{x}=5\)
\(\Rightarrow -2\cos^2{x}=5-6\)
\(\Rightarrow -2\cos^2{x}=-1\)
\(\Rightarrow 2\cos^2{x}=1\)
\(\Rightarrow \cos^2{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=\pm\sqrt{\frac{1}{2}}\)
\(\Rightarrow \cos{x}=\pm\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{x}=\frac{1}{\sqrt{2}}, \ \cos{x}=-\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{4}}, \ \cos{x}=-\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{4}}, \ \cos{x}=\cos{\left(\pi-\frac{\pi}{4}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{4}, \ x=2n\pi\pm\left(\pi-\frac{\pi}{4}\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{4}, \ x=2n\pi\pm\frac{4\pi-\pi}{4}\)
\(\therefore x=2n\pi\pm\frac{\pi}{4}, \ 2n\pi\pm\frac{3\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm\frac{\pi}{4}, \ 2n\pi\pm\frac{3\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(iv)\) \(\sin^2{2\theta}-3\cos^2{\theta}=0\)উত্তরঃ \(n\pi\pm\frac{\pi}{3}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১১; কুঃ২০১০,২০০৯।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin^2{2\theta}-3\cos^2{\theta}=0\)
\(\Rightarrow (\sin{2\theta})^2-3\cos^2{\theta}=0\)
\(\Rightarrow (2\sin{\theta}\cos{\theta})^2-3\cos^2{\theta}=0\)➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 4\sin^2{\theta}\cos^2{\theta}-3\cos^2{\theta}=0\)
\(\Rightarrow \cos^2{\theta}(4\sin^2{\theta}-3)=0\)
\(\Rightarrow \cos^2{\theta}=0, \ 4\sin^2{\theta}-3=0\)
\(\Rightarrow \cos{\theta}=0, \ 4\sin^2{\theta}=3\)
\(\Rightarrow \cos{\theta}=0, \ \sin^2{\theta}=\frac{3}{4}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ -4\cos^2{\theta}+1=0\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \sin{\theta}=\pm\sqrt{\frac{3}{4}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \sin{\theta}=\pm\frac{\sqrt{3}}{2}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \sin{\theta}=\sin{\left(\pm\frac{\pi}{3}\right)}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi+(-1)^n\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\theta=n\pi\pm\frac{\pi}{3}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin^2{2\theta}-3\cos^2{\theta}=0\)
\(\Rightarrow (\sin{2\theta})^2-3\cos^2{\theta}=0\)
\(\Rightarrow (2\sin{\theta}\cos{\theta})^2-3\cos^2{\theta}=0\)➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 4\sin^2{\theta}\cos^2{\theta}-3\cos^2{\theta}=0\)
\(\Rightarrow \cos^2{\theta}(4\sin^2{\theta}-3)=0\)
\(\Rightarrow \cos^2{\theta}=0, \ 4\sin^2{\theta}-3=0\)
\(\Rightarrow \cos{\theta}=0, \ 4\sin^2{\theta}=3\)
\(\Rightarrow \cos{\theta}=0, \ \sin^2{\theta}=\frac{3}{4}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ -4\cos^2{\theta}+1=0\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \sin{\theta}=\pm\sqrt{\frac{3}{4}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \sin{\theta}=\pm\frac{\sqrt{3}}{2}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \sin{\theta}=\sin{\left(\pm\frac{\pi}{3}\right)}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi+(-1)^n\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\theta=n\pi\pm\frac{\pi}{3}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(v)\) \(cosec \ {\theta}+\cot{\theta}=\sqrt{3}\)উত্তরঃ \(2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
ডুয়েটঃ২০১৪-২০১৫;রুয়েটঃ২০০৮-২০০৯; মাঃ২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(cosec \ {\theta}+\cot{\theta}=\sqrt{3}\)
\(\Rightarrow \frac{1}{\sin{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\sqrt{3}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{1+\cos{\theta}}{\sin{\theta}}=\sqrt{3}\)
\(\Rightarrow \frac{2\cos^2{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}=\sqrt{3}\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
এবং \(\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\sqrt{3}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=\frac{1}{\sqrt{3}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=\frac{1}{\sqrt{3}}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\frac{\pi}{6}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi+\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(cosec \ {\theta}+\cot{\theta}=\sqrt{3}\)
\(\Rightarrow \frac{1}{\sin{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\sqrt{3}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{1+\cos{\theta}}{\sin{\theta}}=\sqrt{3}\)
\(\Rightarrow \frac{2\cos^2{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}=\sqrt{3}\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
এবং \(\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\sqrt{3}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=\frac{1}{\sqrt{3}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=\frac{1}{\sqrt{3}}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\frac{\pi}{6}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi+\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(vi)\) \(\frac{\cos{\theta}}{1+\sin{\theta}}+\tan{\theta}=2\)উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১৩ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\frac{\cos{\theta}}{1+\sin{\theta}}+\tan{\theta}=2\)
\(\Rightarrow \frac{\cos{\theta}}{1+\sin{\theta}}+\frac{\sin{\theta}}{\cos{\theta}}=2\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+\sin{\theta}+\sin^2{\theta}}{\cos{\theta}(1+\sin{\theta})}=2\)
\(\Rightarrow \frac{\sin^2{\theta}+\cos^2{\theta}+\sin{\theta}}{\cos{\theta}(1+\sin{\theta})}=2\)
\(\Rightarrow \frac{1+\sin{\theta}}{\cos{\theta}(1+\sin{\theta})}=2\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow \frac{1}{\cos{\theta}}=2, \ \because 1+\sin{\theta}\ne{0}\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\frac{\cos{\theta}}{1+\sin{\theta}}+\tan{\theta}=2\)
\(\Rightarrow \frac{\cos{\theta}}{1+\sin{\theta}}+\frac{\sin{\theta}}{\cos{\theta}}=2\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+\sin{\theta}+\sin^2{\theta}}{\cos{\theta}(1+\sin{\theta})}=2\)
\(\Rightarrow \frac{\sin^2{\theta}+\cos^2{\theta}+\sin{\theta}}{\cos{\theta}(1+\sin{\theta})}=2\)
\(\Rightarrow \frac{1+\sin{\theta}}{\cos{\theta}(1+\sin{\theta})}=2\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow \frac{1}{\cos{\theta}}=2, \ \because 1+\sin{\theta}\ne{0}\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(vii)\) \(cosec \ {\theta}+\cot{\theta}=\frac{1}{\sqrt{2}}\)উত্তরঃ\(\theta=2n\pi+2\alpha\) যেখানে, \(\alpha=\tan^{-1}{(\sqrt{2})}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(cosec \ {\theta}+\cot{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{1}{\sin{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\frac{1}{\sqrt{2}}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{1+\cos{\theta}}{\sin{\theta}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{2\cos^2{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}=\frac{1}{\sqrt{2}}\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
এবং \(\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=\sqrt{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=\sqrt{2}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\alpha}, \ \text{যেখানে,} \ \sqrt{2}=\tan{\alpha}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\alpha\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi+2\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+2\alpha\) যেখানে, \(\alpha=\tan^{-1}{(\sqrt{2})}, \ n\in{\mathbb{Z}}\)
\(cosec \ {\theta}+\cot{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{1}{\sin{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\frac{1}{\sqrt{2}}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{1+\cos{\theta}}{\sin{\theta}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{2\cos^2{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}=\frac{1}{\sqrt{2}}\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
এবং \(\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=\sqrt{2}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=\sqrt{2}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\alpha}, \ \text{যেখানে,} \ \sqrt{2}=\tan{\alpha}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\alpha\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi+2\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+2\alpha\) যেখানে, \(\alpha=\tan^{-1}{(\sqrt{2})}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(viii)\)\(\sec^2{\theta}+3 cosec^2{\theta}=8\)উত্তরঃ \(n\pi\pm\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০০৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sec^2{\theta}+3 cosec^2{\theta}=8\)
\(\Rightarrow 1+\tan^2{\theta}+3(1+\cot^2{\theta})=8\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
এবং \(cosec^2{A}=1+\cot^2{A}\)
\(\Rightarrow 1+\tan^2{\theta}+3+3\cot^2{\theta}=8\)
\(\Rightarrow \tan^2{\theta}+\frac{3}{\tan^2{\theta}}+4=8\) ➜ \(\because \cot^2{A}=\frac{1}{\tan^2{A}}\)
\(\Rightarrow \frac{\tan^4{\theta}+3}{\tan^2{\theta}}=8-4\)
\(\Rightarrow \frac{\tan^4{\theta}+3}{\tan^2{\theta}}=4\)
\(\Rightarrow \tan^4{\theta}+3=4\tan^2{\theta}\)
\(\Rightarrow \tan^4{\theta}-4\tan^2{\theta}+3=0\)
\(\Rightarrow \tan^4{\theta}-3\tan^2{\theta}-\tan^2{\theta}+3=0\)
\(\Rightarrow \tan^2{\theta}(\tan^2{\theta}-3)-1(\tan^2{\theta}-3)=0\)
\(\Rightarrow (\tan^2{\theta}-3)(\tan^2{\theta}-1)=0\)
\(\Rightarrow \tan^2{\theta}-3=0, \ \tan^2{\theta}-1=0\)
\(\Rightarrow \tan^2{\theta}=3, \ \tan^2{\theta}=1\)
\(\Rightarrow \tan{\theta}=\pm\sqrt{3}, \ \tan{\theta}=\pm1\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{3}}, \ \tan{\theta}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{3}\right)}, \ \tan{\theta}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right), \ \theta=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{3}, \ n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sec^2{\theta}+3 cosec^2{\theta}=8\)
\(\Rightarrow 1+\tan^2{\theta}+3(1+\cot^2{\theta})=8\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
এবং \(cosec^2{A}=1+\cot^2{A}\)
\(\Rightarrow 1+\tan^2{\theta}+3+3\cot^2{\theta}=8\)
\(\Rightarrow \tan^2{\theta}+\frac{3}{\tan^2{\theta}}+4=8\) ➜ \(\because \cot^2{A}=\frac{1}{\tan^2{A}}\)
\(\Rightarrow \frac{\tan^4{\theta}+3}{\tan^2{\theta}}=8-4\)
\(\Rightarrow \frac{\tan^4{\theta}+3}{\tan^2{\theta}}=4\)
\(\Rightarrow \tan^4{\theta}+3=4\tan^2{\theta}\)
\(\Rightarrow \tan^4{\theta}-4\tan^2{\theta}+3=0\)
\(\Rightarrow \tan^4{\theta}-3\tan^2{\theta}-\tan^2{\theta}+3=0\)
\(\Rightarrow \tan^2{\theta}(\tan^2{\theta}-3)-1(\tan^2{\theta}-3)=0\)
\(\Rightarrow (\tan^2{\theta}-3)(\tan^2{\theta}-1)=0\)
\(\Rightarrow \tan^2{\theta}-3=0, \ \tan^2{\theta}-1=0\)
\(\Rightarrow \tan^2{\theta}=3, \ \tan^2{\theta}=1\)
\(\Rightarrow \tan{\theta}=\pm\sqrt{3}, \ \tan{\theta}=\pm1\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{3}}, \ \tan{\theta}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{3}\right)}, \ \tan{\theta}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right), \ \theta=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{3}, \ n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(ix)\) \(3\tan{\theta}+\cot{\theta}=5 cosec \ {\theta}\)উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০১৩ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(3\tan{\theta}+\cot{\theta}=5 cosec \ {\theta}\)
\(\Rightarrow \frac{3\sin{\theta}}{\cos{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\frac{5}{\sin{\theta}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{3\sin^2{\theta}+\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{5}{\sin{\theta}}\)
\(\Rightarrow \frac{3\sin^2{\theta}+\cos^2{\theta}}{\cos{\theta}}=5, \ \because \sin{\theta}\ne{0}\)
\(\Rightarrow \frac{3(1-\cos^2{\theta})+\cos^2{\theta}}{\cos{\theta}}=5\)
\(\Rightarrow \frac{3(1-\cos^2{\theta})+\cos^2{\theta}}{\cos{\theta}}=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow \frac{3-3\cos^2{\theta}+\cos^2{\theta}}{\cos{\theta}}=5\)
\(\Rightarrow \frac{3-2\cos^2{\theta}}{\cos{\theta}}=5\)
\(\Rightarrow 5\cos{\theta}=3-2\cos^2{\theta}\)
\(\Rightarrow 2\cos^2{\theta}+5\cos{\theta}-3=0\)
\(\Rightarrow 2\cos^2{\theta}+6\cos{\theta}-\cos{\theta}-3=0\)
\(\Rightarrow 2\cos{\theta}(\cos{\theta}+3)-1(\cos{\theta}+3)=0\)
\(\Rightarrow (\cos{\theta}+3)(2\cos{\theta}-1)=0\)
\(\Rightarrow \cos{\theta}+3\ne{0}, \ 2\cos{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(3\tan{\theta}+\cot{\theta}=5 cosec \ {\theta}\)
\(\Rightarrow \frac{3\sin{\theta}}{\cos{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=\frac{5}{\sin{\theta}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{3\sin^2{\theta}+\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{5}{\sin{\theta}}\)
\(\Rightarrow \frac{3\sin^2{\theta}+\cos^2{\theta}}{\cos{\theta}}=5, \ \because \sin{\theta}\ne{0}\)
\(\Rightarrow \frac{3(1-\cos^2{\theta})+\cos^2{\theta}}{\cos{\theta}}=5\)
\(\Rightarrow \frac{3(1-\cos^2{\theta})+\cos^2{\theta}}{\cos{\theta}}=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow \frac{3-3\cos^2{\theta}+\cos^2{\theta}}{\cos{\theta}}=5\)
\(\Rightarrow \frac{3-2\cos^2{\theta}}{\cos{\theta}}=5\)
\(\Rightarrow 5\cos{\theta}=3-2\cos^2{\theta}\)
\(\Rightarrow 2\cos^2{\theta}+5\cos{\theta}-3=0\)
\(\Rightarrow 2\cos^2{\theta}+6\cos{\theta}-\cos{\theta}-3=0\)
\(\Rightarrow 2\cos{\theta}(\cos{\theta}+3)-1(\cos{\theta}+3)=0\)
\(\Rightarrow (\cos{\theta}+3)(2\cos{\theta}-1)=0\)
\(\Rightarrow \cos{\theta}+3\ne{0}, \ 2\cos{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(x)\) \(2\cos^2{\theta}-3\sin{\theta}=0\)উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\cos^2{\theta}-3\sin{\theta}=0\)
\(\Rightarrow 2(1-\sin^2{\theta})-3\sin{\theta}=0\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2-2\sin^2{\theta}-3\sin{\theta}=0\)
\(\Rightarrow 2\sin^2{\theta}+3\sin{\theta}-2=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\sin^2{\theta}+4\sin{\theta}-\sin{\theta}-2=0\)
\(\Rightarrow 2\sin{\theta}(\sin{\theta}+2)-1(\sin{\theta}+2)=0\)
\(\Rightarrow (\sin{\theta}+2)(2\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}+2\ne{0}, \ 2\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow 2\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(2\cos^2{\theta}-3\sin{\theta}=0\)
\(\Rightarrow 2(1-\sin^2{\theta})-3\sin{\theta}=0\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 2-2\sin^2{\theta}-3\sin{\theta}=0\)
\(\Rightarrow 2\sin^2{\theta}+3\sin{\theta}-2=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\sin^2{\theta}+4\sin{\theta}-\sin{\theta}-2=0\)
\(\Rightarrow 2\sin{\theta}(\sin{\theta}+2)-1(\sin{\theta}+2)=0\)
\(\Rightarrow (\sin{\theta}+2)(2\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}+2\ne{0}, \ 2\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow 2\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xi)\) \(6\cos^2{x}+\sin{x}=5\)উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(6\cos^2{x}+\sin{x}=5\)
\(\Rightarrow 6(1-\sin^2{x})+\sin{x}-5=0\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 6-6\sin^2{x}+\sin{x}-5=0\)
\(\Rightarrow -6\sin^2{x}+\sin{x}+1=0\)
\(\Rightarrow 6\sin^2{x}-\sin{x}-1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 6\sin^2{x}-3\sin{x}+2\sin{x}-1=0\)
\(\Rightarrow 3\sin{x}(2\sin{x}-1)+1(2\sin{x}-1)=0\)
\(\Rightarrow (2\sin{x}-1)(3\sin{x}+1)=0\)
\(\Rightarrow 2\sin{x}=1, \ 3\sin{x}=-1\)
\(\Rightarrow \sin{x}=\frac{1}{2}, \ \sin{x}=-\frac{1}{3}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}, \ \sin{x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}\)
\(\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6}, \ x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}, \ n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}, \ n\in{\mathbb{Z}}\)
\(6\cos^2{x}+\sin{x}=5\)
\(\Rightarrow 6(1-\sin^2{x})+\sin{x}-5=0\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 6-6\sin^2{x}+\sin{x}-5=0\)
\(\Rightarrow -6\sin^2{x}+\sin{x}+1=0\)
\(\Rightarrow 6\sin^2{x}-\sin{x}-1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 6\sin^2{x}-3\sin{x}+2\sin{x}-1=0\)
\(\Rightarrow 3\sin{x}(2\sin{x}-1)+1(2\sin{x}-1)=0\)
\(\Rightarrow (2\sin{x}-1)(3\sin{x}+1)=0\)
\(\Rightarrow 2\sin{x}=1, \ 3\sin{x}=-1\)
\(\Rightarrow \sin{x}=\frac{1}{2}, \ \sin{x}=-\frac{1}{3}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}, \ \sin{x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}\)
\(\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6}, \ x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}, \ n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{1}{3}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xii)\) \(\sqrt{3}\cot^2{x}+4\cot{x}+\sqrt{3}=0\)উত্তরঃ \(n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{3}\cot^2{x}+4\cot{x}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot^2{x}+3\cot{x}+\cot{x}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot{x}(\cot{x}+\sqrt{3})+1(\cot{x}+\sqrt{3})=0\)
\(\Rightarrow (\cot{x}+\sqrt{3})(\sqrt{3}\cot{x}+1)=0\)
\(\Rightarrow \cot{x}+\sqrt{3}=0, \ \sqrt{3}\cot{x}+1=0\)
\(\Rightarrow \cot{x}=-\sqrt{3}, \ \sqrt{3}\cot{x}=-1\)
\(\Rightarrow \cot{x}=-\sqrt{3}, \ \cot{x}=-\frac{1}{\sqrt{3}}\)
\(\Rightarrow \cot{x}=-\cot{\frac{\pi}{6}}, \ \cot{x}=-\cot{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\cot{\frac{\pi}{6}}\)
এবং \(\frac{1}{\sqrt{3}}=\cot{\frac{\pi}{3}}\)
\(\Rightarrow \cot{x}=\cot{\left(\pi-\frac{\pi}{6}\right)}, \ \cot{x}=\cot{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cot{A}=\cot{\left(\pi-A\right)}\)
\(\Rightarrow x=n\pi+\left(\pi-\frac{\pi}{6}\right), \ x=n\pi+\left(\pi-\frac{\pi}{3}\right)\) ➜ \(\because \cot{A}=\cot{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=n\pi+\frac{5\pi}{6}, \ x=n\pi+\frac{2\pi}{3}\)
\(\therefore x=n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sqrt{3}\cot^2{x}+4\cot{x}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot^2{x}+3\cot{x}+\cot{x}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot{x}(\cot{x}+\sqrt{3})+1(\cot{x}+\sqrt{3})=0\)
\(\Rightarrow (\cot{x}+\sqrt{3})(\sqrt{3}\cot{x}+1)=0\)
\(\Rightarrow \cot{x}+\sqrt{3}=0, \ \sqrt{3}\cot{x}+1=0\)
\(\Rightarrow \cot{x}=-\sqrt{3}, \ \sqrt{3}\cot{x}=-1\)
\(\Rightarrow \cot{x}=-\sqrt{3}, \ \cot{x}=-\frac{1}{\sqrt{3}}\)
\(\Rightarrow \cot{x}=-\cot{\frac{\pi}{6}}, \ \cot{x}=-\cot{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\cot{\frac{\pi}{6}}\)
এবং \(\frac{1}{\sqrt{3}}=\cot{\frac{\pi}{3}}\)
\(\Rightarrow \cot{x}=\cot{\left(\pi-\frac{\pi}{6}\right)}, \ \cot{x}=\cot{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cot{A}=\cot{\left(\pi-A\right)}\)
\(\Rightarrow x=n\pi+\left(\pi-\frac{\pi}{6}\right), \ x=n\pi+\left(\pi-\frac{\pi}{3}\right)\) ➜ \(\because \cot{A}=\cot{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=n\pi+\frac{5\pi}{6}, \ x=n\pi+\frac{2\pi}{3}\)
\(\therefore x=n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xiii)\) \(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)উত্তরঃ \(2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
কুঃ২০০৫; রাঃ২০১৪; চুয়েটঃ২০১১-২০১২ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)
\(\Rightarrow \sec^2{\theta}-1-2\sqrt{3}\sec{\theta}+4=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow \sec^2{\theta}-2\sqrt{3}\sec{\theta}+3=0\)
\(\Rightarrow \sec^2{\theta}-2\sqrt{3}\sec{\theta}+(\sqrt{3})^2=0\)
\(\Rightarrow (\sec{\theta}-\sqrt{3})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sec{\theta}-\sqrt{3}=0\)
\(\Rightarrow \sec{\theta}=\sqrt{3}\)
\(\Rightarrow \sec{\theta}=\sec{\alpha}\) যেখানে, \(\sec{\alpha}=\sqrt{3}\)
\(\therefore \theta=2n\pi\pm\alpha\) ➜ \(\because \sec{A}=\sec{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
\(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)
\(\Rightarrow \sec^2{\theta}-1-2\sqrt{3}\sec{\theta}+4=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow \sec^2{\theta}-2\sqrt{3}\sec{\theta}+3=0\)
\(\Rightarrow \sec^2{\theta}-2\sqrt{3}\sec{\theta}+(\sqrt{3})^2=0\)
\(\Rightarrow (\sec{\theta}-\sqrt{3})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sec{\theta}-\sqrt{3}=0\)
\(\Rightarrow \sec{\theta}=\sqrt{3}\)
\(\Rightarrow \sec{\theta}=\sec{\alpha}\) যেখানে, \(\sec{\alpha}=\sqrt{3}\)
\(\therefore \theta=2n\pi\pm\alpha\) ➜ \(\because \sec{A}=\sec{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xiv)\) \(\cot{2\theta}-\cot{4\theta}=\sqrt{2}\)উত্তরঃ \(\frac{n\pi}{4}+(-1)^n\frac{\pi}{16}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{2\theta}-\cot{4\theta}=\sqrt{2}\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{2\theta}}-\frac{\cos{4\theta}}{\sin{4\theta}}=\sqrt{2}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin{4\theta}\cos{2\theta}-\cos{4\theta}\sin{2\theta}}{\sin{2\theta}\sin{4\theta}}=\sqrt{2}\)
\(\Rightarrow \frac{\sin{(4\theta-2\theta)}}{\sin{2\theta}\sin{4\theta}}=\sqrt{2}\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \frac{\sin{2\theta}}{\sin{2\theta}\sin{4\theta}}=\sqrt{2}\)
\(\Rightarrow \frac{1}{\sin{4\theta}}=\sqrt{2}\)
\(\Rightarrow \sin{4\theta}=\frac{1}{\sqrt{2}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \sin{4\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow 4\theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}+(-1)^n\frac{\pi}{16}\)
\(\therefore\) সাধারণ সমাধান, \(\frac{n\pi}{4}+(-1)^n\frac{\pi}{16}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cot{2\theta}-\cot{4\theta}=\sqrt{2}\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{2\theta}}-\frac{\cos{4\theta}}{\sin{4\theta}}=\sqrt{2}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin{4\theta}\cos{2\theta}-\cos{4\theta}\sin{2\theta}}{\sin{2\theta}\sin{4\theta}}=\sqrt{2}\)
\(\Rightarrow \frac{\sin{(4\theta-2\theta)}}{\sin{2\theta}\sin{4\theta}}=\sqrt{2}\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \frac{\sin{2\theta}}{\sin{2\theta}\sin{4\theta}}=\sqrt{2}\)
\(\Rightarrow \frac{1}{\sin{4\theta}}=\sqrt{2}\)
\(\Rightarrow \sin{4\theta}=\frac{1}{\sqrt{2}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \sin{4\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow 4\theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}+(-1)^n\frac{\pi}{16}\)
\(\therefore\) সাধারণ সমাধান, \(\frac{n\pi}{4}+(-1)^n\frac{\pi}{16}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xv)\) \(4\sin^2{x}+\sqrt{3}=2(1+\sqrt{3})\sin{x}\)উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{3}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4\sin^2{x}+\sqrt{3}=2(1+\sqrt{3})\sin{x}\)
\(\Rightarrow 4\sin^2{x}+\sqrt{3}=2\sin{x}+2\sqrt{3}\sin{x}\)
\(\Rightarrow 4\sin^2{x}-2\sin{x}-2\sqrt{3}\sin{x}+\sqrt{3}=0\)
\(\Rightarrow 2\sin{x}(2\sin{x}-1)-\sqrt{3}(2\sin{x}-1)=0\)
\(\Rightarrow (2\sin{x}-1)(2\sin{x}-\sqrt{3})=0\)
\(\Rightarrow 2\sin{x}-1=0, \ 2\sin{x}-\sqrt{3}=0\)
\(\Rightarrow 2\sin{x}=1, \ 2\sin{x}=\sqrt{3}\)
\(\Rightarrow \sin{x}=\frac{1}{2}, \ \sin{x}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}, \ \sin{x}=\sin{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6}, \ x=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+(-1)^n\frac{\pi}{3}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(4\sin^2{x}+\sqrt{3}=2(1+\sqrt{3})\sin{x}\)
\(\Rightarrow 4\sin^2{x}+\sqrt{3}=2\sin{x}+2\sqrt{3}\sin{x}\)
\(\Rightarrow 4\sin^2{x}-2\sin{x}-2\sqrt{3}\sin{x}+\sqrt{3}=0\)
\(\Rightarrow 2\sin{x}(2\sin{x}-1)-\sqrt{3}(2\sin{x}-1)=0\)
\(\Rightarrow (2\sin{x}-1)(2\sin{x}-\sqrt{3})=0\)
\(\Rightarrow 2\sin{x}-1=0, \ 2\sin{x}-\sqrt{3}=0\)
\(\Rightarrow 2\sin{x}=1, \ 2\sin{x}=\sqrt{3}\)
\(\Rightarrow \sin{x}=\frac{1}{2}, \ \sin{x}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}, \ \sin{x}=\sin{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6}, \ x=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(n\pi+(-1)^n\frac{\pi}{3}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xvi)\) \(2\sin{x}\tan{x}+1=\tan{x}+2\sin{x}\)উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
চুয়েটঃ২০০৫-২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin{x}\tan{x}+1=\tan{x}+2\sin{x}\)
\(\Rightarrow 2\sin{x}\tan{x}-\tan{x}-2\sin{x}+1=0\)
\(\Rightarrow \tan{x}(2\sin{x}-1)-1(2\sin{x}-1)=0\)
\(\Rightarrow (2\sin{x}-1)(\tan{x}-1)=0\)
\(\Rightarrow 2\sin{x}-1=0, \ \tan{x}-1=0\)
\(\Rightarrow 2\sin{x}=1, \ \tan{x}=1\)
\(\Rightarrow \sin{x}=\frac{1}{2}, \ \tan{x}=1\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}, \ \tan{x}=\tan{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6}, \ x=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(2\sin{x}\tan{x}+1=\tan{x}+2\sin{x}\)
\(\Rightarrow 2\sin{x}\tan{x}-\tan{x}-2\sin{x}+1=0\)
\(\Rightarrow \tan{x}(2\sin{x}-1)-1(2\sin{x}-1)=0\)
\(\Rightarrow (2\sin{x}-1)(\tan{x}-1)=0\)
\(\Rightarrow 2\sin{x}-1=0, \ \tan{x}-1=0\)
\(\Rightarrow 2\sin{x}=1, \ \tan{x}=1\)
\(\Rightarrow \sin{x}=\frac{1}{2}, \ \tan{x}=1\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}, \ \tan{x}=\tan{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6}, \ x=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xvii)\) \(2\sin^2{\theta}+\sin^2{2\theta}=2\)উত্তরঃ \((2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin^2{\theta}+\sin^2{2\theta}=2\)
\(\Rightarrow 1-\cos{2\theta}+1-\cos^2{2\theta}=2\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
এবং \(\sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-\cos{2\theta}-\cos^2{2\theta}=2\)
\(\Rightarrow -\cos{2\theta}-\cos^2{2\theta}=0\)
\(\Rightarrow -\cos{2\theta}(1+\cos{2\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ 1+\cos{2\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=-1\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=(2n+1)\pi\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(2\sin^2{\theta}+\sin^2{2\theta}=2\)
\(\Rightarrow 1-\cos{2\theta}+1-\cos^2{2\theta}=2\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
এবং \(\sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-\cos{2\theta}-\cos^2{2\theta}=2\)
\(\Rightarrow -\cos{2\theta}-\cos^2{2\theta}=0\)
\(\Rightarrow -\cos{2\theta}(1+\cos{2\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ 1+\cos{2\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=-1\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=(2n+1)\pi\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xviii)\) \(\cos^3{\theta}-\cos{\theta}\sin{\theta}-\sin^3{\theta}=1\)উত্তরঃ \(2n\pi, \ 2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos^3{\theta}-\cos{\theta}\sin{\theta}-\sin^3{\theta}=1\)
\(\Rightarrow \cos^3{\theta}-\sin^3{\theta}-1-\cos{\theta}\sin{\theta}=0\)
\(\Rightarrow (\cos{\theta}-\sin{\theta})(\cos^2{\theta}+\cos{\theta}\sin{\theta}+\sin^2{\theta})-(1+\cos{\theta}\sin{\theta})=0\) ➜ \(\because a^3-b^3=(a-b)(a^2+ab+b^2)\)
\(\Rightarrow (\cos{\theta}-\sin{\theta})(\sin^2{\theta}+\cos^2{\theta}+\cos{\theta}\sin{\theta})-(1+\cos{\theta}\sin{\theta})=0\)
\(\Rightarrow (\cos{\theta}-\sin{\theta})(1+\cos{\theta}\sin{\theta})-1(1+\cos{\theta}\sin{\theta})=0\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow (1+\cos{\theta}\sin{\theta})(\cos{\theta}-\sin{\theta}-1)=0\)
\(\Rightarrow 1+\cos{\theta}\sin{\theta}=0, \ \cos{\theta}-\sin{\theta}-1=0\)
\(\Rightarrow \cos{\theta}\sin{\theta}=-1, \ \cos{\theta}-\sin{\theta}=1\)
প্রথম সমীকরণের ক্ষেত্রে,
\(\Rightarrow 2\sin{\theta}\cos{\theta}=-2\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\therefore \sin{2\theta}=-2, \ \text{যা গ্রহণযোগ্য নয়,} \ \because -1\le{\sin{A}}\le{1}\)
দ্বিতীয় সমীকরণের ক্ষেত্রে,
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \theta+\frac{\pi}{4}=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{4}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{4}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi-\frac{2\pi}{4}\)
\(\therefore \theta=2n\pi, \ 2n\pi-\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ 2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos^3{\theta}-\cos{\theta}\sin{\theta}-\sin^3{\theta}=1\)
\(\Rightarrow \cos^3{\theta}-\sin^3{\theta}-1-\cos{\theta}\sin{\theta}=0\)
\(\Rightarrow (\cos{\theta}-\sin{\theta})(\cos^2{\theta}+\cos{\theta}\sin{\theta}+\sin^2{\theta})-(1+\cos{\theta}\sin{\theta})=0\) ➜ \(\because a^3-b^3=(a-b)(a^2+ab+b^2)\)
\(\Rightarrow (\cos{\theta}-\sin{\theta})(\sin^2{\theta}+\cos^2{\theta}+\cos{\theta}\sin{\theta})-(1+\cos{\theta}\sin{\theta})=0\)
\(\Rightarrow (\cos{\theta}-\sin{\theta})(1+\cos{\theta}\sin{\theta})-1(1+\cos{\theta}\sin{\theta})=0\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow (1+\cos{\theta}\sin{\theta})(\cos{\theta}-\sin{\theta}-1)=0\)
\(\Rightarrow 1+\cos{\theta}\sin{\theta}=0, \ \cos{\theta}-\sin{\theta}-1=0\)
\(\Rightarrow \cos{\theta}\sin{\theta}=-1, \ \cos{\theta}-\sin{\theta}=1\)
প্রথম সমীকরণের ক্ষেত্রে,
\(\Rightarrow 2\sin{\theta}\cos{\theta}=-2\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\therefore \sin{2\theta}=-2, \ \text{যা গ্রহণযোগ্য নয়,} \ \because -1\le{\sin{A}}\le{1}\)
দ্বিতীয় সমীকরণের ক্ষেত্রে,
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \theta+\frac{\pi}{4}=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{4}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{4}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi-\frac{2\pi}{4}\)
\(\therefore \theta=2n\pi, \ 2n\pi-\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ 2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xix)\) \(2\sin{2x}+3\cos{x}=0\)উত্তরঃ \((2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin{2x}+3\cos{x}=0\)
\(\Rightarrow 2\times2\sin{x}\cos{x}+3\cos{x}=0\) ➜ \(\because \sin{2A}=\sin{A}\cos{A}\)
\(\Rightarrow \cos{x}(4\sin{x}+3)=0\)
\(\Rightarrow \cos{x}=0, \ 4\sin{x}+3=0\)
\(\Rightarrow \cos{x}=0, \ 4\sin{x}=-3\)
\(\Rightarrow \cos{x}=0, \ \sin{x}=-\frac{3}{4}\)
\(\Rightarrow \cos{x}=0, \ \sin{x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=-\frac{3}{4}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=n\pi+(-1)^n\alpha\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\alpha\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
\(2\sin{2x}+3\cos{x}=0\)
\(\Rightarrow 2\times2\sin{x}\cos{x}+3\cos{x}=0\) ➜ \(\because \sin{2A}=\sin{A}\cos{A}\)
\(\Rightarrow \cos{x}(4\sin{x}+3)=0\)
\(\Rightarrow \cos{x}=0, \ 4\sin{x}+3=0\)
\(\Rightarrow \cos{x}=0, \ 4\sin{x}=-3\)
\(\Rightarrow \cos{x}=0, \ \sin{x}=-\frac{3}{4}\)
\(\Rightarrow \cos{x}=0, \ \sin{x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=-\frac{3}{4}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=n\pi+(-1)^n\alpha\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\alpha\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xx)\) \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})=-1\)উত্তরঃ \(n\pi+(-1)^n\frac{7\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})=-1\)
\(\Rightarrow 2\times2\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\) ➜ \(\because \sin{2A}=\sin{A}\cos{A}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\)
\(\Rightarrow 2\sin{\theta}(2\cos{\theta}+1)+1(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}+1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}+1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=-1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=-\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \sin{\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \sin{\theta}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
এবং \(-\sin{A}=\sin{\left(\pi+A\right)}\)
\(\Rightarrow \theta=2n\pi\pm\left(\pi-\frac{\pi}{3}\right), \ \theta=n\pi+(-1)^n\left(\pi+\frac{\pi}{6}\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{7\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})=-1\)
\(\Rightarrow 2\times2\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\) ➜ \(\because \sin{2A}=\sin{A}\cos{A}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\)
\(\Rightarrow 2\sin{\theta}(2\cos{\theta}+1)+1(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}+1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}+1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=-1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=-\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \sin{\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \sin{\theta}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
এবং \(-\sin{A}=\sin{\left(\pi+A\right)}\)
\(\Rightarrow \theta=2n\pi\pm\left(\pi-\frac{\pi}{3}\right), \ \theta=n\pi+(-1)^n\left(\pi+\frac{\pi}{6}\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{7\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxi)\) \(2\cos^2{x}+\cos^2{2x}=2\)উত্তরঃ \(n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\cos^2{x}+\cos^2{2x}=2\)
\(\Rightarrow \cos{2x}+1+\cos^2{2x}=2\) ➜ \(\because 2\cos^2{A}=\cos{2A}+1\)
\(\Rightarrow \cos^2{2x}+\cos{2x}+1-2=0\)
\(\Rightarrow \cos^2{2x}+\cos{2x}-1=0\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1+\sqrt{5}}{2}, \ \cos{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \cos{2x}=\cos{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\theta}\)
\(\Rightarrow A=2n\pi\pm\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(2\cos^2{x}+\cos^2{2x}=2\)
\(\Rightarrow \cos{2x}+1+\cos^2{2x}=2\) ➜ \(\because 2\cos^2{A}=\cos{2A}+1\)
\(\Rightarrow \cos^2{2x}+\cos{2x}+1-2=0\)
\(\Rightarrow \cos^2{2x}+\cos{2x}-1=0\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1+\sqrt{5}}{2}, \ \cos{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \cos{2x}=\cos{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\theta}\)
\(\Rightarrow A=2n\pi\pm\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxii)\) \(\tan^2{x}+\sec^2{x}=3\)উত্তরঃ \(n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^2{x}+\sec^2{x}=3\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}=3-1\)
\(\Rightarrow 2\tan^2{x}=2\)
\(\Rightarrow \tan^2{x}=1\)
\(\Rightarrow \tan{x}=\pm\sqrt{1}\)
\(\Rightarrow \tan{x}=\pm1\)
\(\Rightarrow \tan{x}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{x}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow x=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan^2{x}+\sec^2{x}=3\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}=3-1\)
\(\Rightarrow 2\tan^2{x}=2\)
\(\Rightarrow \tan^2{x}=1\)
\(\Rightarrow \tan{x}=\pm\sqrt{1}\)
\(\Rightarrow \tan{x}=\pm1\)
\(\Rightarrow \tan{x}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{x}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow x=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxiii)\) \(\cot^2{\theta}+cosec^2{\theta}=3\)উত্তরঃ \(n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot^2{\theta}+cosec^2{\theta}=3\)
\(\Rightarrow \cot^2{\theta}+1+\cot^2{\theta}=3\) ➜ \(\because cosec^2{A}=1+\cot^2{A}\)
\(\Rightarrow 2\cot^2{\theta}=3-1\)
\(\Rightarrow 2\cot^2{\theta}=2\)
\(\Rightarrow \cot^2{\theta}=1\)
\(\Rightarrow \cot{\theta}=\pm\sqrt{1}\)
\(\Rightarrow \cot{\theta}=\pm1\)
\(\Rightarrow \cot{\theta}=\pm\cot{\frac{\pi}{4}}\) ➜ \(\because 1=\cot{\frac{\pi}{4}}\)
\(\Rightarrow \cot{\theta}=\cot{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \cot{A}=\cot{\alpha}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cot^2{\theta}+cosec^2{\theta}=3\)
\(\Rightarrow \cot^2{\theta}+1+\cot^2{\theta}=3\) ➜ \(\because cosec^2{A}=1+\cot^2{A}\)
\(\Rightarrow 2\cot^2{\theta}=3-1\)
\(\Rightarrow 2\cot^2{\theta}=2\)
\(\Rightarrow \cot^2{\theta}=1\)
\(\Rightarrow \cot{\theta}=\pm\sqrt{1}\)
\(\Rightarrow \cot{\theta}=\pm1\)
\(\Rightarrow \cot{\theta}=\pm\cot{\frac{\pi}{4}}\) ➜ \(\because 1=\cot{\frac{\pi}{4}}\)
\(\Rightarrow \cot{\theta}=\cot{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \cot{A}=\cot{\alpha}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxiv)\) \(\sqrt{2}\sec{x}+\tan{x}=1\)উত্তরঃ \(2n\pi-\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{2}\sec{x}+\tan{x}=1\)
\(\Rightarrow \sqrt{2}\frac{1}{\cos{x}}+\frac{\sin{x}}{\cos{x}}=1\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\sqrt{2}}{\cos{x}}+\frac{\sin{x}}{\cos{x}}=1\)
\(\Rightarrow \frac{\sqrt{2}+\sin{x}}{\cos{x}}=1\)
\(\Rightarrow \cos{x}=\sqrt{2}+\sin{x}\)
\(\Rightarrow \cos{x}-\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{1}{\sqrt{2}}-\sin{x}\frac{1}{\sqrt{2}}=1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{4}}-\sin{x}\sin{\frac{\pi}{4}}=1\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{4}\right)}=1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{4}=2n\pi\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi-\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi-\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sqrt{2}\sec{x}+\tan{x}=1\)
\(\Rightarrow \sqrt{2}\frac{1}{\cos{x}}+\frac{\sin{x}}{\cos{x}}=1\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\sqrt{2}}{\cos{x}}+\frac{\sin{x}}{\cos{x}}=1\)
\(\Rightarrow \frac{\sqrt{2}+\sin{x}}{\cos{x}}=1\)
\(\Rightarrow \cos{x}=\sqrt{2}+\sin{x}\)
\(\Rightarrow \cos{x}-\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{1}{\sqrt{2}}-\sin{x}\frac{1}{\sqrt{2}}=1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{4}}-\sin{x}\sin{\frac{\pi}{4}}=1\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{4}\right)}=1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{4}=2n\pi\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi-\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi-\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxv)\) \(2\cos{x}+3\sin{x}=1\)উত্তরঃ \(2n\pi\pm\alpha+\theta\) যেখানে, \(\cos{\theta}=\frac{2}{\sqrt{13}}, \ \cos{\alpha}=\frac{1}{\sqrt{13}}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\cos{x}+3\sin{x}=1\)
\(\Rightarrow \cos{x}\times2+\sin{x}\times3=1\)
\(\Rightarrow \cos{x}\frac{2}{\sqrt{13}}+\sin{x}\frac{3}{\sqrt{13}}=\frac{1}{\sqrt{13}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{2^2+3^2}=\sqrt{13}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\theta}+\sin{x}\sin{\theta}=\frac{1}{\sqrt{13}}\) ➜ \(\because \cos{\theta}=\frac{2}{\sqrt{13}}\)
\(\Rightarrow \sin{\theta}=\frac{3}{\sqrt{13}}\)
\(\Rightarrow \cos{(x-\theta)}=\cos{\alpha}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{13}}\)
\(\Rightarrow x-\theta=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\pm\alpha+\theta\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm\alpha+\theta\) যেখানে, \(\cos{\theta}=\frac{2}{\sqrt{13}}, \ \cos{\alpha}=\frac{1}{\sqrt{13}}, \ n\in{\mathbb{Z}}\)
\(2\cos{x}+3\sin{x}=1\)
\(\Rightarrow \cos{x}\times2+\sin{x}\times3=1\)
\(\Rightarrow \cos{x}\frac{2}{\sqrt{13}}+\sin{x}\frac{3}{\sqrt{13}}=\frac{1}{\sqrt{13}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{2^2+3^2}=\sqrt{13}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\theta}+\sin{x}\sin{\theta}=\frac{1}{\sqrt{13}}\) ➜ \(\because \cos{\theta}=\frac{2}{\sqrt{13}}\)
\(\Rightarrow \sin{\theta}=\frac{3}{\sqrt{13}}\)
\(\Rightarrow \cos{(x-\theta)}=\cos{\alpha}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{13}}\)
\(\Rightarrow x-\theta=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\pm\alpha+\theta\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm\alpha+\theta\) যেখানে, \(\cos{\theta}=\frac{2}{\sqrt{13}}, \ \cos{\alpha}=\frac{1}{\sqrt{13}}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxvi)\) \(5\tan^2{\theta}-\sec^2{\theta}=11\)উত্তরঃ \(n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(5\tan^2{\theta}-\sec^2{\theta}=11\)
\(\Rightarrow 5\tan^2{\theta}-(1+\tan^2{\theta})=11\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 5\tan^2{\theta}-1-\tan^2{\theta}=11\)
\(\Rightarrow 4\tan^2{\theta}=11+1\)
\(\Rightarrow 4\tan^2{\theta}=12\)
\(\Rightarrow \tan^2{\theta}=3\) ➜ উভয় পার্শে \(4\) ভাগ করে,
\(\Rightarrow \tan{\theta}=\pm\sqrt{3}\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{3}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(5\tan^2{\theta}-\sec^2{\theta}=11\)
\(\Rightarrow 5\tan^2{\theta}-(1+\tan^2{\theta})=11\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 5\tan^2{\theta}-1-\tan^2{\theta}=11\)
\(\Rightarrow 4\tan^2{\theta}=11+1\)
\(\Rightarrow 4\tan^2{\theta}=12\)
\(\Rightarrow \tan^2{\theta}=3\) ➜ উভয় পার্শে \(4\) ভাগ করে,
\(\Rightarrow \tan{\theta}=\pm\sqrt{3}\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{3}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxvii)\) \(\cos{\theta}+\sec{\theta}=\frac{5}{2}\)উত্তরঃ \(2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}+\sec{\theta}=\frac{5}{2}\)
\(\Rightarrow \cos{\theta}+\frac{1}{\cos{\theta}}=\frac{5}{2}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+1}{\cos{\theta}}=\frac{5}{2}\)
\(\Rightarrow 2\cos^2{\theta}+2=5\cos{\theta}\)
\(\Rightarrow 2\cos^2{\theta}-5\cos{\theta}+2=0\)
\(\Rightarrow 2\cos^2{\theta}-4\cos{\theta}-\cos{\theta}+2=0\)
\(\Rightarrow 2\cos{\theta}(\cos{\theta}-2)-1(\cos{\theta}-2)=0\)
\(\Rightarrow \cos{\theta}-2\ne{0}, \ 2\cos{\theta}-1=0, \ \because -1\le{\cos{\theta}}\le{-1}\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{\theta}+\sec{\theta}=\frac{5}{2}\)
\(\Rightarrow \cos{\theta}+\frac{1}{\cos{\theta}}=\frac{5}{2}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+1}{\cos{\theta}}=\frac{5}{2}\)
\(\Rightarrow 2\cos^2{\theta}+2=5\cos{\theta}\)
\(\Rightarrow 2\cos^2{\theta}-5\cos{\theta}+2=0\)
\(\Rightarrow 2\cos^2{\theta}-4\cos{\theta}-\cos{\theta}+2=0\)
\(\Rightarrow 2\cos{\theta}(\cos{\theta}-2)-1(\cos{\theta}-2)=0\)
\(\Rightarrow \cos{\theta}-2\ne{0}, \ 2\cos{\theta}-1=0, \ \because -1\le{\cos{\theta}}\le{-1}\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxviii)\) \(8\sin^2{\theta}-2\cos{\theta}=5\)উত্তরঃ \(2n\pi\pm\frac{\pi}{3}, \ 2n\pi\pm\alpha\) যেখানে, \(\cos{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(8\sin^2{\theta}-2\cos{\theta}=5\)
\(\Rightarrow 8(1-\cos^2{\theta})-2\cos{\theta}=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 8-8\cos^2{\theta}-2\cos{\theta}-5=0\)
\(\Rightarrow -8\cos^2{\theta}-2\cos{\theta}+3=0\)
\(\Rightarrow 8\cos^2{\theta}+2\cos{\theta}-3=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 8\cos^2{\theta}+6\cos{\theta}-4\cos{\theta}-3=0\)
\(\Rightarrow 2\cos{\theta}(4\cos{\theta}+3)-1(4\cos{\theta}+3)=0\)
\(\Rightarrow (4\cos{\theta}+3)(2\cos{\theta}-1)=0\)
\(\Rightarrow 4\cos{\theta}+3=0, \ 2\cos{\theta}-1=0\)
\(\Rightarrow 4\cos{\theta}=-3, \ 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{3}{4}, \ \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\alpha}, \ \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
যেখানে, \(\cos{\alpha}=-\frac{3}{4}\)
\(\Rightarrow \theta=2n\pi\pm\alpha, \ \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\)
যেখানে, \(\cos{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\alpha, \ 2n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}, \ 2n\pi\pm\alpha\) যেখানে, \(\cos{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
\(8\sin^2{\theta}-2\cos{\theta}=5\)
\(\Rightarrow 8(1-\cos^2{\theta})-2\cos{\theta}=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 8-8\cos^2{\theta}-2\cos{\theta}-5=0\)
\(\Rightarrow -8\cos^2{\theta}-2\cos{\theta}+3=0\)
\(\Rightarrow 8\cos^2{\theta}+2\cos{\theta}-3=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 8\cos^2{\theta}+6\cos{\theta}-4\cos{\theta}-3=0\)
\(\Rightarrow 2\cos{\theta}(4\cos{\theta}+3)-1(4\cos{\theta}+3)=0\)
\(\Rightarrow (4\cos{\theta}+3)(2\cos{\theta}-1)=0\)
\(\Rightarrow 4\cos{\theta}+3=0, \ 2\cos{\theta}-1=0\)
\(\Rightarrow 4\cos{\theta}=-3, \ 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{3}{4}, \ \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\alpha}, \ \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
যেখানে, \(\cos{\alpha}=-\frac{3}{4}\)
\(\Rightarrow \theta=2n\pi\pm\alpha, \ \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\)
যেখানে, \(\cos{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\alpha, \ 2n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}, \ 2n\pi\pm\alpha\) যেখানে, \(\cos{\alpha}=-\frac{3}{4}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxix)\) \(\sin{\theta}+cosec \ {\theta}=\frac{3}{\sqrt{2}}\)উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{\theta}+cosec \ {\theta}=\frac{3}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}+\frac{1}{\sin{\theta}}=\frac{3}{\sqrt{2}}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}+1}{\sin{\theta}}=\frac{3}{\sqrt{2}}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}+\sqrt{2}=3\sin{\theta}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-2\sin{\theta}-\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}(\sin{\theta}-\sqrt{2})-1(\sin{\theta}-\sqrt{2})=0\)
\(\Rightarrow (\sin{\theta}-\sqrt{2})(\sqrt{2}\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}-\sqrt{2}\ne{0}, \ \sqrt{2}\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{\theta}+cosec \ {\theta}=\frac{3}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}+\frac{1}{\sin{\theta}}=\frac{3}{\sqrt{2}}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}+1}{\sin{\theta}}=\frac{3}{\sqrt{2}}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}+\sqrt{2}=3\sin{\theta}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-2\sin{\theta}-\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}(\sin{\theta}-\sqrt{2})-1(\sin{\theta}-\sqrt{2})=0\)
\(\Rightarrow (\sin{\theta}-\sqrt{2})(\sqrt{2}\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}-\sqrt{2}\ne{0}, \ \sqrt{2}\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxx)\) \(cosec \ {\theta}\cot{\theta}=2\sqrt{3}\)উত্তরঃ \(2n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(cosec \ {\theta}\cot{\theta}=2\sqrt{3}\)
\(\Rightarrow \frac{1}{\sin{\theta}}\times\frac{\cos{\theta}}{\sin{\theta}}=2\sqrt{3}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin^2{\theta}}=2\sqrt{3}\)
\(\Rightarrow \frac{\cos{\theta}}{1-\cos^2{\theta}}=2\sqrt{3}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow \cos{\theta}=2\sqrt{3}-2\sqrt{3}\cos^2{\theta}\)
\(\Rightarrow 2\sqrt{3}\cos^2{\theta}+\cos{\theta}-2\sqrt{3}=0\)
\(\Rightarrow 2\sqrt{3}\cos^2{\theta}+4\cos{\theta}-3\cos{\theta}-2\sqrt{3}=0\)
\(\Rightarrow 2\cos{\theta}(\sqrt{3}\cos{\theta}+2)-\sqrt{3}(\sqrt{3}\cos{\theta}+2)=0\)
\(\Rightarrow (\sqrt{3}\cos{\theta}+2)(2\cos{\theta}-\sqrt{3})=0\)
\(\Rightarrow \sqrt{3}\cos{\theta}+2=0, \ 2\cos{\theta}-\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cos{\theta}=-2, \ 2\cos{\theta}=\sqrt{3}\)
\(\Rightarrow \cos{\theta}=-\frac{2}{\sqrt{3}}, \ \cos{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{\theta}\ne{-\frac{2}{\sqrt{3}}}, \ \cos{\theta}=\frac{\sqrt{3}}{2}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(cosec \ {\theta}\cot{\theta}=2\sqrt{3}\)
\(\Rightarrow \frac{1}{\sin{\theta}}\times\frac{\cos{\theta}}{\sin{\theta}}=2\sqrt{3}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin^2{\theta}}=2\sqrt{3}\)
\(\Rightarrow \frac{\cos{\theta}}{1-\cos^2{\theta}}=2\sqrt{3}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow \cos{\theta}=2\sqrt{3}-2\sqrt{3}\cos^2{\theta}\)
\(\Rightarrow 2\sqrt{3}\cos^2{\theta}+\cos{\theta}-2\sqrt{3}=0\)
\(\Rightarrow 2\sqrt{3}\cos^2{\theta}+4\cos{\theta}-3\cos{\theta}-2\sqrt{3}=0\)
\(\Rightarrow 2\cos{\theta}(\sqrt{3}\cos{\theta}+2)-\sqrt{3}(\sqrt{3}\cos{\theta}+2)=0\)
\(\Rightarrow (\sqrt{3}\cos{\theta}+2)(2\cos{\theta}-\sqrt{3})=0\)
\(\Rightarrow \sqrt{3}\cos{\theta}+2=0, \ 2\cos{\theta}-\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cos{\theta}=-2, \ 2\cos{\theta}=\sqrt{3}\)
\(\Rightarrow \cos{\theta}=-\frac{2}{\sqrt{3}}, \ \cos{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{\theta}\ne{-\frac{2}{\sqrt{3}}}, \ \cos{\theta}=\frac{\sqrt{3}}{2}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxi)\) \(\sin{5\theta}+\sin{\theta}=\sin{3\theta}\)উত্তরঃ \(\frac{n\pi}{3}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{5\theta}+\sin{\theta}=\sin{3\theta}\)
\(\Rightarrow \sin{5\theta}+\sin{\theta}-\sin{3\theta}=0\)
\(\Rightarrow 2\sin{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}-\sin{3\theta}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\sin{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}-\sin{3\theta}=0\)
\(\Rightarrow 2\sin{3\theta}\cos{2\theta}-\sin{3\theta}=0\)
\(\Rightarrow \sin{3\theta}(2\cos{2\theta}-1)=0\)
\(\Rightarrow \sin{3\theta}=0, \ 2\cos{2\theta}-1=0\)
\(\Rightarrow \sin{3\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \sin{3\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{3\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 3\theta=n\pi, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{3}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=\frac{n\pi}{3}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{3}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{5\theta}+\sin{\theta}=\sin{3\theta}\)
\(\Rightarrow \sin{5\theta}+\sin{\theta}-\sin{3\theta}=0\)
\(\Rightarrow 2\sin{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}-\sin{3\theta}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\sin{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}-\sin{3\theta}=0\)
\(\Rightarrow 2\sin{3\theta}\cos{2\theta}-\sin{3\theta}=0\)
\(\Rightarrow \sin{3\theta}(2\cos{2\theta}-1)=0\)
\(\Rightarrow \sin{3\theta}=0, \ 2\cos{2\theta}-1=0\)
\(\Rightarrow \sin{3\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \sin{3\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{3\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 3\theta=n\pi, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{3}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=\frac{n\pi}{3}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{3}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxii)\) \(\sin{2\theta}+4\cos{\theta}=\sqrt{3}\sin{\theta}+2\sqrt{3}\)উত্তরঃ \(n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{2\theta}+4\cos{\theta}=\sqrt{3}\sin{\theta}+2\sqrt{3}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}+4\cos{\theta}=\sqrt{3}\sin{\theta}+2\sqrt{3}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}+4\cos{\theta}-\sqrt{3}\sin{\theta}-2\sqrt{3}=0\)
\(\Rightarrow 2\cos{\theta}(\sin{\theta}+2)-\sqrt{3}(\sin{\theta}+2)=0\)
\(\Rightarrow (\sin{\theta}+2)(2\cos{\theta}-\sqrt{3})=0\)
\(\Rightarrow \sin{\theta}+2\ne{0}, \ 2\cos{\theta}-\sqrt{3}=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow 2\cos{\theta}-\sqrt{3}=0\)
\(\Rightarrow 2\cos{\theta}=\sqrt{3}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{2\theta}+4\cos{\theta}=\sqrt{3}\sin{\theta}+2\sqrt{3}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}+4\cos{\theta}=\sqrt{3}\sin{\theta}+2\sqrt{3}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}+4\cos{\theta}-\sqrt{3}\sin{\theta}-2\sqrt{3}=0\)
\(\Rightarrow 2\cos{\theta}(\sin{\theta}+2)-\sqrt{3}(\sin{\theta}+2)=0\)
\(\Rightarrow (\sin{\theta}+2)(2\cos{\theta}-\sqrt{3})=0\)
\(\Rightarrow \sin{\theta}+2\ne{0}, \ 2\cos{\theta}-\sqrt{3}=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow 2\cos{\theta}-\sqrt{3}=0\)
\(\Rightarrow 2\cos{\theta}=\sqrt{3}\)
\(\Rightarrow \cos{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\therefore \theta=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxiii)\) \(\tan{\left(\frac{\pi}{4}+\theta\right)}+\tan{\left(\frac{\pi}{4}-\theta\right)}=4\)উত্তরঃ \(n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{\left(\frac{\pi}{4}+\theta\right)}+\tan{\left(\frac{\pi}{4}-\theta\right)}=4\)
\(\Rightarrow \frac{\tan{\frac{\pi}{4}}+\tan{\theta}}{1-\tan{\frac{\pi}{4}}\tan{\theta}}+\frac{\tan{\frac{\pi}{4}}-\tan{\theta}}{1+\tan{\frac{\pi}{4}}\tan{\theta}}=4\) ➜ \(\because \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
এবং \(\tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
\(\Rightarrow \frac{1+\tan{\theta}}{1-1.\tan{\theta}}+\frac{1-\tan{\theta}}{1+1.\tan{\theta}}=4\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(\Rightarrow \frac{1+\tan{\theta}}{1-\tan{\theta}}+\frac{1-\tan{\theta}}{1+\tan{\theta}}=4\)
\(\Rightarrow \frac{(1+\tan{\theta})^2+(1-\tan{\theta})^2}{(1-\tan{\theta})(1+\tan{\theta})}=4\)
\(\Rightarrow \frac{2\times1^2+2\tan^2{\theta}}{1-\tan^2{\theta}}=4\) ➜ \(\because (a+b)^2+(a-b)^2=2a^2+2b^2\)
এবং \((a-b)(a+b)=a^2-b^2\)
\(\Rightarrow \frac{2+2\tan^2{\theta}}{1-\tan^2{\theta}}=4\)
\(\Rightarrow \frac{2(1+\tan^2{\theta})}{1-\tan^2{\theta}}=4\)
\(\Rightarrow \frac{1+\tan^2{\theta}}{1-\tan^2{\theta}}=2\)
\(\Rightarrow 1+\tan^2{\theta}=2-2\tan^2{\theta}\)
\(\Rightarrow \tan^2{\theta}+2\tan^2{\theta}=2-1\)
\(\Rightarrow 3\tan^2{\theta}=1\)
\(\Rightarrow \tan^2{\theta}=\frac{1}{3}\)
\(\Rightarrow \tan{\theta}=\pm\sqrt{\frac{1}{3}}\)
\(\Rightarrow \tan{\theta}=\pm\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{6}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{6}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{\left(\frac{\pi}{4}+\theta\right)}+\tan{\left(\frac{\pi}{4}-\theta\right)}=4\)
\(\Rightarrow \frac{\tan{\frac{\pi}{4}}+\tan{\theta}}{1-\tan{\frac{\pi}{4}}\tan{\theta}}+\frac{\tan{\frac{\pi}{4}}-\tan{\theta}}{1+\tan{\frac{\pi}{4}}\tan{\theta}}=4\) ➜ \(\because \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
এবং \(\tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
\(\Rightarrow \frac{1+\tan{\theta}}{1-1.\tan{\theta}}+\frac{1-\tan{\theta}}{1+1.\tan{\theta}}=4\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(\Rightarrow \frac{1+\tan{\theta}}{1-\tan{\theta}}+\frac{1-\tan{\theta}}{1+\tan{\theta}}=4\)
\(\Rightarrow \frac{(1+\tan{\theta})^2+(1-\tan{\theta})^2}{(1-\tan{\theta})(1+\tan{\theta})}=4\)
\(\Rightarrow \frac{2\times1^2+2\tan^2{\theta}}{1-\tan^2{\theta}}=4\) ➜ \(\because (a+b)^2+(a-b)^2=2a^2+2b^2\)
এবং \((a-b)(a+b)=a^2-b^2\)
\(\Rightarrow \frac{2+2\tan^2{\theta}}{1-\tan^2{\theta}}=4\)
\(\Rightarrow \frac{2(1+\tan^2{\theta})}{1-\tan^2{\theta}}=4\)
\(\Rightarrow \frac{1+\tan^2{\theta}}{1-\tan^2{\theta}}=2\)
\(\Rightarrow 1+\tan^2{\theta}=2-2\tan^2{\theta}\)
\(\Rightarrow \tan^2{\theta}+2\tan^2{\theta}=2-1\)
\(\Rightarrow 3\tan^2{\theta}=1\)
\(\Rightarrow \tan^2{\theta}=\frac{1}{3}\)
\(\Rightarrow \tan{\theta}=\pm\sqrt{\frac{1}{3}}\)
\(\Rightarrow \tan{\theta}=\pm\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{6}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{6}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxiv)\) \(\sin^2{2\theta}-\sin^2{\theta}=\frac{1}{3}\sin{3\theta}\)উত্তরঃ \(n\pi, \ 2n\pi\pm\frac{\pi}{3}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=\frac{1}{3}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin^2{2\theta}-\sin^2{\theta}=\frac{1}{3}\sin{3\theta}\)
\(\Rightarrow 4\sin^2{\theta}\cos^2{\theta}-\sin^2{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
এবং \(\sin{3A}=3\sin{A}-4\sin^3{A}\)
\(\Rightarrow 4\sin^2{\theta}(1-\sin^2{\theta})-\sin^2{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 4\sin^2{\theta}-4\sin^4{\theta}-\sin^2{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\)
\(\Rightarrow 3\sin^2{\theta}-4\sin^4{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\)
\(\Rightarrow -9\sin^2{\theta}+12\sin^4{\theta}=-3\sin{\theta}+4\sin^3{\theta}\) ➜ উভয় পার্শে \(-3\) গুণ করে,
\(\Rightarrow 12\sin^4{\theta}-4\sin^3{\theta}-9\sin^2{\theta}+3\sin{\theta}=0\)
\(\Rightarrow 4\sin^3{\theta}(3\sin{\theta}-1)-3\sin{\theta}(3\sin{\theta}-1)=0\)
\(\Rightarrow (3\sin{\theta}-1)(4\sin^3{\theta}-3\sin{\theta})=0\)
\(\Rightarrow 3\sin{\theta}-1=0, \ 4\sin^3{\theta}-3\sin{\theta}=0\)
\(\Rightarrow 3\sin{\theta}=1, \ \sin{\theta}(4\sin^2{\theta}-3)=0\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ 4\sin^2{\theta}-3=0\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ 4\sin^2{\theta}=3\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ \sin^2{\theta}=\frac{3}{4}\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ \sin{\theta}=\pm\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\alpha}, \ \sin{\theta}=0, \ \sin{\theta}=\sin{\left(\pm\frac{\pi}{3}\right)}\) যেখানে, \(\sin{\alpha}=\frac{1}{3}\) ➜ \(\because \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha, \ \theta=n\pi, \ \theta=n\pi+(-1)^n\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha, \ n\pi, \ \theta=2n\pi\pm\frac{\pi}{3}\)
\(\therefore \theta=n\pi, \ 2n\pi\pm\frac{\pi}{3}, \ n\pi+(-1)^n\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi, \ 2n\pi\pm\frac{\pi}{3}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=\frac{1}{3}, \ n\in{\mathbb{Z}}\)
\(\sin^2{2\theta}-\sin^2{\theta}=\frac{1}{3}\sin{3\theta}\)
\(\Rightarrow 4\sin^2{\theta}\cos^2{\theta}-\sin^2{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
এবং \(\sin{3A}=3\sin{A}-4\sin^3{A}\)
\(\Rightarrow 4\sin^2{\theta}(1-\sin^2{\theta})-\sin^2{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 4\sin^2{\theta}-4\sin^4{\theta}-\sin^2{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\)
\(\Rightarrow 3\sin^2{\theta}-4\sin^4{\theta}=\frac{1}{3}(3\sin{\theta}-4\sin^3{\theta})\)
\(\Rightarrow -9\sin^2{\theta}+12\sin^4{\theta}=-3\sin{\theta}+4\sin^3{\theta}\) ➜ উভয় পার্শে \(-3\) গুণ করে,
\(\Rightarrow 12\sin^4{\theta}-4\sin^3{\theta}-9\sin^2{\theta}+3\sin{\theta}=0\)
\(\Rightarrow 4\sin^3{\theta}(3\sin{\theta}-1)-3\sin{\theta}(3\sin{\theta}-1)=0\)
\(\Rightarrow (3\sin{\theta}-1)(4\sin^3{\theta}-3\sin{\theta})=0\)
\(\Rightarrow 3\sin{\theta}-1=0, \ 4\sin^3{\theta}-3\sin{\theta}=0\)
\(\Rightarrow 3\sin{\theta}=1, \ \sin{\theta}(4\sin^2{\theta}-3)=0\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ 4\sin^2{\theta}-3=0\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ 4\sin^2{\theta}=3\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ \sin^2{\theta}=\frac{3}{4}\)
\(\Rightarrow \sin{\theta}=\frac{1}{3}, \ \sin{\theta}=0, \ \sin{\theta}=\pm\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\alpha}, \ \sin{\theta}=0, \ \sin{\theta}=\sin{\left(\pm\frac{\pi}{3}\right)}\) যেখানে, \(\sin{\alpha}=\frac{1}{3}\) ➜ \(\because \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha, \ \theta=n\pi, \ \theta=n\pi+(-1)^n\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha, \ n\pi, \ \theta=2n\pi\pm\frac{\pi}{3}\)
\(\therefore \theta=n\pi, \ 2n\pi\pm\frac{\pi}{3}, \ n\pi+(-1)^n\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi, \ 2n\pi\pm\frac{\pi}{3}, \ n\pi+(-1)^n\alpha\) যেখানে, \(\sin{\alpha}=\frac{1}{3}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxv)\) \(\tan^3{\theta}-\sec^2{\theta}=4\tan^2{\theta}-5\tan{\theta}\)উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+\alpha, \ n\pi+\beta\) যেখানে, \(\tan{\alpha}=2+\sqrt{3}, \ \tan{\beta}=2-\sqrt{3}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^3{\theta}-\sec^2{\theta}=4\tan^2{\theta}-5\tan{\theta}\)
\(\Rightarrow \tan^3{\theta}-1-\tan^2{\theta}=4\tan^2{\theta}-5\tan{\theta}\) ➜ \(\because \sec^2{A}=1-\tan^2{A}\)
\(\Rightarrow \tan^3{\theta}-1-\tan^2{\theta}-4\tan^2{\theta}+5\tan{\theta}=0\)
\(\Rightarrow \tan^3{\theta}-5\tan^2{\theta}+5\tan{\theta}-1=0\)
\(\Rightarrow \tan^3{\theta}-\tan^2{\theta}-4\tan^2{\theta}+4\tan{\theta}+\tan{\theta}-1=0\)
\(\Rightarrow \tan^2{\theta}(\tan{\theta}-1)-4\tan{\theta}(\tan{\theta}-1)+1(\tan{\theta}-1)=0\)
\(\Rightarrow (\tan{\theta}-1)(\tan^2{\theta}-4\tan{\theta}+1)=0\)
\(\Rightarrow \tan{\theta}-1=0, \ \tan^2{\theta}-4\tan{\theta}+1=0\)
প্রথম সমীকরণের ক্ষেত্রে,
\(\Rightarrow \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
দ্বিতীয় সমীকরণের ক্ষেত্রে,
\(\Rightarrow \tan{\theta}=\frac{4\pm{\sqrt{(-4)^2-4\times1\times1}}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}\)
\(\Rightarrow \tan{\theta}=\frac{4\pm{\sqrt{16-4}}}{2}\)
\(\Rightarrow \tan{\theta}=\frac{4\pm{\sqrt{12}}}{2}\)
\(\Rightarrow \tan{\theta}=\frac{4\pm{2\sqrt{3}}}{2}\)
\(\Rightarrow \tan{\theta}=\frac{2(2\pm{\sqrt{3}})}{2}\)
\(\Rightarrow \tan{\theta}=2\pm{\sqrt{3}}\)
\(\Rightarrow \tan{\theta}=2+\sqrt{3}, \ \tan{\theta}=2-\sqrt{3}\)
\(\Rightarrow \tan{\theta}=\tan{\alpha}, \ \tan{\theta}=\tan{\beta}\) যেখানে, \(\tan{\alpha}=2+\sqrt{3}, \ \tan{\beta}=2-\sqrt{3}\)
\(\Rightarrow \theta=n\pi+\alpha, \ \theta=n\pi+\beta\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+\alpha, \ n\pi+\beta\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}, \ n\pi+\alpha, \ n\pi+\beta\) যেখানে, \(\tan{\alpha}=2+\sqrt{3}, \ \tan{\beta}=2-\sqrt{3}, \ n\in{\mathbb{Z}}\)
\(\tan^3{\theta}-\sec^2{\theta}=4\tan^2{\theta}-5\tan{\theta}\)
\(\Rightarrow \tan^3{\theta}-1-\tan^2{\theta}=4\tan^2{\theta}-5\tan{\theta}\) ➜ \(\because \sec^2{A}=1-\tan^2{A}\)
\(\Rightarrow \tan^3{\theta}-1-\tan^2{\theta}-4\tan^2{\theta}+5\tan{\theta}=0\)
\(\Rightarrow \tan^3{\theta}-5\tan^2{\theta}+5\tan{\theta}-1=0\)
\(\Rightarrow \tan^3{\theta}-\tan^2{\theta}-4\tan^2{\theta}+4\tan{\theta}+\tan{\theta}-1=0\)
\(\Rightarrow \tan^2{\theta}(\tan{\theta}-1)-4\tan{\theta}(\tan{\theta}-1)+1(\tan{\theta}-1)=0\)
\(\Rightarrow (\tan{\theta}-1)(\tan^2{\theta}-4\tan{\theta}+1)=0\)
\(\Rightarrow \tan{\theta}-1=0, \ \tan^2{\theta}-4\tan{\theta}+1=0\)
প্রথম সমীকরণের ক্ষেত্রে,
\(\Rightarrow \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
দ্বিতীয় সমীকরণের ক্ষেত্রে,
\(\Rightarrow \tan{\theta}=\frac{4\pm{\sqrt{(-4)^2-4\times1\times1}}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}\)
\(\Rightarrow \tan{\theta}=\frac{4\pm{\sqrt{16-4}}}{2}\)
\(\Rightarrow \tan{\theta}=\frac{4\pm{\sqrt{12}}}{2}\)
\(\Rightarrow \tan{\theta}=\frac{4\pm{2\sqrt{3}}}{2}\)
\(\Rightarrow \tan{\theta}=\frac{2(2\pm{\sqrt{3}})}{2}\)
\(\Rightarrow \tan{\theta}=2\pm{\sqrt{3}}\)
\(\Rightarrow \tan{\theta}=2+\sqrt{3}, \ \tan{\theta}=2-\sqrt{3}\)
\(\Rightarrow \tan{\theta}=\tan{\alpha}, \ \tan{\theta}=\tan{\beta}\) যেখানে, \(\tan{\alpha}=2+\sqrt{3}, \ \tan{\beta}=2-\sqrt{3}\)
\(\Rightarrow \theta=n\pi+\alpha, \ \theta=n\pi+\beta\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+\alpha, \ n\pi+\beta\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}, \ n\pi+\alpha, \ n\pi+\beta\) যেখানে, \(\tan{\alpha}=2+\sqrt{3}, \ \tan{\beta}=2-\sqrt{3}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxvi)\) \(cosec \ {\theta}+\sec{\theta}=2\sqrt{2}\)উত্তরঃ \((8n+3)\frac{\pi}{12}, \ (8n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(cosec \ {\theta}+\sec{\theta}=2\sqrt{2}\)
\(\Rightarrow \frac{1}{\sin{\theta}}+\frac{1}{\cos{\theta}}=2\sqrt{2}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{\theta}+\sin{\theta}}{\sin{\theta}\cos{\theta}}=2\sqrt{2}\)
\(\Rightarrow \cos{\theta}+\sin{\theta}=2\sqrt{2}\sin{\theta}\cos{\theta}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{2}\times2\sin{\theta}\cos{\theta}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{2}\sin{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{\theta}\frac{1}{\sqrt{2}}+\cos{\theta}\frac{1}{\sqrt{2}}=\sin{2\theta}\) ➜ উভয় পার্শে \(\sqrt{2}\) ভাগ করে,
\(\Rightarrow \sin{\theta}\cos{\frac{\pi}{4}}+\cos{\theta}\sin{\frac{\pi}{4}}=\sin{2\theta}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \sin{\left(\theta+\frac{\pi}{4}\right)}=\sin{2\theta}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \sin{2\theta}=\sin{\left(\theta+\frac{\pi}{4}\right)}\)
\(\Rightarrow 2\theta=m\pi+(-1)^m\left(\theta+\frac{\pi}{4}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=m\pi+(-1)^m\alpha\) যেখানে, \(m\in{\mathbb{Z}}\)
\(\Rightarrow 2\theta=2n\pi+(-1)^{2n}\left(\theta+\frac{\pi}{4}\right), \ 2\theta=(2n+1)\pi+(-1)^{2n+1}\left(\theta+\frac{\pi}{4}\right)\) যেখানে, \(m=2n, \ 2n+1\)
\(\Rightarrow 2\theta=2n\pi+\theta+\frac{\pi}{4}, \ 2\theta=(2n+1)\pi-\theta-\frac{\pi}{4}\)
\(\Rightarrow 2\theta-\theta=2n\pi+\frac{\pi}{4}, \ 2\theta+\theta=(2n+1)\pi-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}, \ 3\theta=(2n+1)\pi-\frac{\pi}{4}\)
\(\Rightarrow \theta=(8n+1)\frac{\pi}{4}, \ 3\theta=(8n+4-1)\frac{\pi}{4}\)
\(\Rightarrow \theta=(8n+1)\frac{\pi}{4}, \ \theta=(8n+3)\frac{\pi}{12}\)
\(\therefore \theta=(8n+1)\frac{\pi}{4}, \ (8n+3)\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(8n+3)\frac{\pi}{12}, \ (8n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(cosec \ {\theta}+\sec{\theta}=2\sqrt{2}\)
\(\Rightarrow \frac{1}{\sin{\theta}}+\frac{1}{\cos{\theta}}=2\sqrt{2}\) ➜ \(\because cosec \ {A}=\frac{1}{\sin{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{\theta}+\sin{\theta}}{\sin{\theta}\cos{\theta}}=2\sqrt{2}\)
\(\Rightarrow \cos{\theta}+\sin{\theta}=2\sqrt{2}\sin{\theta}\cos{\theta}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{2}\times2\sin{\theta}\cos{\theta}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{2}\sin{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{\theta}\frac{1}{\sqrt{2}}+\cos{\theta}\frac{1}{\sqrt{2}}=\sin{2\theta}\) ➜ উভয় পার্শে \(\sqrt{2}\) ভাগ করে,
\(\Rightarrow \sin{\theta}\cos{\frac{\pi}{4}}+\cos{\theta}\sin{\frac{\pi}{4}}=\sin{2\theta}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \sin{\left(\theta+\frac{\pi}{4}\right)}=\sin{2\theta}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \sin{2\theta}=\sin{\left(\theta+\frac{\pi}{4}\right)}\)
\(\Rightarrow 2\theta=m\pi+(-1)^m\left(\theta+\frac{\pi}{4}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=m\pi+(-1)^m\alpha\) যেখানে, \(m\in{\mathbb{Z}}\)
\(\Rightarrow 2\theta=2n\pi+(-1)^{2n}\left(\theta+\frac{\pi}{4}\right), \ 2\theta=(2n+1)\pi+(-1)^{2n+1}\left(\theta+\frac{\pi}{4}\right)\) যেখানে, \(m=2n, \ 2n+1\)
\(\Rightarrow 2\theta=2n\pi+\theta+\frac{\pi}{4}, \ 2\theta=(2n+1)\pi-\theta-\frac{\pi}{4}\)
\(\Rightarrow 2\theta-\theta=2n\pi+\frac{\pi}{4}, \ 2\theta+\theta=(2n+1)\pi-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}, \ 3\theta=(2n+1)\pi-\frac{\pi}{4}\)
\(\Rightarrow \theta=(8n+1)\frac{\pi}{4}, \ 3\theta=(8n+4-1)\frac{\pi}{4}\)
\(\Rightarrow \theta=(8n+1)\frac{\pi}{4}, \ \theta=(8n+3)\frac{\pi}{12}\)
\(\therefore \theta=(8n+1)\frac{\pi}{4}, \ (8n+3)\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(8n+3)\frac{\pi}{12}, \ (8n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxvii)\) \(\cos{3\theta}=\cos{2\theta}\)উত্তরঃ \(2n\pi, \ \frac{2}{5}n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{3\theta}=\cos{2\theta}\)
\(\Rightarrow 3\theta=2n\pi\pm2\theta\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=2n\pi+2\theta, \ 3\theta=2n\pi-2\theta\)
\(\Rightarrow 3\theta-2\theta=2n\pi, \ 3\theta+2\theta=2n\pi\)
\(\Rightarrow \theta=2n\pi, \ 5\theta=2n\pi\)
\(\Rightarrow \theta=2n\pi, \ \theta=\frac{2}{5}n\pi\)
\(\therefore \theta=2n\pi, \ \frac{2}{5}n\pi\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ \frac{2}{5}n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{3\theta}=\cos{2\theta}\)
\(\Rightarrow 3\theta=2n\pi\pm2\theta\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=2n\pi+2\theta, \ 3\theta=2n\pi-2\theta\)
\(\Rightarrow 3\theta-2\theta=2n\pi, \ 3\theta+2\theta=2n\pi\)
\(\Rightarrow \theta=2n\pi, \ 5\theta=2n\pi\)
\(\Rightarrow \theta=2n\pi, \ \theta=\frac{2}{5}n\pi\)
\(\therefore \theta=2n\pi, \ \frac{2}{5}n\pi\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ \frac{2}{5}n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxviii)\) \(\sin{5\theta}\cos{\theta}=\sin{6\theta}\cos{2\theta}\)উত্তরঃ \(n\pi, \ (2n+1)\frac{\pi}{14}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{5\theta}\cos{\theta}=\sin{6\theta}\cos{2\theta}\)
\(\Rightarrow 2\sin{5\theta}\cos{\theta}=2\sin{6\theta}\cos{2\theta}\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \sin{(5\theta+\theta)}+\sin{(5\theta-\theta)}=\sin{(6\theta+2\theta)}+\sin{(6\theta-2\theta)}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(\Rightarrow \sin{6\theta}+\sin{4\theta}=\sin{8\theta}+\sin{4\theta}\)
\(\Rightarrow \sin{6\theta}+\sin{4\theta}-\sin{8\theta}-\sin{4\theta}=0\)
\(\Rightarrow \sin{6\theta}-\sin{8\theta}=0\)
\(\Rightarrow \sin{8\theta}-\sin{6\theta}=0\)
\(\Rightarrow 2\cos{\frac{8\theta+6\theta}{2}}\sin{\frac{8\theta-6\theta}{2}}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \cos{\frac{14\theta}{2}}\sin{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{7\theta}\sin{\theta}=0\)
\(\Rightarrow \cos{7\theta}=0, \ \sin{\theta}=0\)
\(\Rightarrow 7\theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{14}, \ \theta=n\pi\)
\(\therefore \theta=(2n+1)\frac{\pi}{14}, \ n\pi\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi, \ (2n+1)\frac{\pi}{14}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{5\theta}\cos{\theta}=\sin{6\theta}\cos{2\theta}\)
\(\Rightarrow 2\sin{5\theta}\cos{\theta}=2\sin{6\theta}\cos{2\theta}\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \sin{(5\theta+\theta)}+\sin{(5\theta-\theta)}=\sin{(6\theta+2\theta)}+\sin{(6\theta-2\theta)}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(\Rightarrow \sin{6\theta}+\sin{4\theta}=\sin{8\theta}+\sin{4\theta}\)
\(\Rightarrow \sin{6\theta}+\sin{4\theta}-\sin{8\theta}-\sin{4\theta}=0\)
\(\Rightarrow \sin{6\theta}-\sin{8\theta}=0\)
\(\Rightarrow \sin{8\theta}-\sin{6\theta}=0\)
\(\Rightarrow 2\cos{\frac{8\theta+6\theta}{2}}\sin{\frac{8\theta-6\theta}{2}}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \cos{\frac{14\theta}{2}}\sin{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{7\theta}\sin{\theta}=0\)
\(\Rightarrow \cos{7\theta}=0, \ \sin{\theta}=0\)
\(\Rightarrow 7\theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{14}, \ \theta=n\pi\)
\(\therefore \theta=(2n+1)\frac{\pi}{14}, \ n\pi\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi, \ (2n+1)\frac{\pi}{14}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xxxix)\) \(4\sin{x}=\sec{x}\)উত্তরঃ \(\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4\sin{x}=\sec{x}\)
\(\Rightarrow 4\sin{x}=\frac{1}{\cos{x}}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow 4\sin{x}\cos{x}=1\)
\(\Rightarrow 2\times2\sin{x}\cos{x}=1\)
\(\Rightarrow 2\sin{2x}=1\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{2x}=\frac{1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 2x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(4\sin{x}=\sec{x}\)
\(\Rightarrow 4\sin{x}=\frac{1}{\cos{x}}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow 4\sin{x}\cos{x}=1\)
\(\Rightarrow 2\times2\sin{x}\cos{x}=1\)
\(\Rightarrow 2\sin{2x}=1\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{2x}=\frac{1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 2x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xL)\) \(\sqrt{2}(\cos^2{x}-\sin^2{x})=1\)উত্তরঃ \(n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{2}(\cos^2{x}-\sin^2{x})=1\)
\(\Rightarrow \sqrt{2}(\cos^2{x}-1+\cos^2{x})=1\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow \sqrt{2}(2\cos^2{x}-1)=1\)
\(\Rightarrow \sqrt{2}\cos{2x}=1\) ➜ \(\because 2\cos^2{A}-1=\cos{2A}\)
\(\Rightarrow \cos{2x}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{2x}=\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow 2x=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sqrt{2}(\cos^2{x}-\sin^2{x})=1\)
\(\Rightarrow \sqrt{2}(\cos^2{x}-1+\cos^2{x})=1\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow \sqrt{2}(2\cos^2{x}-1)=1\)
\(\Rightarrow \sqrt{2}\cos{2x}=1\) ➜ \(\because 2\cos^2{A}-1=\cos{2A}\)
\(\Rightarrow \cos{2x}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{2x}=\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow 2x=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xLi)\) \(\cos{2x}=\cos{x}\sin{x}\)উত্তরঃ \(\frac{1}{2}(n\pi+\alpha)\) যেখানে, \(\tan{\alpha}=2, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{2x}=\cos{x}\sin{x}\)
\(\Rightarrow \cos{2x}=\frac{1}{2}\times2\sin{x}\cos{x}\)
\(\Rightarrow \cos{2x}=\frac{1}{2}\sin{2x}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\frac{1}{2}\)
\(\Rightarrow \frac{\sin{2x}}{\cos{2x}}=2\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{2x}=2\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{2x}=\tan{\alpha}\) যেখানে, \(\tan{\alpha}=2\)
\(\Rightarrow 2x=n\pi+\alpha\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{1}{2}(n\pi+\alpha)\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{1}{2}(n\pi+\alpha)\) যেখানে, \(\tan{\alpha}=2, \ n\in{\mathbb{Z}}\)
\(\cos{2x}=\cos{x}\sin{x}\)
\(\Rightarrow \cos{2x}=\frac{1}{2}\times2\sin{x}\cos{x}\)
\(\Rightarrow \cos{2x}=\frac{1}{2}\sin{2x}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\frac{1}{2}\)
\(\Rightarrow \frac{\sin{2x}}{\cos{2x}}=2\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{2x}=2\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{2x}=\tan{\alpha}\) যেখানে, \(\tan{\alpha}=2\)
\(\Rightarrow 2x=n\pi+\alpha\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{1}{2}(n\pi+\alpha)\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{1}{2}(n\pi+\alpha)\) যেখানে, \(\tan{\alpha}=2, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xLii)\) \(\sin{4\theta}\cos{\theta}=\frac{1}{4}+\sin{\frac{5\theta}{2}}\cos{\frac{5\theta}{2}}\)উত্তরঃ \(\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{4\theta}\cos{\theta}=\frac{1}{4}+\sin{\frac{5\theta}{2}}\cos{\frac{5\theta}{2}}\)
\(\Rightarrow 2\sin{4\theta}\cos{\theta}=\frac{1}{2}+2\sin{\frac{5\theta}{2}}\cos{\frac{5\theta}{2}}\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \sin{(4\theta+\theta)}+\sin{(4\theta-\theta)}=\frac{1}{2}+\sin{5\theta}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{5\theta}+\sin{3\theta}=\frac{1}{2}+\sin{5\theta}\)
\(\Rightarrow \sin{3\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{3\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 3\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{4\theta}\cos{\theta}=\frac{1}{4}+\sin{\frac{5\theta}{2}}\cos{\frac{5\theta}{2}}\)
\(\Rightarrow 2\sin{4\theta}\cos{\theta}=\frac{1}{2}+2\sin{\frac{5\theta}{2}}\cos{\frac{5\theta}{2}}\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \sin{(4\theta+\theta)}+\sin{(4\theta-\theta)}=\frac{1}{2}+\sin{5\theta}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{5\theta}+\sin{3\theta}=\frac{1}{2}+\sin{5\theta}\)
\(\Rightarrow \sin{3\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{3\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 3\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xLiii)\) \(\tan{2\theta}-\tan{\theta}=\frac{1}{2}\sec{\theta}\sec{2\theta}\)উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{2\theta}-\tan{\theta}=\frac{1}{2}\sec{\theta}\sec{2\theta}\)
\(\Rightarrow \frac{\sin{2\theta}}{\cos{2\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=\frac{1}{2\cos{\theta}\cos{2\theta}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \sin{2\theta}\cos{\theta}-\cos{2\theta}\sin{\theta}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{2\theta}\cos{\theta}\) গুণ করে,
\(\Rightarrow \sin{(2\theta-\theta)}=\frac{1}{2}\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{2\theta}-\tan{\theta}=\frac{1}{2}\sec{\theta}\sec{2\theta}\)
\(\Rightarrow \frac{\sin{2\theta}}{\cos{2\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=\frac{1}{2\cos{\theta}\cos{2\theta}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \sin{2\theta}\cos{\theta}-\cos{2\theta}\sin{\theta}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{2\theta}\cos{\theta}\) গুণ করে,
\(\Rightarrow \sin{(2\theta-\theta)}=\frac{1}{2}\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.1.(xLiv)\) \(\sin{2\theta}=\cos{3\theta}\)উত্তরঃ \((4n+1)\frac{\pi}{10}, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{2\theta}=\cos{3\theta}\)
\(\Rightarrow \cos{3\theta}=\sin{2\theta}\)
\(\Rightarrow \cos{3\theta}=\cos{\left(\frac{\pi}{2}-2\theta\right)}\) ➜ \(\because \sin(x)=\cos{\left(\frac{\pi}{2}-x\right)}\)
\(\Rightarrow 3\theta=2n\pi\pm\left(\frac{\pi}{2}-2\theta\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=2n\pi+\left(\frac{\pi}{2}-2\theta\right), \ 3\theta=2n\pi-\left(\frac{\pi}{2}-2\theta\right)\)
\(\Rightarrow 3\theta=2n\pi+\frac{\pi}{2}-2\theta, \ 3\theta=2n\pi-\frac{\pi}{2}+2\theta\)
\(\Rightarrow 3\theta+2\theta=2n\pi+\frac{\pi}{2}, \ 3\theta-2\theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=2n\pi+\frac{\pi}{2}, \ \theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=(4n+1)\frac{\pi}{2}, \ \theta=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n+1)\frac{\pi}{10}, \ (4n-1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{10}, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{2\theta}=\cos{3\theta}\)
\(\Rightarrow \cos{3\theta}=\sin{2\theta}\)
\(\Rightarrow \cos{3\theta}=\cos{\left(\frac{\pi}{2}-2\theta\right)}\) ➜ \(\because \sin(x)=\cos{\left(\frac{\pi}{2}-x\right)}\)
\(\Rightarrow 3\theta=2n\pi\pm\left(\frac{\pi}{2}-2\theta\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=2n\pi+\left(\frac{\pi}{2}-2\theta\right), \ 3\theta=2n\pi-\left(\frac{\pi}{2}-2\theta\right)\)
\(\Rightarrow 3\theta=2n\pi+\frac{\pi}{2}-2\theta, \ 3\theta=2n\pi-\frac{\pi}{2}+2\theta\)
\(\Rightarrow 3\theta+2\theta=2n\pi+\frac{\pi}{2}, \ 3\theta-2\theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=2n\pi+\frac{\pi}{2}, \ \theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=(4n+1)\frac{\pi}{2}, \ \theta=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n+1)\frac{\pi}{10}, \ (4n-1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{10}, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
অধ্যায় \(7H\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
সমাধান করঃ
\(Q.2.(i)\) \(2\sin^2{\theta}=3\cos{\theta}, \ 0\lt{\theta}\lt{2\pi}\) উত্তরঃ \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
যঃ২০০৯।
\(Q.2.(ii)\) \(4\sin{\theta}\cos{\theta}=1-2\sin{\theta}+2\cos{\theta}, \ 0\lt{\theta}\lt{180^{o}}\)
উত্তরঃ \(30^{o}, \ 120^{o}\) এবং \(150^{o}\)
সিঃ২০১৩; রাঃ২০০৩; মাঃ২০০৮ ।
\(Q.2.(iii)\) \(\sin{x}-\cos{x}=0, \ 0\lt{x}\lt{\frac{\pi}{2}}\)
উত্তরঃ \(\frac{\pi}{4}\)
রাঃ২০১৩ ।
\(Q.2.(iv)\) \(\sin{x}+\cos{x}=\sqrt{2}, \ -\pi\lt{x}\lt{\pi}\)
উত্তরঃ \(\frac{\pi}{4}\)
দিঃ২০১০; বঃ২০০৬; বুয়েটঃ২০০২-২০০৩; রুয়েটঃ২০০৪-২০০৫; চুয়েটঃ২০১০-২০১১ ।
\(Q.2.(v)\) \(\sqrt{3}\sin{\theta}-\cos{\theta}=2, \ -\pi\lt{\theta}\lt{\pi}\)
উত্তরঃ \(-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
ঢাঃ২০১৭,২০১৩,২০০৯,২০০৬; যঃ২০১৫; চঃ২০১২; রাঃ২০১১,২০০৮; মাঃ২০১৩; বুয়েটঃ,রুয়েটঃ,চুয়েটঃ২০০৭-২০০৮ ।
\(Q.2.(vi)\) \(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -2\pi\lt{\theta}\lt{\pi}\)
উত্তরঃ \(-\frac{23\pi}{12}, \ -\frac{7\pi}{12}, \ \frac{\pi}{12}\)
কুঃ২০১৩,২০০৯; দিঃ২০১৩; মাঃ২০১১ ।
\(Q.2.(vii)\) \(\sqrt{2}\cos{x}-\sqrt{2}\sin{x}=1, \ -\pi\lt{x}\lt{\pi}\)
উত্তরঃ \(-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
কুঃ২০১৩,২০০৯; দিঃ২০১৩; মাঃ২০১১ ।
\(Q.2.(viii)\) \(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi}\)
উত্তরঃ \(\frac{\pi}{8}, \ \frac{3\pi}{8}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\) এবং \(\frac{\pi}{2}\)
\(Q.2.(ix)\) \(2\sin^2{x}+\sin^2{2x}=2, \ -\pi\lt{x}\lt{\pi}\)
উত্তরঃ \(\pm\frac{\pi}{2}, \ \pm\frac{\pi}{4}\) এবং \(\pm\frac{3\pi}{4}\)
মাঃ২০১৩ ।
\(Q.2.(x)\) \(\sin{\theta}-2=\cos{2\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
উত্তরঃ \(-\frac{3\pi}{2}, \ \frac{\pi}{2}\)
দিঃ২০১২; চঃ২০১০; যঃ২০০৬ ।
\(Q.2.(xi)\) \(\sec^2{\frac{x}{2}}=2\sqrt{2}\tan{\frac{x}{2}}, \ 0\lt{x}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}\)
কুঃ২০১৫,২০০৩; বঃ২০১৩; সিঃ২০০৩; বুটেক্সঃ২০০৫-২০০৬ ।
\(Q.2.(xii)\) \(1+\sqrt{3}\tan^2{\theta}=(1+\sqrt{3})\tan{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(30^{o}, \ 45^{o}, \ 210^{o}\) এবং \(225^{o}\)
বঃ২০১৪,২০০৭; যঃ২০০৫ ।
\(Q.2.(xiii)\) \(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}, \ 0\lt{\theta}\lt{\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{2\pi}{3}\) এবং \(\frac{5\pi}{6}\)
চঃ২০০৮; বুয়েটঃ২০০৮-২০০৯ ।
\(Q.2.(xiv)\) \(4\cos{x}\cos{2x}\cos{3x}=1, \ 0\lt{x}\lt{\pi}\)
উত্তরঃ \(\frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
দিঃ২০১৬,২০১৪,২০১১; রাঃ২০১৫,২০১২,২০০৯; কুঃ২০১২; সিঃ২০১১,২০০৮; ঢাঃ২০০৭; মাঃ২০১৪; চুয়েটঃ২০০৮-২০০৯; কুয়েটঃ২০০৬-২০০৭।
\(Q.2.(xv)\) \(\cos{9x}\cos{7x}=\cos{5x}\cos{3x}, \ -\frac{\pi}{4}\lt{x}\lt{\frac{\pi}{4}}\)
উত্তরঃ \(0, \ \pm\frac{\pi}{12}, \ \pm\frac{\pi}{6}\)
ঢাঃ২০১২ ।
সমাধান করঃ
\(Q.2.(xvi)\) \(2\sin{x}\sin{3x}=1, \ 0\lt{x}\lt{2\pi}\)উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
ঢাঃ২০১৯,২০১৪; যঃ২০১৩,২০০৮; দিঃ২০১২; রাঃ২০১০; চঃ,সিঃ২০০৯ ।
\(Q.2.(xvii)\) \(4(\cos^2{x}+\sin{x})=5, \ 0\lt{x}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{5\pi}{6}\)
\(Q.2.(xviii)\) \(2\sin^2{x}-5\cos{x}+1=0, \ 0\lt{x}\lt{360^{o}}\)
উত্তরঃ \(60^{o}, \ 300^{o}\)
\(Q.2.(xix)\) \(3\tan^2{x}-4\sqrt{3}\sec{x}+7=0, \ 0\lt{x}\lt{360^{o}}\)
উত্তরঃ \(30^{o}, \ 330^{o}\)
\(Q.2.(xx)\) \(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
উত্তরঃ \(\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) এবং \(\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(Q.2.(xxi)\) \(\cot{\theta}-\tan{\theta}=2, \ 0\le{\theta}\le{\pi}\)
উত্তরঃ \(\frac{\pi}{8}, \ \frac{5\pi}{8}\)
\(Q.2.(xxii)\) \(\cos{7x}=\cos{3x}+\sin{5x}, \ -\frac{\pi}{2}\le{x}\le{\frac{\pi}{2}}\)
উত্তরঃ \(-\frac{5\pi}{12}, \ -\frac{2\pi}{5}, \ -\frac{\pi}{5}, \ -\frac{\pi}{12}, \ 0, \ \frac{\pi}{5}\) এবং \(\frac{2\pi}{5}\)
\(Q.2.(xxiii)\) \(\cos{2x}+\cos{x}+1=0, \ 0\le{x}\le{360^{o}}\)
উত্তরঃ \(90^{o}, \ 120^{o}, \ 240^{o}\) এবং \(270^{o}\)
\(Q.2.(xxiv)\) \(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{360^{o}}\)
উত্তরঃ \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 198^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\) এবং \(342^{o}\)
ঢাঃ২০০৮; কুঃ২০১১; চুয়েটঃ২০০৩-২০০৪ ।
\(Q.2.(xxv)\) \(2(\sin{\theta}\cos{\theta}+\sqrt{3})=\sqrt{3}\cos{\theta}+4\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(60^{o}, \ 120^{o}\)
\(Q.2.(xxvi)\) \(3\tan^2{\theta}+1=\frac{2\sqrt{3}}{\cot{\theta}}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(30^{o}, \ 210^{o}\)
\(Q.2.(xxvii)\) \(\tan^2{\theta}+\sec{\theta}=-1, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(180^{o}\)
\(Q.2.(xxviii)\) \(\cot^2{\theta}-2\sqrt{2} cosec \ {\theta}+3=0, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(45^{o}, \ 135^{o}\)
\(Q.2.(xxix)\) \(1-2\sin{\theta}-2\cos{\theta}+\cot{\theta}=0, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(30^{o}, \ 135^{o}, \ 150^{o}\) এবং \(315^{o}\)
চঃ২০১৬ ।
\(Q.2.(xxx)\) \((2+\sqrt{3})\cos{\theta}=1-\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(90^{o}, \ 300^{o}\)
\(Q.2.(xxxi)\) \(4\cot{2\theta}=\cot^2{\theta}-\tan^2{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(45^{o}, \ 135^{o}, \ 225^{o}\) এবং \(315^{o}\)
\(Q.2.(xxxii)\) \(1+\cos{x}+\cos{2x}=0, \ 0\lt{x}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{2}, \ \frac{2\pi}{3}, \ \frac{3\pi}{2}\) এবং \(\frac{4\pi}{3}\)
\(Q.2.(xxxiii)\) \(4 \ cosec^2{\theta}-7\cot{\theta} \ cosec \ {\theta}-2=0, \ 0\lt{\theta}\lt{\frac{\pi}{2}}\)
উত্তরঃ \(\cos^{-1}{(0.314)}\)
সমাধান করঃ
\(Q.2.(i)\) \(2\sin^2{\theta}=3\cos{\theta}, \ 0\lt{\theta}\lt{2\pi}\) উত্তরঃ \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
যঃ২০০৯।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin^2{\theta}=3\cos{\theta}, \ 0\lt{\theta}\lt{2\pi}\)
\(\Rightarrow 2(1-\cos^2{\theta})=3\cos{\theta}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-2\cos^2{\theta}-3\cos{\theta}=0\)
\(\Rightarrow 2\cos^2{\theta}+3\cos{\theta}-2=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos^2{\theta}+4\cos{\theta}-\cos{\theta}-2=0\)
\(\Rightarrow 2\cos{\theta}(\cos{\theta}+2)-1(\cos{\theta}+2)=0\)
\(\Rightarrow (\cos{\theta}+2)(2\cos{\theta}-1)=0\)
\(\Rightarrow \cos{\theta}+2\ne{0}, \ 2\cos{\theta}-1=0\) ➜ \(\because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{3}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
\(2\sin^2{\theta}=3\cos{\theta}, \ 0\lt{\theta}\lt{2\pi}\)
\(\Rightarrow 2(1-\cos^2{\theta})=3\cos{\theta}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-2\cos^2{\theta}-3\cos{\theta}=0\)
\(\Rightarrow 2\cos^2{\theta}+3\cos{\theta}-2=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos^2{\theta}+4\cos{\theta}-\cos{\theta}-2=0\)
\(\Rightarrow 2\cos{\theta}(\cos{\theta}+2)-1(\cos{\theta}+2)=0\)
\(\Rightarrow (\cos{\theta}+2)(2\cos{\theta}-1)=0\)
\(\Rightarrow \cos{\theta}+2\ne{0}, \ 2\cos{\theta}-1=0\) ➜ \(\because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{3}, \ -\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{3}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\frac{\pi}{3}, \ \frac{5\pi}{3}\)
সমাধান করঃ
\(Q.2.(ii)\) \(4\sin{\theta}\cos{\theta}=1-2\sin{\theta}+2\cos{\theta}, \ 0\lt{\theta}\lt{180^{o}}\) উত্তরঃ \(30^{o}, \ 120^{o}, \ 150^{o}\)
সিঃ২০১৩; রাঃ২০০৩; মাঃ২০০৮ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4\sin{\theta}\cos{\theta}=1-2\sin{\theta}+2\cos{\theta}, \ 0\lt{\theta}\lt{180^{o}}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}-2\cos{\theta}-1=0\)
\(\Rightarrow 2\sin{\theta}(2\cos{\theta}+1)-1(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}-1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\left(\pi-\frac{\pi}{3}\right), \ \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{3\pi-\pi}{3}, \ \theta=n\pi+(-1)^n\frac{\pi}{6}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{\pi}{6}\)
\(\therefore \theta=2n\pi\pm120^{o}, \ n\pi+(-1)^n30^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm120^{o}, \ n\pi+(-1)^n30^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=30^{o}, \ 120^{o}\)
যখন, \(n=1,\) \(\theta=2\pi+120^{o}, \ 2\pi-120^{o}, \ n\pi-30^{o}\)
\(\Rightarrow \theta=360^{o}+120^{o}, \ 360^{o}-120^{o}, \ 180^{o}-30^{o}\)
\(\Rightarrow \theta=480^{o}, \ 240^{o}, \ 150^{o}\) শেষ মানটি গ্রহনযোগ্য।
\(\therefore \theta=150^{o}\)
যখন, \(n=-1,\) \(\theta=-2\pi+120^{o}, \ -2\pi-120^{o}, \ -n\pi-30^{o}\)
\(\Rightarrow \theta=-360^{o}+120^{o}, \ -360^{o}-120^{o}, \ -180^{o}-30^{o}\)
\(\Rightarrow \theta=-240^{o}, \ -480^{o}, \ -210^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{\theta}\lt{180^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=30^{o}, \ 120^{o}, \ 150^{o}\)
\(4\sin{\theta}\cos{\theta}=1-2\sin{\theta}+2\cos{\theta}, \ 0\lt{\theta}\lt{180^{o}}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}-2\cos{\theta}-1=0\)
\(\Rightarrow 2\sin{\theta}(2\cos{\theta}+1)-1(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}-1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\left(\pi-\frac{\pi}{3}\right), \ \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{3\pi-\pi}{3}, \ \theta=n\pi+(-1)^n\frac{\pi}{6}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{\pi}{6}\)
\(\therefore \theta=2n\pi\pm120^{o}, \ n\pi+(-1)^n30^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm120^{o}, \ n\pi+(-1)^n30^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{180^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=120^{o}, \ -120^{o}, \ 30^{o}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=30^{o}, \ 120^{o}\)
যখন, \(n=1,\) \(\theta=2\pi+120^{o}, \ 2\pi-120^{o}, \ n\pi-30^{o}\)
\(\Rightarrow \theta=360^{o}+120^{o}, \ 360^{o}-120^{o}, \ 180^{o}-30^{o}\)
\(\Rightarrow \theta=480^{o}, \ 240^{o}, \ 150^{o}\) শেষ মানটি গ্রহনযোগ্য।
\(\therefore \theta=150^{o}\)
যখন, \(n=-1,\) \(\theta=-2\pi+120^{o}, \ -2\pi-120^{o}, \ -n\pi-30^{o}\)
\(\Rightarrow \theta=-360^{o}+120^{o}, \ -360^{o}-120^{o}, \ -180^{o}-30^{o}\)
\(\Rightarrow \theta=-240^{o}, \ -480^{o}, \ -210^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{\theta}\lt{180^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=30^{o}, \ 120^{o}, \ 150^{o}\)
সমাধান করঃ
\(Q.2.(iii)\) \(\sin{x}-\cos{x}=0, \ 0\lt{x}\lt{\frac{\pi}{2}}\) উত্তরঃ \(\frac{\pi}{4}\)
রাঃ২০১৩ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}-\cos{x}=0, \ 0\lt{x}\lt{\frac{\pi}{2}}\)
\(\Rightarrow \sin{x}=\cos{x}\)
\(\Rightarrow \frac{\sin{x}}{\cos{x}}=1\) ➜ উভয় পার্শে \(\cos{x}\) ভাগ করে,
\(\Rightarrow \tan{x}=1\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\therefore x=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}\)
\(\Rightarrow x=\frac{5\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}\)
\(\Rightarrow x=-\frac{3\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{\frac{\pi}{2}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}\)
\(\sin{x}-\cos{x}=0, \ 0\lt{x}\lt{\frac{\pi}{2}}\)
\(\Rightarrow \sin{x}=\cos{x}\)
\(\Rightarrow \frac{\sin{x}}{\cos{x}}=1\) ➜ উভয় পার্শে \(\cos{x}\) ভাগ করে,
\(\Rightarrow \tan{x}=1\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\therefore x=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{\frac{\pi}{2}}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}\) মানটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}\)
\(\Rightarrow x=\frac{5\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}\)
\(\Rightarrow x=-\frac{3\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{\frac{\pi}{2}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}\)
সমাধান করঃ
\(Q.2.(iv)\) \(\sin{x}+\cos{x}=\sqrt{2}, \ -\pi\lt{x}\lt{\pi}\) উত্তরঃ \(\frac{\pi}{4}\)
দিঃ২০১০; বঃ২০০৬; বুয়েটঃ২০০২-২০০৩; রুয়েটঃ২০০৪-২০০৫; চুয়েটঃ২০১০-২০১১ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}+\cos{x}=\sqrt{2}, \ -\pi\lt{x}\lt{\pi}\)
\(\Rightarrow \cos{x}+\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{1}{\sqrt{2}}+\sin{x}\frac{1}{\sqrt{2}}=1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{4}}+\sin{x}\sin{\frac{\pi}{4}}=1\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{4}\right)}=1\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow x-\frac{\pi}{4}=2n\pi\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi+\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{4}\)
\(\Rightarrow x=\frac{9\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{4}\)
\(\Rightarrow x=-\frac{7\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -\pi\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}\)
\(\sin{x}+\cos{x}=\sqrt{2}, \ -\pi\lt{x}\lt{\pi}\)
\(\Rightarrow \cos{x}+\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{1}{\sqrt{2}}+\sin{x}\frac{1}{\sqrt{2}}=1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{4}}+\sin{x}\sin{\frac{\pi}{4}}=1\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{4}\right)}=1\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow x-\frac{\pi}{4}=2n\pi\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi+\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-\pi\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}\) মানটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{4}\)
\(\Rightarrow x=\frac{9\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{4}\)
\(\Rightarrow x=-\frac{7\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -\pi\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}\)
সমাধান করঃ
\(Q.2.(v)\) \(\sqrt{3}\sin{\theta}-\cos{\theta}=2, \ -\pi\lt{\theta}\lt{\pi}\) উত্তরঃ \(-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
ঢাঃ২০১৭,২০১৩,২০০৯,২০০৬; যঃ২০১৫; চঃ২০১২; রাঃ২০১১,২০০৮; মাঃ২০১৩; বুয়েটঃ,রুয়েটঃ,চুয়েটঃ২০০৭-২০০৮ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{3}\sin{\theta}-\cos{\theta}=2, \ -\pi\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{\theta}-\sqrt{3}\sin{\theta}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{\theta}\frac{1}{2}-\sin{\theta}\frac{\sqrt{3}}{2}=-1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}-\sin{\theta}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \theta+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\{3(2n+1)-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=\{6n+3-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=(6n+2)\frac{\pi}{3}\)
\(\therefore \theta=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(\theta=4\frac{2\pi}{3}\)
\(\Rightarrow \theta=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\frac{2\pi}{3}\)
\(\Rightarrow \theta=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{4\pi}{3}\)
\(\therefore -\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(\sqrt{3}\sin{\theta}-\cos{\theta}=2, \ -\pi\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{\theta}-\sqrt{3}\sin{\theta}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{\theta}\frac{1}{2}-\sin{\theta}\frac{\sqrt{3}}{2}=-1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}-\sin{\theta}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \theta+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\{3(2n+1)-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=\{6n+3-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=(6n+2)\frac{\pi}{3}\)
\(\therefore \theta=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{2\pi}{3}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(\theta=4\frac{2\pi}{3}\)
\(\Rightarrow \theta=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\frac{2\pi}{3}\)
\(\Rightarrow \theta=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{4\pi}{3}\)
\(\therefore -\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
সমাধান করঃ
\(Q.2.(vi)\) \(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -2\pi\lt{\theta}\lt{\pi}\) উত্তরঃ \(-\frac{23\pi}{12}, \ -\frac{7\pi}{12}, \ \frac{\pi}{12}\)
কুঃ২০১৩,২০০৯; দিঃ২০১৩; মাঃ২০১১ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -2\pi\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}\times\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta+\frac{\pi}{4}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{3}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{4\pi-3\pi}{12}, \ 2n\pi-\frac{4\pi+3\pi}{12}\)
\(\therefore \theta=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{12}, \ -\frac{7\pi}{12}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{\pi}{12}, \ 2\pi-\frac{7\pi}{12}\)
\(\Rightarrow \theta=\frac{25\pi}{12}, \ \frac{17\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{\pi}{12}, \ -2\pi-\frac{7\pi}{12}\)
\(\Rightarrow \theta=-\frac{23\pi}{12}, \ -\frac{31\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{23\pi}{12}\)
\(\therefore -2\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{23\pi}{12}, \ -\frac{7\pi}{12}, \ \frac{\pi}{12}\)
\(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -2\pi\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}\times\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta+\frac{\pi}{4}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{3}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{4\pi-3\pi}{12}, \ 2n\pi-\frac{4\pi+3\pi}{12}\)
\(\therefore \theta=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{12}, \ -\frac{7\pi}{12}\) মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{12}, \ -\frac{7\pi}{12}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{\pi}{12}, \ 2\pi-\frac{7\pi}{12}\)
\(\Rightarrow \theta=\frac{25\pi}{12}, \ \frac{17\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{\pi}{12}, \ -2\pi-\frac{7\pi}{12}\)
\(\Rightarrow \theta=-\frac{23\pi}{12}, \ -\frac{31\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{23\pi}{12}\)
\(\therefore -2\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{23\pi}{12}, \ -\frac{7\pi}{12}, \ \frac{\pi}{12}\)
সমাধান করঃ
\(Q.2.(vii)\) \(\sqrt{2}\cos{x}-\sqrt{2}\sin{x}=1, \ -\pi\lt{x}\lt{\pi}\) উত্তরঃ \(-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
কুঃ২০১৩,২০০৯; দিঃ২০১৩; মাঃ২০১১ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{2}\cos{x}-\sqrt{2}\sin{x}=1, \ -\pi\lt{x}\lt{\pi}\)
\(\Rightarrow \cos{x}\frac{\sqrt{2}}{2}-\sin{x}\frac{\sqrt{2}}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}-\sin{x}\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{1}{2}\)
\(\Rightarrow \cos{x}\frac{1}{\sqrt{2}}-\sin{x}\frac{1}{\sqrt{2}}=\frac{1}{2}\)
\(\Rightarrow \cos{x}\cos{\frac{\pi}{4}}-\sin{x}\sin{\frac{\pi}{4}}=\frac{1}{2}\)➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x+\frac{\pi}{4}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow x=2n\pi+\frac{4\pi-3\pi}{12}, \ 2n\pi-\frac{4\pi+3\pi}{12}\)
\(\therefore x=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{12}, \ -\frac{7\pi}{12}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{12}, \ 2\pi-\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{25\pi}{12}, \ \frac{17\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{12}, \ -2\pi-\frac{7\pi}{12}\)
\(\Rightarrow x=-\frac{23\pi}{12}, \ -\frac{31\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore -\pi\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
\(\sqrt{2}\cos{x}-\sqrt{2}\sin{x}=1, \ -\pi\lt{x}\lt{\pi}\)
\(\Rightarrow \cos{x}\frac{\sqrt{2}}{2}-\sin{x}\frac{\sqrt{2}}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}-\sin{x}\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{1}{2}\)
\(\Rightarrow \cos{x}\frac{1}{\sqrt{2}}-\sin{x}\frac{1}{\sqrt{2}}=\frac{1}{2}\)
\(\Rightarrow \cos{x}\cos{\frac{\pi}{4}}-\sin{x}\sin{\frac{\pi}{4}}=\frac{1}{2}\)➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x+\frac{\pi}{4}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{3}-\frac{\pi}{4}\)
\(\Rightarrow x=2n\pi+\frac{4\pi-3\pi}{12}, \ 2n\pi-\frac{4\pi+3\pi}{12}\)
\(\therefore x=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{12}, \ 2n\pi-\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-\pi\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{12}, \ -\frac{7\pi}{12}\) মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{12}, \ -\frac{7\pi}{12}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{12}, \ 2\pi-\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{25\pi}{12}, \ \frac{17\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{12}, \ -2\pi-\frac{7\pi}{12}\)
\(\Rightarrow x=-\frac{23\pi}{12}, \ -\frac{31\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore -\pi\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
সমাধান করঃ
\(Q.2.(viii)\) \(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi}\)উত্তরঃ \(\frac{\pi}{8}, \ \frac{3\pi}{8}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\) এবং \(\frac{\pi}{2}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{5\theta}+\cos{\theta}+\cos{7\theta}+\cos{3\theta}=0\)
\(\Rightarrow 2\cos{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}+2\cos{\frac{7\theta+3\theta}{2}}\cos{\frac{7\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}+2\cos{\frac{10\theta}{2}}\cos{\frac{4\theta}{2}}=0\)
\(\Rightarrow 2\cos{3\theta}\cos{2\theta}+2\cos{5\theta}\cos{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}(\cos{3\theta}+\cos{5\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}+\cos{5\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{5\theta+3\theta}{2}}\cos{\frac{5\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{8\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 4\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{8}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{8}, \ \frac{\pi}{4}, \ \frac{\pi}{2}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{8}, \ \frac{3\pi}{4}, \ \frac{3\pi}{2}\) প্রথম মান দুইটি গ্রহনযোগ্য
\(\therefore \theta=\frac{3\pi}{8}, \ \frac{3\pi}{4}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{8}, \ -\frac{\pi}{4}, \ -\frac{\pi}{2}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{8}, \ \frac{5\pi}{4}, \ \frac{5\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore \theta=\frac{5\pi}{8}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{8}, \ \frac{7\pi}{4}, \ \frac{7\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore \theta=\frac{7\pi}{8}\)
\(\therefore 0\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{8}, \ \frac{3\pi}{8}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\) এবং \(\frac{\pi}{2}\)
\(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{5\theta}+\cos{\theta}+\cos{7\theta}+\cos{3\theta}=0\)
\(\Rightarrow 2\cos{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}+2\cos{\frac{7\theta+3\theta}{2}}\cos{\frac{7\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}+2\cos{\frac{10\theta}{2}}\cos{\frac{4\theta}{2}}=0\)
\(\Rightarrow 2\cos{3\theta}\cos{2\theta}+2\cos{5\theta}\cos{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}(\cos{3\theta}+\cos{5\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}+\cos{5\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{5\theta+3\theta}{2}}\cos{\frac{5\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{8\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 4\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{8}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{8}, \ \frac{\pi}{4}, \ \frac{\pi}{2}\) মানগুলি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{8}, \ \frac{\pi}{4}, \ \frac{\pi}{2}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{8}, \ \frac{3\pi}{4}, \ \frac{3\pi}{2}\) প্রথম মান দুইটি গ্রহনযোগ্য
\(\therefore \theta=\frac{3\pi}{8}, \ \frac{3\pi}{4}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{8}, \ -\frac{\pi}{4}, \ -\frac{\pi}{2}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{8}, \ \frac{5\pi}{4}, \ \frac{5\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore \theta=\frac{5\pi}{8}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{8}, \ \frac{7\pi}{4}, \ \frac{7\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore \theta=\frac{7\pi}{8}\)
\(\therefore 0\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{8}, \ \frac{3\pi}{8}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\) এবং \(\frac{\pi}{2}\)
সমাধান করঃ
\(Q.2.(ix)\) \(2\sin^2{x}+\sin^2{2x}=2, \ -\pi\lt{x}\lt{\pi}\)উত্তরঃ \(\pm\frac{\pi}{2}, \ \pm\frac{\pi}{4}\) এবং \(\pm\frac{3\pi}{4}\)
মাঃ২০১৩ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin^2{x}+\sin^2{2x}=2, \ -\pi\lt{x}\lt{\pi}\)
\(\Rightarrow 1-\cos{2x}+1-\cos^2{2x}=2\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
এবং \(\sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-\cos{2x}-\cos^2{2x}=2\)
\(\Rightarrow -\cos{2x}-\cos^2{2x}=0\)
\(\Rightarrow -\cos{2x}(1+\cos{2x})=0\)
\(\Rightarrow \cos{2x}=0, \ 1+\cos{2x}=0\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=-1\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ 2x=(2n+1)\pi\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ x=(2n+1)\frac{\pi}{2}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}, \ \frac{\pi}{2}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{4}, \ \frac{3\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore x=\frac{3\pi}{4}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{4}, \ -\frac{\pi}{2}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{4}, \ -\frac{\pi}{2}\)
যখন, \(n=2,\) \(x=\frac{5\pi}{4}, \ \frac{5\pi}{2}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(x=-\frac{3\pi}{4}, \ -\frac{3\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore x=-\frac{3\pi}{4}\)
যখন, \(n=3,\) \(x=\frac{7\pi}{4}, \ \frac{7\pi}{2}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore -\pi\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\pm\frac{\pi}{2}, \ \pm\frac{\pi}{4}\) এবং \(\pm\frac{3\pi}{4}\)
\(2\sin^2{x}+\sin^2{2x}=2, \ -\pi\lt{x}\lt{\pi}\)
\(\Rightarrow 1-\cos{2x}+1-\cos^2{2x}=2\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
এবং \(\sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-\cos{2x}-\cos^2{2x}=2\)
\(\Rightarrow -\cos{2x}-\cos^2{2x}=0\)
\(\Rightarrow -\cos{2x}(1+\cos{2x})=0\)
\(\Rightarrow \cos{2x}=0, \ 1+\cos{2x}=0\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=-1\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ 2x=(2n+1)\pi\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ x=(2n+1)\frac{\pi}{2}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-\pi\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}, \ \frac{\pi}{2}\) মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}, \ \frac{\pi}{2}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{4}, \ \frac{3\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore x=\frac{3\pi}{4}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{4}, \ -\frac{\pi}{2}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{4}, \ -\frac{\pi}{2}\)
যখন, \(n=2,\) \(x=\frac{5\pi}{4}, \ \frac{5\pi}{2}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(x=-\frac{3\pi}{4}, \ -\frac{3\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য
\(\therefore x=-\frac{3\pi}{4}\)
যখন, \(n=3,\) \(x=\frac{7\pi}{4}, \ \frac{7\pi}{2}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore -\pi\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\pm\frac{\pi}{2}, \ \pm\frac{\pi}{4}\) এবং \(\pm\frac{3\pi}{4}\)
সমাধান করঃ
\(Q.2.(x)\) \(\sin{\theta}-2=\cos{2\theta}, \ -2\pi\le{\theta}\le{2\pi}\)উত্তরঃ \(-\frac{3\pi}{2}, \ \frac{\pi}{2}\)
দিঃ২০১২; চঃ২০১০; যঃ২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{\theta}-2=\cos{2\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow \sin{\theta}-2=1-2\sin^2{\theta}\) ➜ \(\because \cos{2A}=1-2\sin^2{A}\)
\(\Rightarrow 2\sin^2{\theta}+\sin{\theta}-2-1=0\)
\(\Rightarrow 2\sin^2{\theta}+3\sin{\theta}-2\sin{\theta}-3=0\)
\(\Rightarrow \sin{\theta}(2\sin{\theta}+3)-1(2\sin{\theta}+3)=0\)
\(\Rightarrow (2\sin{\theta}+3)(\sin{\theta}-1)=0\)
\(\Rightarrow 2\sin{\theta}+3\ne{0}, \ \sin{\theta}-1=0\) ➜ \(\because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sin{\theta}=1\)
\(\therefore \theta=(4n+1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=1\)
\(\Rightarrow A=(4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{2}\)
যখন, \(n=1,\) \(\theta=\frac{5\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-\frac{3\pi}{2}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{3\pi}{2}\)
যখন, \(n=2,\) \(\theta=\frac{9\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(\theta=-\frac{7\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{3\pi}{2}, \ \frac{\pi}{2}\)
\(\sin{\theta}-2=\cos{2\theta}, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow \sin{\theta}-2=1-2\sin^2{\theta}\) ➜ \(\because \cos{2A}=1-2\sin^2{A}\)
\(\Rightarrow 2\sin^2{\theta}+\sin{\theta}-2-1=0\)
\(\Rightarrow 2\sin^2{\theta}+3\sin{\theta}-2\sin{\theta}-3=0\)
\(\Rightarrow \sin{\theta}(2\sin{\theta}+3)-1(2\sin{\theta}+3)=0\)
\(\Rightarrow (2\sin{\theta}+3)(\sin{\theta}-1)=0\)
\(\Rightarrow 2\sin{\theta}+3\ne{0}, \ \sin{\theta}-1=0\) ➜ \(\because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sin{\theta}=1\)
\(\therefore \theta=(4n+1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=1\)
\(\Rightarrow A=(4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{2}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{2}\)
যখন, \(n=1,\) \(\theta=\frac{5\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-\frac{3\pi}{2}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{3\pi}{2}\)
যখন, \(n=2,\) \(\theta=\frac{9\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(\theta=-\frac{7\pi}{2}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{3\pi}{2}, \ \frac{\pi}{2}\)
সমাধান করঃ
\(Q.2.(xi)\) \(\sec^2{\frac{x}{2}}=2\sqrt{2}\tan{\frac{x}{2}}, \ 0\lt{x}\lt{2\pi}\)উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}\)
কুঃ২০১৫,২০০৩; বঃ২০১৩; সিঃ২০০৩; বুটেক্সঃ২০০৫-২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sec^2{\frac{x}{2}}=2\sqrt{2}\tan{\frac{x}{2}}, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow \frac{1}{\cos^2{\frac{x}{2}}}=2\sqrt{2}\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{1}{\cos{\frac{x}{2}}}=2\sqrt{2}\sin{\frac{x}{2}}\) ➜ \(\because \cos{\frac{x}{2}}\ne{0}\)
\(\Rightarrow 2\sqrt{2}\sin{\frac{x}{2}}\cos{\frac{x}{2}}=1\)
\(\Rightarrow \sqrt{2}\times2\sin{\frac{x}{2}}\cos{\frac{x}{2}}=1\) ➜ \(\because 2\sin{\frac{A}{2}}\cos{\frac{A}{2}}=\sin{A}\)
\(\Rightarrow \sqrt{2}\sin{x}=1\)
\(\Rightarrow \sin{x}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}\)
যখন, \(n=1,\) \(x=\pi-\frac{\pi}{4}\)
\(\Rightarrow x=\frac{3\pi}{4}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{4}\)
যখন, \(n=-1,\) \(x=-\pi-\frac{\pi}{4}\)
\(\Rightarrow x=-\frac{5\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}\)
\(\Rightarrow x=\frac{9\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{3\pi}{4}\)
\(\sec^2{\frac{x}{2}}=2\sqrt{2}\tan{\frac{x}{2}}, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow \frac{1}{\cos^2{\frac{x}{2}}}=2\sqrt{2}\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{1}{\cos{\frac{x}{2}}}=2\sqrt{2}\sin{\frac{x}{2}}\) ➜ \(\because \cos{\frac{x}{2}}\ne{0}\)
\(\Rightarrow 2\sqrt{2}\sin{\frac{x}{2}}\cos{\frac{x}{2}}=1\)
\(\Rightarrow \sqrt{2}\times2\sin{\frac{x}{2}}\cos{\frac{x}{2}}=1\) ➜ \(\because 2\sin{\frac{A}{2}}\cos{\frac{A}{2}}=\sin{A}\)
\(\Rightarrow \sqrt{2}\sin{x}=1\)
\(\Rightarrow \sin{x}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}\) মানটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}\)
যখন, \(n=1,\) \(x=\pi-\frac{\pi}{4}\)
\(\Rightarrow x=\frac{3\pi}{4}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{4}\)
যখন, \(n=-1,\) \(x=-\pi-\frac{\pi}{4}\)
\(\Rightarrow x=-\frac{5\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}\)
\(\Rightarrow x=\frac{9\pi}{4}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{3\pi}{4}\)
সমাধান করঃ
\(Q.2.(xii)\) \(1+\sqrt{3}\tan^2{\theta}=(1+\sqrt{3})\tan{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)উত্তরঃ \(30^{o}, \ 45^{o}, \ 210^{o}\) এবং \(225^{o}\)
বঃ২০১৪,২০০৭; যঃ২০০৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(1+\sqrt{3}\tan^2{\theta}=(1+\sqrt{3})\tan{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 1+\sqrt{3}\tan^2{\theta}=\tan{\theta}+\sqrt{3}\tan{\theta}\)
\(\Rightarrow \tan{\theta}(\sqrt{3}\tan{\theta}-1)-1(\sqrt{3}\tan{\theta}-1)=0\)
\(\Rightarrow (\sqrt{3}\tan{\theta}-1)(\tan{\theta}-1)=0\)
\(\Rightarrow \sqrt{3}\tan{\theta}-1=0, \ \tan{\theta}-1=0\)
\(\Rightarrow \sqrt{3}\tan{\theta}=1, \ \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\frac{1}{\sqrt{3}}, \ \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{6}}, \ \tan{\theta}=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{6}, \ \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+30^{o}, \ n\pi+45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+30^{o}, \ n\pi+45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=30^{o}, \ 45^{o}\)
যখন, \(n=1,\) \(\theta=180^{o}+30^{o}, \ 180^{o}+45^{o}\)
\(\Rightarrow \theta=210^{o}, \ 225^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=210^{o}, \ 225^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}+30^{o}, \ -180^{o}+45^{o}\)
\(\Rightarrow \theta=-150^{o}, \ -135^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=360^{o}+30^{o}, \ 360^{o}+45^{o}\)
\(\Rightarrow \theta=390^{o}, \ 405^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore 0\lt{\theta}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=30^{o}, \ 45^{o}, \ 210^{o}\) এবং \(225^{o}\)
\(1+\sqrt{3}\tan^2{\theta}=(1+\sqrt{3})\tan{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 1+\sqrt{3}\tan^2{\theta}=\tan{\theta}+\sqrt{3}\tan{\theta}\)
\(\Rightarrow \tan{\theta}(\sqrt{3}\tan{\theta}-1)-1(\sqrt{3}\tan{\theta}-1)=0\)
\(\Rightarrow (\sqrt{3}\tan{\theta}-1)(\tan{\theta}-1)=0\)
\(\Rightarrow \sqrt{3}\tan{\theta}-1=0, \ \tan{\theta}-1=0\)
\(\Rightarrow \sqrt{3}\tan{\theta}=1, \ \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\frac{1}{\sqrt{3}}, \ \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{6}}, \ \tan{\theta}=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{6}, \ \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+30^{o}, \ n\pi+45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+30^{o}, \ n\pi+45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=30^{o}, \ 45^{o}\) মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=30^{o}, \ 45^{o}\)
যখন, \(n=1,\) \(\theta=180^{o}+30^{o}, \ 180^{o}+45^{o}\)
\(\Rightarrow \theta=210^{o}, \ 225^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=210^{o}, \ 225^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}+30^{o}, \ -180^{o}+45^{o}\)
\(\Rightarrow \theta=-150^{o}, \ -135^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=360^{o}+30^{o}, \ 360^{o}+45^{o}\)
\(\Rightarrow \theta=390^{o}, \ 405^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore 0\lt{\theta}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=30^{o}, \ 45^{o}, \ 210^{o}\) এবং \(225^{o}\)
সমাধান করঃ
\(Q.2.(xiii)\) \(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}, \ 0\lt{\theta}\lt{\pi}\)উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{2\pi}{3}\) এবং \(\frac{5\pi}{6}\)
চঃ২০০৮; বুয়েটঃ২০০৮-২০০৯ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \sin{3\theta}+\sin{\theta}+\sin{2\theta}=1+\cos{2\theta}+\cos{\theta}\)
\(\Rightarrow 2\sin{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}+\sin{2\theta}=2\cos^2{\theta}+\cos{\theta}\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow 2\sin{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\sin{2\theta}\cos{\theta}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow \sin{2\theta}(2\cos{\theta}+1)-\cos{\theta}(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(\sin{2\theta}-\cos{\theta})=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \sin{2\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ \cos{\theta}=0, \ 2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \cos{\theta}=0, \ \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\cos{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ (2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{2\pi}{3}, \ (2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{2\pi}{3}, \ \frac{\pi}{2}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{2\pi}{3}, \ 2\pi-\frac{2\pi}{3}, \ \frac{3\pi}{2}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{8\pi}{3}, \ \frac{4\pi}{3}, \ \frac{3\pi}{2}, \ \frac{5\pi}{6}\) শেষ মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{2\pi}{3}, \ -2\pi-\frac{2\pi}{3}, \ -\frac{\pi}{2}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{4\pi}{3}, \ -\frac{8\pi}{3}, \ -\frac{\pi}{2}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{2\pi}{3}, \ \frac{\pi}{2}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
\(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \sin{3\theta}+\sin{\theta}+\sin{2\theta}=1+\cos{2\theta}+\cos{\theta}\)
\(\Rightarrow 2\sin{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}+\sin{2\theta}=2\cos^2{\theta}+\cos{\theta}\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow 2\sin{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\sin{2\theta}\cos{\theta}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow \sin{2\theta}(2\cos{\theta}+1)-\cos{\theta}(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(\sin{2\theta}-\cos{\theta})=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \sin{2\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ \cos{\theta}=0, \ 2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \cos{\theta}=0, \ \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\cos{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ (2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{2\pi}{3}, \ (2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{2\pi}{3}, \ \frac{\pi}{2}, \ \frac{\pi}{6}\) সবগুলি মান গ্রহনযোগ্য।\(\therefore \theta=\frac{2\pi}{3}, \ \frac{\pi}{2}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=2\pi+\frac{2\pi}{3}, \ 2\pi-\frac{2\pi}{3}, \ \frac{3\pi}{2}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{8\pi}{3}, \ \frac{4\pi}{3}, \ \frac{3\pi}{2}, \ \frac{5\pi}{6}\) শেষ মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-2\pi+\frac{2\pi}{3}, \ -2\pi-\frac{2\pi}{3}, \ -\frac{\pi}{2}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{4\pi}{3}, \ -\frac{8\pi}{3}, \ -\frac{\pi}{2}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{2\pi}{3}, \ \frac{\pi}{2}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
সমাধান করঃ
\(Q.2.(xiv)\) \(4\cos{x}\cos{2x}\cos{3x}=1, \ 0\lt{x}\lt{\pi}\)উত্তরঃ \(\frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
দিঃ২০১৬,২০১৪,২০১১; রাঃ২০১৫,২০১২,২০০৯; কুঃ২০১২; সিঃ২০১১,২০০৮; ঢাঃ২০০৭; মাঃ২০১৪; চুয়েটঃ২০০৮-২০০৯; কুয়েটঃ২০০৬-২০০৭।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4\cos{x}\cos{2x}\cos{3x}=1, \ 0\lt{x}\lt{\pi}\)
\(\Rightarrow 2\cos{2x}\times2\cos{3x}\cos{x}=1\)
\(\Rightarrow 2\cos{2x}\{\cos{(3x-x)}+\cos{(3x+x)}\}=1\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow 2\cos{2x}\{\cos{2x}+\cos{4x}\}-1=0\)
\(\Rightarrow 2\cos^2{2x}+2\cos{4x}\cos{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+2\cos^2{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+\cos{4x}=0\) ➜ \(\because 2\cos^2{A}-1=\cos{2A}\)
\(\Rightarrow \cos{4x}(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 4x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{8}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{8}, \ \frac{\pi}{3}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{8}, \ \pi+\frac{\pi}{3}, \ \pi-\frac{\pi}{3}\)
\(\Rightarrow x=\frac{3\pi}{8}, \ \frac{4\pi}{3}, \ \frac{2\pi}{3}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{8}, \ \frac{2\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{8}, \ -\pi+\frac{\pi}{3}, \ -\pi-\frac{\pi}{3}\)
\(\Rightarrow x=-\frac{\pi}{8}, \ -\frac{2\pi}{3}, \ -\frac{4\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{5\pi}{8}, \ 2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{8}\)
যখন, \(n=3,\) \(x=\frac{7\pi}{8}, \ 3\pi+\frac{\pi}{3}, \ 3\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{7\pi}{8}\)
যখন, \(n=4,\) \(x=\frac{9\pi}{8}, \ 4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
\(4\cos{x}\cos{2x}\cos{3x}=1, \ 0\lt{x}\lt{\pi}\)
\(\Rightarrow 2\cos{2x}\times2\cos{3x}\cos{x}=1\)
\(\Rightarrow 2\cos{2x}\{\cos{(3x-x)}+\cos{(3x+x)}\}=1\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow 2\cos{2x}\{\cos{2x}+\cos{4x}\}-1=0\)
\(\Rightarrow 2\cos^2{2x}+2\cos{4x}\cos{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+2\cos^2{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+\cos{4x}=0\) ➜ \(\because 2\cos^2{A}-1=\cos{2A}\)
\(\Rightarrow \cos{4x}(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 4x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{8}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ -\frac{\pi}{3}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{8}, \ \frac{\pi}{3}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{8}, \ \pi+\frac{\pi}{3}, \ \pi-\frac{\pi}{3}\)
\(\Rightarrow x=\frac{3\pi}{8}, \ \frac{4\pi}{3}, \ \frac{2\pi}{3}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{8}, \ \frac{2\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{8}, \ -\pi+\frac{\pi}{3}, \ -\pi-\frac{\pi}{3}\)
\(\Rightarrow x=-\frac{\pi}{8}, \ -\frac{2\pi}{3}, \ -\frac{4\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{5\pi}{8}, \ 2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{8}\)
যখন, \(n=3,\) \(x=\frac{7\pi}{8}, \ 3\pi+\frac{\pi}{3}, \ 3\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{7\pi}{8}\)
যখন, \(n=4,\) \(x=\frac{9\pi}{8}, \ 4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
সমাধান করঃ
\(Q.2.(xv)\) \(\cos{9x}\cos{7x}=\cos{5x}\cos{3x}, \ -\frac{\pi}{4}\lt{x}\lt{\frac{\pi}{4}}\)উত্তরঃ \(0, \ \pm\frac{\pi}{12}, \ \pm\frac{\pi}{6}\)
ঢাঃ২০১২ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{9x}\cos{7x}=\cos{5x}\cos{3x}, \ -\frac{\pi}{4}\lt{x}\lt{\frac{\pi}{4}}\)
\(\Rightarrow 2\cos{9x}\cos{7x}=2\cos{5x}\cos{3x}\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \cos{(9x-7x)}+\cos{(9x+7x)}=\cos{(5x-3x)}+\cos{(5x+3x)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2x}+\cos{16x}=\cos{2x}+\cos{8x}\)
\(\Rightarrow \cos{2x}+\cos{16x}-\cos{2x}-\cos{8x}=0\)
\(\Rightarrow \cos{16x}-\cos{8x}=0\)
\(\Rightarrow \cos{8x}-\cos{16x}=0\)
\(\Rightarrow 2\sin{\frac{8x+16x}{2}}\sin{\frac{16x-8x}{2}}=0\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{24x}{2}}\sin{\frac{8x}{2}}=0\)
\(\Rightarrow \sin{12x}\sin{4x}=0\)
\(\Rightarrow \sin{12x}=0, \ \sin{4x}=0\)
\(\Rightarrow 12x=n\pi, \ 4x=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{12}, \ \frac{n\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{12}, \ \frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=0\)
যখন, \(n=1,\) \(x=\frac{\pi}{12}, \ \frac{\pi}{4}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{12}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{12}, \ -\frac{\pi}{4}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{12}\)
যখন, \(n=2,\) \(x=\frac{2\pi}{12}, \ \frac{2\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{6}, \ \frac{\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{6}\)
যখন, \(n=-2,\) \(x=\frac{-2\pi}{12}, \ \frac{-2\pi}{4}\)
\(\Rightarrow x=-\frac{\pi}{6}, \ -\frac{\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{6}\)
যখন, \(n=3,\) \(x=\frac{3\pi}{12}, \ \frac{3\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{4}, \ \frac{3\pi}{4}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ -\frac{\pi}{4}\lt{x}\lt{\frac{\pi}{4}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=0, \ \pm\frac{\pi}{12}, \ \pm\frac{\pi}{6}\)
\(\cos{9x}\cos{7x}=\cos{5x}\cos{3x}, \ -\frac{\pi}{4}\lt{x}\lt{\frac{\pi}{4}}\)
\(\Rightarrow 2\cos{9x}\cos{7x}=2\cos{5x}\cos{3x}\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \cos{(9x-7x)}+\cos{(9x+7x)}=\cos{(5x-3x)}+\cos{(5x+3x)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2x}+\cos{16x}=\cos{2x}+\cos{8x}\)
\(\Rightarrow \cos{2x}+\cos{16x}-\cos{2x}-\cos{8x}=0\)
\(\Rightarrow \cos{16x}-\cos{8x}=0\)
\(\Rightarrow \cos{8x}-\cos{16x}=0\)
\(\Rightarrow 2\sin{\frac{8x+16x}{2}}\sin{\frac{16x-8x}{2}}=0\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{24x}{2}}\sin{\frac{8x}{2}}=0\)
\(\Rightarrow \sin{12x}\sin{4x}=0\)
\(\Rightarrow \sin{12x}=0, \ \sin{4x}=0\)
\(\Rightarrow 12x=n\pi, \ 4x=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{12}, \ \frac{n\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{12}, \ \frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-\frac{\pi}{4}\lt{x}\lt{\frac{\pi}{4}}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=0\) মানটি গ্রহনযোগ্য।\(\therefore x=0\)
যখন, \(n=1,\) \(x=\frac{\pi}{12}, \ \frac{\pi}{4}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{12}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{12}, \ -\frac{\pi}{4}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{12}\)
যখন, \(n=2,\) \(x=\frac{2\pi}{12}, \ \frac{2\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{6}, \ \frac{\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{6}\)
যখন, \(n=-2,\) \(x=\frac{-2\pi}{12}, \ \frac{-2\pi}{4}\)
\(\Rightarrow x=-\frac{\pi}{6}, \ -\frac{\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{6}\)
যখন, \(n=3,\) \(x=\frac{3\pi}{12}, \ \frac{3\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{4}, \ \frac{3\pi}{4}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ -\frac{\pi}{4}\lt{x}\lt{\frac{\pi}{4}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=0, \ \pm\frac{\pi}{12}, \ \pm\frac{\pi}{6}\)
সমাধান করঃ
\(Q.2.(xvi)\) \(2\sin{x}\sin{3x}=1, \ 0\lt{x}\lt{2\pi}\)উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
ঢাঃ২০১৯,২০১৪; যঃ২০১৩,২০০৮; দিঃ২০১২; রাঃ২০১০; চঃ,সিঃ২০০৯ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin{x}\sin{3x}=1, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow \cos{(3x-x)}-\cos{(3x+x)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2x}-\cos{4x}=1\)
\(\Rightarrow \cos{2x}=1+\cos{4x}\)
\(\Rightarrow \cos{2x}=2\cos^2{2x}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2x}-2\cos^2{2x}=0\)
\(\Rightarrow \cos{2x}(1-2\cos{2x})=0\)
\(\Rightarrow \cos{2x}=0, \ 1-2\cos{2x}=0\)
\(\Rightarrow \cos{2x}=0, \ -2\cos{2x}=-1\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{2x}=1\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ x=n\pi\pm\frac{\pi}{6}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{\pi}{4}, \ -\frac{5\pi}{6}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=-2,\) \(x=-\frac{3\pi}{4}, \ -2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\frac{11\pi}{6}, \ -\frac{13\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=3,\) \(x=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(x=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\)কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
\(2\sin{x}\sin{3x}=1, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow \cos{(3x-x)}-\cos{(3x+x)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2x}-\cos{4x}=1\)
\(\Rightarrow \cos{2x}=1+\cos{4x}\)
\(\Rightarrow \cos{2x}=2\cos^2{2x}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2x}-2\cos^2{2x}=0\)
\(\Rightarrow \cos{2x}(1-2\cos{2x})=0\)
\(\Rightarrow \cos{2x}=0, \ 1-2\cos{2x}=0\)
\(\Rightarrow \cos{2x}=0, \ -2\cos{2x}=-1\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{2x}=1\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ x=n\pi\pm\frac{\pi}{6}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}, \ \frac{\pi}{6}, \ -\frac{\pi}{6}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{\pi}{4}, \ -\frac{5\pi}{6}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=-2,\) \(x=-\frac{3\pi}{4}, \ -2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\frac{11\pi}{6}, \ -\frac{13\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=3,\) \(x=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(x=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\)কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
সমাধান করঃ
\(Q.2.(xvii)\) \(4(\cos^2{x}+\sin{x})=5, \ 0\lt{x}\lt{2\pi}\)উত্তরঃ \(\frac{\pi}{6}, \ \frac{5\pi}{6}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4(\cos^2{x}+\sin{x})=5, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow 4(1-\sin^2{x}+\sin{x})=5\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 4-4\sin^2{x}+4\sin{x}-5=0\)
\(\Rightarrow -4\sin^2{x}+4\sin{x}-1=0\)
\(\Rightarrow 4\sin^2{x}-4\sin{x}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow (2\sin{x}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\sin{x}-1=0\)
\(\Rightarrow 2\sin{x}=1\)
\(\Rightarrow \sin{x}=\frac{1}{2}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{6}\)
যখন, \(n=1,\) \(x=\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{5\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(x=-\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{7\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{6}\)
\(\Rightarrow x=\frac{13\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{6}, \ \frac{5\pi}{6}\)
\(4(\cos^2{x}+\sin{x})=5, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow 4(1-\sin^2{x}+\sin{x})=5\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
\(\Rightarrow 4-4\sin^2{x}+4\sin{x}-5=0\)
\(\Rightarrow -4\sin^2{x}+4\sin{x}-1=0\)
\(\Rightarrow 4\sin^2{x}-4\sin{x}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow (2\sin{x}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\sin{x}-1=0\)
\(\Rightarrow 2\sin{x}=1\)
\(\Rightarrow \sin{x}=\frac{1}{2}\)
\(\Rightarrow \sin{x}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\therefore x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{6}\) মানটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{6}\)
যখন, \(n=1,\) \(x=\pi-\frac{\pi}{6}\)
\(\Rightarrow x=\frac{5\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(x=-\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{7\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{6}\)
\(\Rightarrow x=\frac{13\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{6}, \ \frac{5\pi}{6}\)
সমাধান করঃ
\(Q.2.(xviii)\) \(2\sin^2{x}-5\cos{x}+1=0, \ 0\lt{x}\lt{360^{o}}\)উত্তরঃ \(60^{o}, \ 300^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin^2{x}-5\cos{x}+1=0, \ 0\lt{x}\lt{360^{o}}\)
\(\Rightarrow 2(1-\cos^2{x})-5\cos{x}+1=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-2\cos^2{x}-5\cos{x}+1=0\)
\(\Rightarrow -2\cos^2{x}-5\cos{x}+3=0\)
\(\Rightarrow 2\cos^2{x}+5\cos{x}-3=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos^2{x}+6\cos{x}-\cos{x}-3=0\)
\(\Rightarrow 2\cos{x}(\cos{x}+3)-1(\cos{x}+3)=0\)
\(\Rightarrow (\cos{x}+3)(2\cos{x}-1)=0\)
\(\Rightarrow \cos{x}+3\ne{0}, \ 2\cos{x}-1=0, \ \because -1\le{\cos{x}}\le{1}\)
\(\Rightarrow 2\cos{x}=1\)
\(\Rightarrow \cos{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\pm60^{o}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm60^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=60^{o}\)
যখন, \(n=1,\) \(x=2\pi+60^{o}, \ 2\pi-60^{o}\)
\(\Rightarrow x=360^{o}+60^{o}, \ 360^{o}-60^{o}\)
\(\Rightarrow x=420^{o}, \ 300^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=300^{o}\)
যখন, \(n=-1,\) \(x=-2\pi+60^{o}, \ -2\pi-60^{o}\)
\(\Rightarrow x=-360^{o}+60^{o}, \ -360^{o}-60^{o}\)
\(\Rightarrow x=-300^{o}, \ -420^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=60^{o}, \ 300^{o}\)
\(2\sin^2{x}-5\cos{x}+1=0, \ 0\lt{x}\lt{360^{o}}\)
\(\Rightarrow 2(1-\cos^2{x})-5\cos{x}+1=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 2-2\cos^2{x}-5\cos{x}+1=0\)
\(\Rightarrow -2\cos^2{x}-5\cos{x}+3=0\)
\(\Rightarrow 2\cos^2{x}+5\cos{x}-3=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos^2{x}+6\cos{x}-\cos{x}-3=0\)
\(\Rightarrow 2\cos{x}(\cos{x}+3)-1(\cos{x}+3)=0\)
\(\Rightarrow (\cos{x}+3)(2\cos{x}-1)=0\)
\(\Rightarrow \cos{x}+3\ne{0}, \ 2\cos{x}-1=0, \ \because -1\le{\cos{x}}\le{1}\)
\(\Rightarrow 2\cos{x}=1\)
\(\Rightarrow \cos{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\pm60^{o}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm60^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{360^{o}}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=60^{o}, \ -60^{o}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore x=60^{o}\)
যখন, \(n=1,\) \(x=2\pi+60^{o}, \ 2\pi-60^{o}\)
\(\Rightarrow x=360^{o}+60^{o}, \ 360^{o}-60^{o}\)
\(\Rightarrow x=420^{o}, \ 300^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=300^{o}\)
যখন, \(n=-1,\) \(x=-2\pi+60^{o}, \ -2\pi-60^{o}\)
\(\Rightarrow x=-360^{o}+60^{o}, \ -360^{o}-60^{o}\)
\(\Rightarrow x=-300^{o}, \ -420^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=60^{o}, \ 300^{o}\)
সমাধান করঃ
\(Q.2.(xix)\) \(3\tan^2{x}-4\sqrt{3}\sec{x}+7=0, \ 0\lt{x}\lt{360^{o}}\)উত্তরঃ \(30^{o}, \ 330^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(3\tan^2{x}-4\sqrt{3}\sec{x}+7=0, \ 0\lt{x}\lt{360^{o}}\)
\(\Rightarrow 3(\sec^2{x}-1)-4\sqrt{3}\sec{x}+7=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow 3\sec^2{x}-3-4\sqrt{3}\sec{x}+7=0\)
\(\Rightarrow 3\sec^2{x}-4\sqrt{3}\sec{x}+4=0\)
\(\Rightarrow (\sqrt{3}\sec{x})^2-2\sqrt{3}\sec{x}.2+2^2=0\)
\(\Rightarrow (\sqrt{3}\sec{x}-2)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{3}\sec{x}-2=0\)
\(\Rightarrow \sqrt{3}\sec{x}=2\)
\(\Rightarrow \sec{x}=\frac{2}{\sqrt{3}}\)
\(\Rightarrow \sec{x}=\sec{\frac{\pi}{6}}\) ➜ \(\because \frac{2}{\sqrt{3}}=\sec{\frac{\pi}{6}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \sec{A}=\sec{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\pm30^{o}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm30^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=30^{o}\)
যখন, \(n=1,\) \(x=2\pi+30^{o}, \ 2\pi-30^{o}\)
\(\Rightarrow x=360^{o}+30^{o}, \ 360^{o}-30^{o}\)
\(\Rightarrow x=390^{o}, \ 330^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=330^{o}\)
যখন, \(n=-1,\) \(x=-2\pi+30^{o}, \ -2\pi-30^{o}\)
\(\Rightarrow x=-360^{o}+30^{o}, \ -360^{o}-30^{o}\)
\(\Rightarrow x=-330^{o}, \ -390^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=30^{o}, \ 330^{o}\)
\(3\tan^2{x}-4\sqrt{3}\sec{x}+7=0, \ 0\lt{x}\lt{360^{o}}\)
\(\Rightarrow 3(\sec^2{x}-1)-4\sqrt{3}\sec{x}+7=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow 3\sec^2{x}-3-4\sqrt{3}\sec{x}+7=0\)
\(\Rightarrow 3\sec^2{x}-4\sqrt{3}\sec{x}+4=0\)
\(\Rightarrow (\sqrt{3}\sec{x})^2-2\sqrt{3}\sec{x}.2+2^2=0\)
\(\Rightarrow (\sqrt{3}\sec{x}-2)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{3}\sec{x}-2=0\)
\(\Rightarrow \sqrt{3}\sec{x}=2\)
\(\Rightarrow \sec{x}=\frac{2}{\sqrt{3}}\)
\(\Rightarrow \sec{x}=\sec{\frac{\pi}{6}}\) ➜ \(\because \frac{2}{\sqrt{3}}=\sec{\frac{\pi}{6}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \sec{A}=\sec{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi\pm30^{o}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi\pm30^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{360^{o}}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=30^{o}, \ -30^{o}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore x=30^{o}\)
যখন, \(n=1,\) \(x=2\pi+30^{o}, \ 2\pi-30^{o}\)
\(\Rightarrow x=360^{o}+30^{o}, \ 360^{o}-30^{o}\)
\(\Rightarrow x=390^{o}, \ 330^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=330^{o}\)
যখন, \(n=-1,\) \(x=-2\pi+30^{o}, \ -2\pi-30^{o}\)
\(\Rightarrow x=-360^{o}+30^{o}, \ -360^{o}-30^{o}\)
\(\Rightarrow x=-330^{o}, \ -390^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=30^{o}, \ 330^{o}\)
সমাধান করঃ
\(Q.2.(xx)\) \(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)উত্তরঃ \(\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) এবং \(\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{x}\le{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
সমাধান করঃ
\(Q.2.(xxi)\) \(\cot{\theta}-\tan{\theta}=2, \ 0\le{\theta}\le{\pi}\)উত্তরঃ \(\frac{\pi}{8}, \ \frac{5\pi}{8}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{\theta}-\tan{\theta}=2, \ 0\le{\theta}\le{\pi}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=2\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=2\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{2\theta}\)
\(\Rightarrow \sin{2\theta}=\cos{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\cos{2A}\)
\(\Rightarrow \frac{\sin{2\theta}}{\cos{2\theta}}=1\)
\(\Rightarrow \tan{2\theta}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{2\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow 2\theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 2\theta=(4n+1)\frac{\pi}{4}\)
\(\therefore \theta=(4n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{8}\)
যখন, \(n=1,\) \(\theta=5\frac{\pi}{8}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{8}\)
যখন, \(n=-1,\) \(\theta=-3\frac{\pi}{8}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=9\frac{\pi}{8}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{8}, \ \frac{5\pi}{8}\)
\(\cot{\theta}-\tan{\theta}=2, \ 0\le{\theta}\le{\pi}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=2\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=2\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}=\cos{2\theta}\)
\(\Rightarrow \sin{2\theta}=\cos{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\cos{2A}\)
\(\Rightarrow \frac{\sin{2\theta}}{\cos{2\theta}}=1\)
\(\Rightarrow \tan{2\theta}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{2\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow 2\theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 2\theta=(4n+1)\frac{\pi}{4}\)
\(\therefore \theta=(4n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{8}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{8}\)
যখন, \(n=1,\) \(\theta=5\frac{\pi}{8}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{8}\)
যখন, \(n=-1,\) \(\theta=-3\frac{\pi}{8}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=9\frac{\pi}{8}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{8}, \ \frac{5\pi}{8}\)
সমাধান করঃ
\(Q.2.(xxii)\) \(\cos{7x}=\cos{3x}+\sin{5x}, \ -\frac{\pi}{2}\le{x}\le{\frac{\pi}{2}}\)উত্তরঃ \(-\frac{5\pi}{12}, \ -\frac{2\pi}{5}, \ -\frac{\pi}{5}, \ -\frac{\pi}{12}, \ 0, \ \frac{\pi}{5}\) এবং \(\frac{2\pi}{5}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{7x}=\cos{3x}+\sin{5x}, \ -\frac{\pi}{2}\le{x}\le{\frac{\pi}{2}}\)
\(\Rightarrow \cos{7x}-\cos{3x}=\sin{5x}\)
\(\Rightarrow -(\cos{3x}-\cos{7x})=\sin{5x}\)
\(\Rightarrow -2\sin{\frac{3x+7x}{2}}\sin{\frac{7x-3x}{2}}=\sin{5x}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow -2\sin{\frac{10x}{2}}\sin{\frac{4x}{2}}=\sin{5x}\)
\(\Rightarrow -2\sin{5x}\sin{2x}-\sin{5x}=0\)
\(\Rightarrow -\sin{5x}(2\sin{2x}+1)=0\)
\(\Rightarrow -\sin{5x}=0, \ 2\sin{2x}+1=0\)
\(\Rightarrow \sin{5x}=0, \ 2\sin{2x}=-1\)
\(\Rightarrow \sin{5x}=0, \ \sin{2x}=-\frac{1}{2}\)
\(\Rightarrow \sin{5x}=0, \ \sin{2x}=-\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \sin{5x}=0, \ \sin{2x}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\sin{\frac{\pi}{6}}=\sin{\left(\pi+\frac{\pi}{6}\right)}\)
\(\Rightarrow 5x=n\pi, \ 2x=n\pi+(-1)^n\left(\pi+\frac{\pi}{6}\right)\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=\frac{n\pi}{5}, \ 2x=n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore x=\frac{n\pi}{5}, \ x=\frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{5}, \ \frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=0\)
যখন, \(n=1,\) \(x=\frac{\pi}{5}, \ \frac{\pi}{2}-\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{\pi}{5}, \ -\frac{\pi}{12}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{5}, \ -\frac{\pi}{12}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{5}, \ -\frac{\pi}{2}-\frac{7\pi}{12}\)
\(\Rightarrow x=-\frac{\pi}{5}, \ -\frac{13\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{5}\)
যখন, \(n=2,\) \(x=\frac{2\pi}{5}, \ \pi+\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{2\pi}{5}, \ \frac{19\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{2\pi}{5}\)
যখন, \(n=-2,\) \(x=-\frac{2\pi}{5}, \ -\pi+\frac{7\pi}{12}\)
\(\Rightarrow x=-\frac{2\pi}{5}, \ -\frac{5\pi}{12}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=-\frac{2\pi}{5}, \ -\frac{5\pi}{12}\)
যখন, \(n=3,\) \(x=\frac{3\pi}{5}, \ \frac{3\pi}{2}-\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{3\pi}{5}, \ \frac{11\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore \ -\frac{\pi}{2}\le{x}\le{\frac{\pi}{2}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{5\pi}{12}, \ -\frac{2\pi}{5}, \ -\frac{\pi}{5}, \ -\frac{\pi}{12}, \ 0, \ \frac{\pi}{5}\) এবং \(\frac{2\pi}{5}\)
\(\cos{7x}=\cos{3x}+\sin{5x}, \ -\frac{\pi}{2}\le{x}\le{\frac{\pi}{2}}\)
\(\Rightarrow \cos{7x}-\cos{3x}=\sin{5x}\)
\(\Rightarrow -(\cos{3x}-\cos{7x})=\sin{5x}\)
\(\Rightarrow -2\sin{\frac{3x+7x}{2}}\sin{\frac{7x-3x}{2}}=\sin{5x}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow -2\sin{\frac{10x}{2}}\sin{\frac{4x}{2}}=\sin{5x}\)
\(\Rightarrow -2\sin{5x}\sin{2x}-\sin{5x}=0\)
\(\Rightarrow -\sin{5x}(2\sin{2x}+1)=0\)
\(\Rightarrow -\sin{5x}=0, \ 2\sin{2x}+1=0\)
\(\Rightarrow \sin{5x}=0, \ 2\sin{2x}=-1\)
\(\Rightarrow \sin{5x}=0, \ \sin{2x}=-\frac{1}{2}\)
\(\Rightarrow \sin{5x}=0, \ \sin{2x}=-\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \sin{5x}=0, \ \sin{2x}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\sin{\frac{\pi}{6}}=\sin{\left(\pi+\frac{\pi}{6}\right)}\)
\(\Rightarrow 5x=n\pi, \ 2x=n\pi+(-1)^n\left(\pi+\frac{\pi}{6}\right)\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=\frac{n\pi}{5}, \ 2x=n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore x=\frac{n\pi}{5}, \ x=\frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{5}, \ \frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-\frac{\pi}{2}\le{x}\le{\frac{\pi}{2}}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=0, \ \frac{7\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore x=0\)
যখন, \(n=1,\) \(x=\frac{\pi}{5}, \ \frac{\pi}{2}-\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{\pi}{5}, \ -\frac{\pi}{12}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{5}, \ -\frac{\pi}{12}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{5}, \ -\frac{\pi}{2}-\frac{7\pi}{12}\)
\(\Rightarrow x=-\frac{\pi}{5}, \ -\frac{13\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{\pi}{5}\)
যখন, \(n=2,\) \(x=\frac{2\pi}{5}, \ \pi+\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{2\pi}{5}, \ \frac{19\pi}{12}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{2\pi}{5}\)
যখন, \(n=-2,\) \(x=-\frac{2\pi}{5}, \ -\pi+\frac{7\pi}{12}\)
\(\Rightarrow x=-\frac{2\pi}{5}, \ -\frac{5\pi}{12}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=-\frac{2\pi}{5}, \ -\frac{5\pi}{12}\)
যখন, \(n=3,\) \(x=\frac{3\pi}{5}, \ \frac{3\pi}{2}-\frac{7\pi}{12}\)
\(\Rightarrow x=\frac{3\pi}{5}, \ \frac{11\pi}{12}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore \ -\frac{\pi}{2}\le{x}\le{\frac{\pi}{2}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{5\pi}{12}, \ -\frac{2\pi}{5}, \ -\frac{\pi}{5}, \ -\frac{\pi}{12}, \ 0, \ \frac{\pi}{5}\) এবং \(\frac{2\pi}{5}\)
সমাধান করঃ
\(Q.2.(xxiii)\) \(\cos{2x}+\cos{x}+1=0, \ 0\le{x}\le{360^{o}}\)উত্তরঃ \(90^{o}, \ 120^{o}, \ 240^{o}\) এবং \(270^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{2x}+\cos{x}+1=0, \ 0\le{x}\le{360^{o}}\)
\(\Rightarrow 2\cos^2{x}-1+\cos{x}+1=0\) ➜ \(\because \cos{2x}=2\cos^2{x}-1\)
\(\Rightarrow 2\cos^2{x}+\cos{x}=0\)
\(\Rightarrow \cos{x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=\cos{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\left(\pi-\frac{\pi}{3}\right)\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ 2n\pi\pm\frac{2\pi}{3}\)
\(\therefore x=(2n+1)90^{o}, \ 2n\pi\pm120^{o}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)90^{o}, \ 2n\pi\pm120^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=90^{o}, \ 120^{o}\)
যখন, \(n=1,\) \(x=3\times90^{o}, \ 2\pi+120^{o}, 2\pi-\ 120^{o}\)
\(\Rightarrow x=270^{o}, \ 360^{o}+120^{o}, 360^{o}-\ 120^{o}\)
\(\Rightarrow x=270^{o}, \ 480^{o}, 240^{o}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=270^{o}, 240^{o}\)
যখন, \(n=-1,\) \(x=-90^{o}, \ -2\pi+120^{o}, -2\pi-\ 120^{o}\)
\(\Rightarrow x=-90^{o}, \ -360^{o}+120^{o}, -360^{o}-\ 120^{o}\)
\(\Rightarrow x=-90^{o}, \ -240^{o}, -480^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=90^{o}, \ 120^{o}, \ 240^{o}\) এবং \(270^{o}\)
\(\cos{2x}+\cos{x}+1=0, \ 0\le{x}\le{360^{o}}\)
\(\Rightarrow 2\cos^2{x}-1+\cos{x}+1=0\) ➜ \(\because \cos{2x}=2\cos^2{x}-1\)
\(\Rightarrow 2\cos^2{x}+\cos{x}=0\)
\(\Rightarrow \cos{x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=\cos{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\left(\pi-\frac{\pi}{3}\right)\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ 2n\pi\pm\frac{2\pi}{3}\)
\(\therefore x=(2n+1)90^{o}, \ 2n\pi\pm120^{o}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)90^{o}, \ 2n\pi\pm120^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{x}\le{360^{o}}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=90^{o}, \ 120^{o}, -\ 120^{o}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore x=90^{o}, \ 120^{o}\)
যখন, \(n=1,\) \(x=3\times90^{o}, \ 2\pi+120^{o}, 2\pi-\ 120^{o}\)
\(\Rightarrow x=270^{o}, \ 360^{o}+120^{o}, 360^{o}-\ 120^{o}\)
\(\Rightarrow x=270^{o}, \ 480^{o}, 240^{o}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=270^{o}, 240^{o}\)
যখন, \(n=-1,\) \(x=-90^{o}, \ -2\pi+120^{o}, -2\pi-\ 120^{o}\)
\(\Rightarrow x=-90^{o}, \ -360^{o}+120^{o}, -360^{o}-\ 120^{o}\)
\(\Rightarrow x=-90^{o}, \ -240^{o}, -480^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=90^{o}, \ 120^{o}, \ 240^{o}\) এবং \(270^{o}\)
\(Q.2.(xxiv)\) সমাধান করঃ \(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{360^{o}}\)
উত্তরঃ \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 198^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\) এবং \(342^{o}\)
উত্তরঃ \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 198^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\) এবং \(342^{o}\)
ঢাঃ২০০৮; কুঃ২০১১; চুয়েটঃ২০০৩-২০০৪ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{180^{o}}\)
\(\Rightarrow \frac{1}{\cos{4\theta}}-\frac{1}{\cos{2\theta}}=2\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{2\theta}-\cos{4\theta}}{\cos{4\theta}\cos{2\theta}}=2\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=2\cos{4\theta}\cos{2\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{(4\theta-2\theta)}+\cos{(4\theta+2\theta)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{2\theta}+\cos{6\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}-\cos{2\theta}-\cos{6\theta}=0\)
\(\Rightarrow -\cos{4\theta}-\cos{6\theta}=0\)
\(\Rightarrow \cos{6\theta}+\cos{4\theta}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{\frac{6\theta+4\theta}{2}}\cos{\frac{6\theta-4\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{5\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{5\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 5\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=18^{o}, \ 90^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=18^{o}, \ 90^{o}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{10}, \ \frac{3\pi}{2}\)
\(\Rightarrow \theta=54^{o}, \ 270^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=54^{o}, \ 270^{o}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{10}, \ -\frac{\pi}{2}\)
\(\Rightarrow \theta=-18^{o}, \ -90^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{10}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=\frac{\pi}{2}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=90^{o}, \ 450^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=90^{o}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{10}, \ \frac{7\pi}{2}\)
\(\Rightarrow \theta=126^{o}, \ 630^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=126^{o}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{10}, \ \frac{9\pi}{2}\)
\(\Rightarrow \theta=162^{o}, \ 810^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=5,\) \(\theta=\frac{11\pi}{10}, \ \frac{11\pi}{2}\)
\(\Rightarrow \theta=198^{o}, \ 990^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=198^{o}\)
যখন, \(n=6,\) \(\theta=\frac{13\pi}{10}, \ \frac{13\pi}{2}\)
\(\Rightarrow \theta=234^{o}, \ 1170^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=234^{o}\)
যখন, \(n=7,\) \(\theta=\frac{15\pi}{10}, \ \frac{15\pi}{2}\)
\(\Rightarrow \theta=270^{o}, \ 1350^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=270^{o}\)
যখন, \(n=8,\) \(\theta=\frac{17\pi}{10}, \ \frac{17\pi}{2}\)
\(\Rightarrow \theta=306^{o}, \ 1530^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=306^{o}\)
যখন, \(n=9,\) \(\theta=\frac{19\pi}{10}, \ \frac{19\pi}{2}\)
\(\Rightarrow \theta=342^{o}, \ 1710^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=342^{o}\)
যখন, \(n=10,\) \(\theta=\frac{21\pi}{10}, \ \frac{21\pi}{2}\)
\(\Rightarrow \theta=378^{o}, \ 1890^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 162^{o}, \ 198^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\) এবং \(342^{o}\)
\(\sec{4\theta}-\sec{2\theta}=2, \ 0\le{\theta}\le{180^{o}}\)
\(\Rightarrow \frac{1}{\cos{4\theta}}-\frac{1}{\cos{2\theta}}=2\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{2\theta}-\cos{4\theta}}{\cos{4\theta}\cos{2\theta}}=2\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=2\cos{4\theta}\cos{2\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{(4\theta-2\theta)}+\cos{(4\theta+2\theta)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{2\theta}+\cos{6\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}-\cos{2\theta}-\cos{6\theta}=0\)
\(\Rightarrow -\cos{4\theta}-\cos{6\theta}=0\)
\(\Rightarrow \cos{6\theta}+\cos{4\theta}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{\frac{6\theta+4\theta}{2}}\cos{\frac{6\theta-4\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{5\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{5\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 5\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{10}, \ \frac{\pi}{2}\) \(\Rightarrow \theta=18^{o}, \ 90^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=18^{o}, \ 90^{o}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{10}, \ \frac{3\pi}{2}\)
\(\Rightarrow \theta=54^{o}, \ 270^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=54^{o}, \ 270^{o}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{10}, \ -\frac{\pi}{2}\)
\(\Rightarrow \theta=-18^{o}, \ -90^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{10}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=\frac{\pi}{2}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=90^{o}, \ 450^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=90^{o}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{10}, \ \frac{7\pi}{2}\)
\(\Rightarrow \theta=126^{o}, \ 630^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=126^{o}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{10}, \ \frac{9\pi}{2}\)
\(\Rightarrow \theta=162^{o}, \ 810^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=5,\) \(\theta=\frac{11\pi}{10}, \ \frac{11\pi}{2}\)
\(\Rightarrow \theta=198^{o}, \ 990^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=198^{o}\)
যখন, \(n=6,\) \(\theta=\frac{13\pi}{10}, \ \frac{13\pi}{2}\)
\(\Rightarrow \theta=234^{o}, \ 1170^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=234^{o}\)
যখন, \(n=7,\) \(\theta=\frac{15\pi}{10}, \ \frac{15\pi}{2}\)
\(\Rightarrow \theta=270^{o}, \ 1350^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=270^{o}\)
যখন, \(n=8,\) \(\theta=\frac{17\pi}{10}, \ \frac{17\pi}{2}\)
\(\Rightarrow \theta=306^{o}, \ 1530^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=306^{o}\)
যখন, \(n=9,\) \(\theta=\frac{19\pi}{10}, \ \frac{19\pi}{2}\)
\(\Rightarrow \theta=342^{o}, \ 1710^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=342^{o}\)
যখন, \(n=10,\) \(\theta=\frac{21\pi}{10}, \ \frac{21\pi}{2}\)
\(\Rightarrow \theta=378^{o}, \ 1890^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}, \ 162^{o}, \ 198^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\) এবং \(342^{o}\)
\(Q.2.(xxv)\) \(2(\sin{\theta}\cos{\theta}+\sqrt{3})=\sqrt{3}\cos{\theta}+4\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(60^{o}, \ 120^{o}\)
উত্তরঃ \(60^{o}, \ 120^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2(\sin{\theta}\cos{\theta}+\sqrt{3})=\sqrt{3}\cos{\theta}+4\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}+2\sqrt{3}-\sqrt{3}\cos{\theta}-4\sin{\theta}=0\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}-4\sin{\theta}-\sqrt{3}\cos{\theta}+2\sqrt{3}=0\)
\(\Rightarrow 2\sin{\theta}(\cos{\theta}-2)-\sqrt{3}(\cos{\theta}-2)=0\)
\(\Rightarrow (\cos{\theta}-2)(2\sin{\theta}-\sqrt{3})=0\)
\(\Rightarrow \cos{\theta}-2\ne{0}, \ 2\sin{\theta}-\sqrt{3}=0, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow 2\sin{\theta}=\sqrt{3}\)
\(\Rightarrow \sin{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n60^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n60^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=60^{o}\)
যখন, \(n=1,\) \(\theta=\pi-60^{o}\)
\(\Rightarrow \theta=180^{o}-60^{o}\)
\(\Rightarrow \theta=120^{o}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=120^{o}\)
যখন, \(n=-1,\) \(\theta=-\pi-60^{o}\)
\(\Rightarrow \theta=-180^{o}-60^{o}\)
\(\therefore \theta=-240^{o}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+60^{o}\)
\(\Rightarrow \theta=360^{o}+60^{o}\)
\(\Rightarrow \theta=420^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(60^{o}, \ 120^{o}\)
\(2(\sin{\theta}\cos{\theta}+\sqrt{3})=\sqrt{3}\cos{\theta}+4\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}+2\sqrt{3}-\sqrt{3}\cos{\theta}-4\sin{\theta}=0\)
\(\Rightarrow 2\sin{\theta}\cos{\theta}-4\sin{\theta}-\sqrt{3}\cos{\theta}+2\sqrt{3}=0\)
\(\Rightarrow 2\sin{\theta}(\cos{\theta}-2)-\sqrt{3}(\cos{\theta}-2)=0\)
\(\Rightarrow (\cos{\theta}-2)(2\sin{\theta}-\sqrt{3})=0\)
\(\Rightarrow \cos{\theta}-2\ne{0}, \ 2\sin{\theta}-\sqrt{3}=0, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow 2\sin{\theta}=\sqrt{3}\)
\(\Rightarrow \sin{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n60^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n60^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=60^{o}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=60^{o}\)
যখন, \(n=1,\) \(\theta=\pi-60^{o}\)
\(\Rightarrow \theta=180^{o}-60^{o}\)
\(\Rightarrow \theta=120^{o}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=120^{o}\)
যখন, \(n=-1,\) \(\theta=-\pi-60^{o}\)
\(\Rightarrow \theta=-180^{o}-60^{o}\)
\(\therefore \theta=-240^{o}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+60^{o}\)
\(\Rightarrow \theta=360^{o}+60^{o}\)
\(\Rightarrow \theta=420^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(60^{o}, \ 120^{o}\)
\(Q.2.(xxvi)\) \(3\tan^2{\theta}+1=\frac{2\sqrt{3}}{\cot{\theta}}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(30^{o}, \ 210^{o}\)
উত্তরঃ \(30^{o}, \ 210^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(3\tan^2{\theta}+1=\frac{2\sqrt{3}}{\cot{\theta}}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 3\tan^2{\theta}+1=2\sqrt{3}\tan{\theta}\) ➜ \(\because \frac{1}{\cot{A}}=\tan{A}\)
\(\Rightarrow 3\tan^2{\theta}-2\sqrt{3}\tan{\theta}+1=0\)
\(\Rightarrow (\sqrt{3}\tan{\theta})^2-2\sqrt{3}\tan{\theta}+1^2=0\)
\(\Rightarrow (\sqrt{3}\tan{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{3}\tan{\theta}-1=0\)
\(\Rightarrow \sqrt{3}\tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{6}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+30^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+30^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=30^{o}\)
যখন, \(n=1,\) \(\theta=\pi+30^{o}\)
\(\Rightarrow \theta=180^{o}+30^{o}\)
\(\Rightarrow \theta=210^{o}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=210^{o}\)
যখন, \(n=-1,\) \(\theta=-\pi+30^{o}\)
\(\Rightarrow \theta=-180^{o}+30^{o}\)
\(\therefore \theta=-150^{o}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+30^{o}\)
\(\Rightarrow \theta=360^{o}+30^{o}\)
\(\Rightarrow \theta=390^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(30^{o}, \ 210^{o}\)
\(3\tan^2{\theta}+1=\frac{2\sqrt{3}}{\cot{\theta}}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 3\tan^2{\theta}+1=2\sqrt{3}\tan{\theta}\) ➜ \(\because \frac{1}{\cot{A}}=\tan{A}\)
\(\Rightarrow 3\tan^2{\theta}-2\sqrt{3}\tan{\theta}+1=0\)
\(\Rightarrow (\sqrt{3}\tan{\theta})^2-2\sqrt{3}\tan{\theta}+1^2=0\)
\(\Rightarrow (\sqrt{3}\tan{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{3}\tan{\theta}-1=0\)
\(\Rightarrow \sqrt{3}\tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{6}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+30^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+30^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=30^{o}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=30^{o}\)
যখন, \(n=1,\) \(\theta=\pi+30^{o}\)
\(\Rightarrow \theta=180^{o}+30^{o}\)
\(\Rightarrow \theta=210^{o}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=210^{o}\)
যখন, \(n=-1,\) \(\theta=-\pi+30^{o}\)
\(\Rightarrow \theta=-180^{o}+30^{o}\)
\(\therefore \theta=-150^{o}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+30^{o}\)
\(\Rightarrow \theta=360^{o}+30^{o}\)
\(\Rightarrow \theta=390^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(30^{o}, \ 210^{o}\)
\(Q.2.(xxvii)\) \(\tan^2{\theta}+\sec{\theta}=-1, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(180^{o}\)
উত্তরঃ \(180^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^2{\theta}+\sec{\theta}=-1, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow \sec^2{\theta}-1+\sec{\theta}+1=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow \sec^2{\theta}+\sec{\theta}=0\)
\(\Rightarrow \sec{\theta}(\sec{\theta}+1)=0\)
\(\Rightarrow \sec{\theta}\ne{0}, \ \sec{\theta}+1=0, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \sec{\theta}=-1\)
\(\Rightarrow \frac{1}{\cos{\theta}}=-1\)
\(\Rightarrow \cos{\theta}=-1\)
\(\Rightarrow \theta=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)180^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)180^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=180^{o}\)
যখন, \(n=1,\) \(\theta=3\times180^{o}\)
\(\Rightarrow \theta=540^{o}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-180^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(180^{o}\)
\(\tan^2{\theta}+\sec{\theta}=-1, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow \sec^2{\theta}-1+\sec{\theta}+1=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow \sec^2{\theta}+\sec{\theta}=0\)
\(\Rightarrow \sec{\theta}(\sec{\theta}+1)=0\)
\(\Rightarrow \sec{\theta}\ne{0}, \ \sec{\theta}+1=0, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \sec{\theta}=-1\)
\(\Rightarrow \frac{1}{\cos{\theta}}=-1\)
\(\Rightarrow \cos{\theta}=-1\)
\(\Rightarrow \theta=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)180^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)180^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=180^{o}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=180^{o}\)
যখন, \(n=1,\) \(\theta=3\times180^{o}\)
\(\Rightarrow \theta=540^{o}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-180^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(180^{o}\)
\(Q.2.(xxviii)\) \(\cot^2{\theta}-2\sqrt{2} cosec \ {\theta}+3=0, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(45^{o}, \ 135^{o}\)
উত্তরঃ \(45^{o}, \ 135^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot^2{\theta}-2\sqrt{2} cosec \ {\theta}+3=0, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow cosec^2{\theta}-1-2\sqrt{2} cosec \ {\theta}+3=0\) ➜ \(\because \cot^2{A}=cosec^2{A}-1\)
\(\Rightarrow cosec^2{\theta}-2\sqrt{2} cosec \ {\theta}+2=0\)
\(\Rightarrow cosec^2{\theta}-2\sqrt{2} cosec \ {\theta}+(\sqrt{2})^2=0\)
\(\Rightarrow (cosec \ {\theta}-\sqrt{2})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow cosec \ {\theta}-\sqrt{2}=0\)
\(\Rightarrow cosec \ {\theta}=\sqrt{2}\)
\(\Rightarrow \frac{1}{cosec \ {\theta}}=\frac{1}{\sqrt{2}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{cosec \ {A}}=\sin{A}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=45^{o}\)
যখন, \(n=1,\) \(\theta=\pi-45^{o}\)
\(\Rightarrow \theta=180^{o}-45^{o}\)
\(\Rightarrow \theta=135^{o}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=135^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}-45^{o}\)
\(\Rightarrow \theta=-225^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(45^{o}, \ 135^{o}\)
\(\cot^2{\theta}-2\sqrt{2} cosec \ {\theta}+3=0, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow cosec^2{\theta}-1-2\sqrt{2} cosec \ {\theta}+3=0\) ➜ \(\because \cot^2{A}=cosec^2{A}-1\)
\(\Rightarrow cosec^2{\theta}-2\sqrt{2} cosec \ {\theta}+2=0\)
\(\Rightarrow cosec^2{\theta}-2\sqrt{2} cosec \ {\theta}+(\sqrt{2})^2=0\)
\(\Rightarrow (cosec \ {\theta}-\sqrt{2})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow cosec \ {\theta}-\sqrt{2}=0\)
\(\Rightarrow cosec \ {\theta}=\sqrt{2}\)
\(\Rightarrow \frac{1}{cosec \ {\theta}}=\frac{1}{\sqrt{2}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{cosec \ {A}}=\sin{A}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=45^{o}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=45^{o}\)
যখন, \(n=1,\) \(\theta=\pi-45^{o}\)
\(\Rightarrow \theta=180^{o}-45^{o}\)
\(\Rightarrow \theta=135^{o}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=135^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}-45^{o}\)
\(\Rightarrow \theta=-225^{o}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(45^{o}, \ 135^{o}\)
\(Q.2.(xxix)\) \(1-2\sin{\theta}-2\cos{\theta}+\cot{\theta}=0, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(30^{o}, \ 135^{o}, \ 150^{o}\) এবং \(315^{o}\)
উত্তরঃ \(30^{o}, \ 135^{o}, \ 150^{o}\) এবং \(315^{o}\)
চঃ২০১৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(1-2\sin{\theta}-2\cos{\theta}+\cot{\theta}=0, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 1-2\sin{\theta}-2\cos{\theta}+\frac{\cos{\theta}}{\sin{\theta}}=0\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow -\sin{\theta}+2\sin^2{\theta}+2\sin{\theta}\cos{\theta}-\cos{\theta}=0\) ➜ উভয় পার্শে \(-\sin{\theta}\) গুণ করে,
\(\Rightarrow 2\sin^2{\theta}-\sin{\theta}+2\sin{\theta}\cos{\theta}-\cos{\theta}=0\)
\(\Rightarrow \sin{\theta}(2\sin{\theta}-1)+\cos{\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow (2\sin{\theta}-1)(\sin{\theta}+\cos{\theta})=0\)
\(\Rightarrow 2\sin{\theta}-1=0, \ \sin{\theta}+\cos{\theta}=0\)
\(\Rightarrow 2\sin{\theta}=1, \ \sin{\theta}=-\cos{\theta}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}, \ \frac{\sin{\theta}}{\cos{\theta}}=-1\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \tan{\theta}=-\tan{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\frac{\sin{A}}{\cos{A}}=\tan{A}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \tan{\theta}=\tan{\left(-\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi+\left(-\frac{\pi}{4}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi-\frac{\pi}{4}\)
\(\therefore \theta=n\pi+(-1)^n30^{o}, \ \theta=n\pi-45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n30^{o}, \ n\pi-45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=30^{o}\)
যখন, \(n=1,\) \(\theta=\pi-30^{o}, \ \pi-45^{o}\)
\(\Rightarrow \theta=180^{o}-30^{o}, \ 180^{o}-45^{o}\)
\(\Rightarrow \theta=150^{o}, \ 135^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=150^{o}, \ 135^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}-30^{o}, \ -180^{o}-45^{o}\)
\(\Rightarrow \theta=-210^{o}, \ -225^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+30^{o}, \ 2\pi-45^{o}\)
\(\Rightarrow \theta=360^{o}+30^{o}, \ 360^{o}-45^{o}\)
\(\Rightarrow \theta=390^{o}, \ 315^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=315^{o}\)
যখন, \(n=-2,\) \(\theta=-360^{o}-30^{o}, \ -360^{o}-45^{o}\)
\(\Rightarrow \theta=-390^{o}, \ -405^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(30^{o}, \ 135^{o}, \ 150^{o}\) এবং \(315^{o}\)
\(1-2\sin{\theta}-2\cos{\theta}+\cot{\theta}=0, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow 1-2\sin{\theta}-2\cos{\theta}+\frac{\cos{\theta}}{\sin{\theta}}=0\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow -\sin{\theta}+2\sin^2{\theta}+2\sin{\theta}\cos{\theta}-\cos{\theta}=0\) ➜ উভয় পার্শে \(-\sin{\theta}\) গুণ করে,
\(\Rightarrow 2\sin^2{\theta}-\sin{\theta}+2\sin{\theta}\cos{\theta}-\cos{\theta}=0\)
\(\Rightarrow \sin{\theta}(2\sin{\theta}-1)+\cos{\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow (2\sin{\theta}-1)(\sin{\theta}+\cos{\theta})=0\)
\(\Rightarrow 2\sin{\theta}-1=0, \ \sin{\theta}+\cos{\theta}=0\)
\(\Rightarrow 2\sin{\theta}=1, \ \sin{\theta}=-\cos{\theta}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}, \ \frac{\sin{\theta}}{\cos{\theta}}=-1\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \tan{\theta}=-\tan{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\frac{\sin{A}}{\cos{A}}=\tan{A}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \tan{\theta}=\tan{\left(-\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi+\left(-\frac{\pi}{4}\right)\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi-\frac{\pi}{4}\)
\(\therefore \theta=n\pi+(-1)^n30^{o}, \ \theta=n\pi-45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n30^{o}, \ n\pi-45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=30^{o}, \ -45^{o}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=30^{o}\)
যখন, \(n=1,\) \(\theta=\pi-30^{o}, \ \pi-45^{o}\)
\(\Rightarrow \theta=180^{o}-30^{o}, \ 180^{o}-45^{o}\)
\(\Rightarrow \theta=150^{o}, \ 135^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=150^{o}, \ 135^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}-30^{o}, \ -180^{o}-45^{o}\)
\(\Rightarrow \theta=-210^{o}, \ -225^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+30^{o}, \ 2\pi-45^{o}\)
\(\Rightarrow \theta=360^{o}+30^{o}, \ 360^{o}-45^{o}\)
\(\Rightarrow \theta=390^{o}, \ 315^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=315^{o}\)
যখন, \(n=-2,\) \(\theta=-360^{o}-30^{o}, \ -360^{o}-45^{o}\)
\(\Rightarrow \theta=-390^{o}, \ -405^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(30^{o}, \ 135^{o}, \ 150^{o}\) এবং \(315^{o}\)
\(Q.2.(xxx)\) \((2+\sqrt{3})\cos{\theta}=1-\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(90^{o}, \ 300^{o}\)
উত্তরঃ \(90^{o}, \ 300^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\((2+\sqrt{3})\cos{\theta}=1-\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow (2+\sqrt{3})\cos{\theta}-(1-\sin{\theta})=0\)
\(\Rightarrow (2+\sqrt{3})\cos{\theta}-\left(\sin^2{\frac{\theta}{2}}+\cos^2{\frac{\theta}{2}}-2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}\right)=0\) ➜ \(\because 1=\sin^2{A}+\cos^2{A}\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow (2+\sqrt{3})\left(\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)^2=0\) ➜ \(\because \cos{2A}=\cos^2{A}-\sin^2{A}\)
এবং \(a^2+b^2-2ab=(a-b)^2\)
\(\Rightarrow -(2+\sqrt{3})\left(\sin^2{\frac{\theta}{2}}-\cos^2{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)^2=0\)
\(\Rightarrow \left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)\left\{-(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)\right\}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}=0, \ -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)=0\)
প্রথম সমীকরণের ক্ষেত্রে,
\(\Rightarrow \sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=\cos{\frac{\theta}{2}}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=1\) ➜ উভয় পার্শে \(\cos{\frac{\theta}{2}}\) ভাগ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{2}\)
\(\therefore \theta=2n\pi+90^{o}\)
দ্বিতীয় সমীকরণের ক্ষেত্রে,
\(\Rightarrow -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)=0\)
\(\Rightarrow -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)=\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\)
\(\Rightarrow -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)=-\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right)\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}}=\frac{1}{2+\sqrt{3}}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}}=\frac{1+2+\sqrt{3}}{1-2-\sqrt{3}}\) ➜ যোজন-বিয়োজন করে,
\(\Rightarrow \frac{2\cos{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}}=\frac{3+\sqrt{3}}{-1-\sqrt{3}}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\frac{\sqrt{3}(\sqrt{3}+1)}{-(\sqrt{3}+1)}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=-\sqrt{3}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=-\frac{1}{\sqrt{3}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=-\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
এবং \(\frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\left(-\frac{\pi}{6}\right)}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\left(-\frac{\pi}{6}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=2n\pi-\frac{\pi}{3}\)
\(\therefore \theta=2n\pi-60^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+90^{o}, \ 2n\pi-60^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=90^{o}\)
যখন, \(n=1,\) \(\theta=2\pi+90^{o}, \ 2\pi-60^{o}\)
\(\Rightarrow \theta=360^{o}+90^{o}, \ 360^{o}-60^{o}\)
\(\Rightarrow \theta=450^{o}, \ 300^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=300^{o}\)
যখন, \(n=-1,\) \(\theta=-360^{o}+90^{o}, \ -360^{o}-60^{o}\)
\(\Rightarrow \theta=-270^{o}, \ -420^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(90^{o}, \ 300^{o}\)
\((2+\sqrt{3})\cos{\theta}=1-\sin{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow (2+\sqrt{3})\cos{\theta}-(1-\sin{\theta})=0\)
\(\Rightarrow (2+\sqrt{3})\cos{\theta}-\left(\sin^2{\frac{\theta}{2}}+\cos^2{\frac{\theta}{2}}-2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}\right)=0\) ➜ \(\because 1=\sin^2{A}+\cos^2{A}\)
এবং \(\sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow (2+\sqrt{3})\left(\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)^2=0\) ➜ \(\because \cos{2A}=\cos^2{A}-\sin^2{A}\)
এবং \(a^2+b^2-2ab=(a-b)^2\)
\(\Rightarrow -(2+\sqrt{3})\left(\sin^2{\frac{\theta}{2}}-\cos^2{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)^2=0\)
\(\Rightarrow \left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)\left\{-(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)\right\}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}=0, \ -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)=0\)
প্রথম সমীকরণের ক্ষেত্রে,
\(\Rightarrow \sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=\cos{\frac{\theta}{2}}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=1\) ➜ উভয় পার্শে \(\cos{\frac{\theta}{2}}\) ভাগ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{2}\)
\(\therefore \theta=2n\pi+90^{o}\)
দ্বিতীয় সমীকরণের ক্ষেত্রে,
\(\Rightarrow -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)-\left(\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\right)=0\)
\(\Rightarrow -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)=\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}\)
\(\Rightarrow -(2+\sqrt{3})\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right)=-\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right)\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}}=\frac{1}{2+\sqrt{3}}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}}=\frac{1+2+\sqrt{3}}{1-2-\sqrt{3}}\) ➜ যোজন-বিয়োজন করে,
\(\Rightarrow \frac{2\cos{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}}=\frac{3+\sqrt{3}}{-1-\sqrt{3}}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\frac{\sqrt{3}(\sqrt{3}+1)}{-(\sqrt{3}+1)}\)
\(\Rightarrow \frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=-\sqrt{3}\)
\(\Rightarrow \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=-\frac{1}{\sqrt{3}}\) ➜ ব্যাস্তকরণ করে,
\(\Rightarrow \tan{\frac{\theta}{2}}=-\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
এবং \(\frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \tan{\frac{\theta}{2}}=\tan{\left(-\frac{\pi}{6}\right)}\)
\(\Rightarrow \frac{\theta}{2}=n\pi+\left(-\frac{\pi}{6}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=2n\pi-\frac{\pi}{3}\)
\(\therefore \theta=2n\pi-60^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+90^{o}, \ 2n\pi-60^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=90^{o}, \ -60^{o}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=90^{o}\)
যখন, \(n=1,\) \(\theta=2\pi+90^{o}, \ 2\pi-60^{o}\)
\(\Rightarrow \theta=360^{o}+90^{o}, \ 360^{o}-60^{o}\)
\(\Rightarrow \theta=450^{o}, \ 300^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=300^{o}\)
যখন, \(n=-1,\) \(\theta=-360^{o}+90^{o}, \ -360^{o}-60^{o}\)
\(\Rightarrow \theta=-270^{o}, \ -420^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(90^{o}, \ 300^{o}\)
\(Q.2.(xxxi)\) \(4\cot{2\theta}=\cot^2{\theta}-\tan^2{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
উত্তরঃ \(45^{o}, \ 135^{o}, \ 225^{o}\) এবং \(315^{o}\)
উত্তরঃ \(45^{o}, \ 135^{o}, \ 225^{o}\) এবং \(315^{o}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4\cot{2\theta}=\cot^2{\theta}-\tan^2{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow \frac{4}{\tan{2\theta}}=\frac{1}{\tan^2{\theta}}-\tan^2{\theta}\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(\Rightarrow \frac{4}{\frac{2\tan{\theta}}{1-\tan^2{\theta}}}=\frac{1-\tan^4{\theta}}{\tan^2{\theta}}\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\Rightarrow \frac{4(1-\tan^2{\theta})}{2\tan{\theta}}=\frac{1-\tan^4{\theta}}{\tan^2{\theta}}\)
\(\Rightarrow \frac{2(1-\tan^2{\theta})}{\tan{\theta}}=\frac{1-\tan^4{\theta}}{\tan^2{\theta}}\)
\(\Rightarrow 2(1-\tan^2{\theta})=\frac{1-\tan^4{\theta}}{\tan{\theta}}, \ \because \tan{\theta}\ne{0}\)
\(\Rightarrow 2\tan{\theta}(1-\tan^2{\theta})=1-\tan^4{\theta}\)
\(\Rightarrow 2\tan{\theta}(1-\tan^2{\theta})-(1-\tan^4{\theta})=0\)
\(\Rightarrow (1-\tan^2{\theta})\left\{2\tan{\theta}-(1+\tan^2{\theta})\right\}=0\)
\(\Rightarrow (1-\tan^2{\theta})\left\{2\tan{\theta}-1-\tan^2{\theta}\right\}=0\)
\(\Rightarrow -(1-\tan^2{\theta})\left\{\tan^2{\theta}-2\tan{\theta}+1\right\}=0\)
\(\Rightarrow (\tan^2{\theta}-1)(\tan{\theta}-1)^2=0\)
\(\Rightarrow (\tan{\theta}+1)(\tan{\theta}-1)(\tan{\theta}-1)^2=0\)
\(\Rightarrow (\tan{\theta}+1)(\tan{\theta}-1)^3=0\)
\(\Rightarrow \tan{\theta}+1=0, \ (\tan{\theta}-1)^3=0\)
\(\Rightarrow \tan{\theta}+1=0, \ \tan{\theta}-1=0\)
\(\Rightarrow \tan{\theta}=-1, \ \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\pm1\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi\pm\frac{\pi}{4}\)
\(\therefore \theta=n\pi\pm45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=45^{o}\)
যখন, \(n=1,\) \(\theta=\pi+45^{o}, \ \pi-45^{o}\)
\(\Rightarrow \theta=180^{o}+45^{o}, \ 180^{o}-45^{o}\)
\(\Rightarrow \theta=225^{o}, \ 135^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=225^{o}, \ 135^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}+45^{o}, \ -180^{o}-45^{o}\)
\(\Rightarrow \theta=-135^{o}, \ -225^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+45^{o}, \ 2\pi-45^{o}\)
\(\Rightarrow \theta=360^{o}+45^{o}, \ 360^{o}-45^{o}\)
\(\Rightarrow \theta=405^{o}, \ 315^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=315^{o}\)
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(45^{o}, \ 135^{o}, \ 225^{o}\) এবং \(315^{o}\)
\(4\cot{2\theta}=\cot^2{\theta}-\tan^2{\theta}, \ 0\lt{\theta}\lt{360^{o}}\)
\(\Rightarrow \frac{4}{\tan{2\theta}}=\frac{1}{\tan^2{\theta}}-\tan^2{\theta}\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(\Rightarrow \frac{4}{\frac{2\tan{\theta}}{1-\tan^2{\theta}}}=\frac{1-\tan^4{\theta}}{\tan^2{\theta}}\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\Rightarrow \frac{4(1-\tan^2{\theta})}{2\tan{\theta}}=\frac{1-\tan^4{\theta}}{\tan^2{\theta}}\)
\(\Rightarrow \frac{2(1-\tan^2{\theta})}{\tan{\theta}}=\frac{1-\tan^4{\theta}}{\tan^2{\theta}}\)
\(\Rightarrow 2(1-\tan^2{\theta})=\frac{1-\tan^4{\theta}}{\tan{\theta}}, \ \because \tan{\theta}\ne{0}\)
\(\Rightarrow 2\tan{\theta}(1-\tan^2{\theta})=1-\tan^4{\theta}\)
\(\Rightarrow 2\tan{\theta}(1-\tan^2{\theta})-(1-\tan^4{\theta})=0\)
\(\Rightarrow (1-\tan^2{\theta})\left\{2\tan{\theta}-(1+\tan^2{\theta})\right\}=0\)
\(\Rightarrow (1-\tan^2{\theta})\left\{2\tan{\theta}-1-\tan^2{\theta}\right\}=0\)
\(\Rightarrow -(1-\tan^2{\theta})\left\{\tan^2{\theta}-2\tan{\theta}+1\right\}=0\)
\(\Rightarrow (\tan^2{\theta}-1)(\tan{\theta}-1)^2=0\)
\(\Rightarrow (\tan{\theta}+1)(\tan{\theta}-1)(\tan{\theta}-1)^2=0\)
\(\Rightarrow (\tan{\theta}+1)(\tan{\theta}-1)^3=0\)
\(\Rightarrow \tan{\theta}+1=0, \ (\tan{\theta}-1)^3=0\)
\(\Rightarrow \tan{\theta}+1=0, \ \tan{\theta}-1=0\)
\(\Rightarrow \tan{\theta}=-1, \ \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\pm1\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=n\pi\pm\frac{\pi}{4}\)
\(\therefore \theta=n\pi\pm45^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm45^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{360^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=45^{o}, \ -45^{o}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=45^{o}\)
যখন, \(n=1,\) \(\theta=\pi+45^{o}, \ \pi-45^{o}\)
\(\Rightarrow \theta=180^{o}+45^{o}, \ 180^{o}-45^{o}\)
\(\Rightarrow \theta=225^{o}, \ 135^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=225^{o}, \ 135^{o}\)
যখন, \(n=-1,\) \(\theta=-180^{o}+45^{o}, \ -180^{o}-45^{o}\)
\(\Rightarrow \theta=-135^{o}, \ -225^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=2\pi+45^{o}, \ 2\pi-45^{o}\)
\(\Rightarrow \theta=360^{o}+45^{o}, \ 360^{o}-45^{o}\)
\(\Rightarrow \theta=405^{o}, \ 315^{o}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=315^{o}\)
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(45^{o}, \ 135^{o}, \ 225^{o}\) এবং \(315^{o}\)
\(Q.2.(xxxii)\) \(1+\cos{x}+\cos{2x}=0, \ 0\lt{x}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{2}, \ \frac{2\pi}{3}, \ \frac{3\pi}{2}\) এবং \(\frac{4\pi}{3}\)
উত্তরঃ \(\frac{\pi}{2}, \ \frac{2\pi}{3}, \ \frac{3\pi}{2}\) এবং \(\frac{4\pi}{3}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(1+\cos{x}+\cos{2x}=0, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow 1+\cos{2x}+\cos{x}=0\)
\(\Rightarrow 2\cos^2{x}+\cos{x}=0\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=\cos{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cos{A}=\cos{(\pi-A)}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=2n\pi+\frac{2\pi}{3}, \ 2n\pi-\frac{2\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{2}, \ 2n\pi+\frac{2\pi}{3}, \ 2n\pi-\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{2}, \ 2n\pi+\frac{2\pi}{3}, \ 2n\pi-\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{2}, \ \frac{2\pi}{3}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{2}, \ 2\pi+\frac{2\pi}{3}, \ 2\pi-\frac{2\pi}{3}\)
\(\Rightarrow x=\frac{3\pi}{2}, \ \frac{8\pi}{3}, \ \frac{4\pi}{3}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{2}, \ \frac{4\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{2}, \ -2\pi+\frac{2\pi}{3}, \ -2\pi-\frac{2\pi}{3}\)
\(\Rightarrow x=-\frac{3\pi}{2}, \ -\frac{4\pi}{3}, \ -\frac{8\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\frac{\pi}{2}, \ \frac{2\pi}{3}, \ \frac{3\pi}{2}\) এবং \(\frac{4\pi}{3}\)
\(1+\cos{x}+\cos{2x}=0, \ 0\lt{x}\lt{2\pi}\)
\(\Rightarrow 1+\cos{2x}+\cos{x}=0\)
\(\Rightarrow 2\cos^2{x}+\cos{x}=0\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=-\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=\cos{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cos{A}=\cos{(\pi-A)}\)
\(\Rightarrow \cos{x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=2n\pi+\frac{2\pi}{3}, \ 2n\pi-\frac{2\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{2}, \ 2n\pi+\frac{2\pi}{3}, \ 2n\pi-\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{2}, \ 2n\pi+\frac{2\pi}{3}, \ 2n\pi-\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{2}, \ \frac{2\pi}{3}, \ -\frac{2\pi}{3}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{2}, \ \frac{2\pi}{3}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{2}, \ 2\pi+\frac{2\pi}{3}, \ 2\pi-\frac{2\pi}{3}\)
\(\Rightarrow x=\frac{3\pi}{2}, \ \frac{8\pi}{3}, \ \frac{4\pi}{3}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{2}, \ \frac{4\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{2}, \ -2\pi+\frac{2\pi}{3}, \ -2\pi-\frac{2\pi}{3}\)
\(\Rightarrow x=-\frac{3\pi}{2}, \ -\frac{4\pi}{3}, \ -\frac{8\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\frac{\pi}{2}, \ \frac{2\pi}{3}, \ \frac{3\pi}{2}\) এবং \(\frac{4\pi}{3}\)
\(Q.2.(xxxiii)\) \(4cosec^2{\theta}-7\cot{\theta} cosec \ {\theta}-2=0, \ 0\lt{\theta}\lt{\frac{\pi}{2}}\)
উত্তরঃ \(\cos^{-1}{(0.314)}\)
উত্তরঃ \(\cos^{-1}{(0.314)}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4cosec^2{\theta}-7\cot{\theta} cosec \ {\theta}-2=0, \ 0\lt{\theta}\lt{\frac{\pi}{2}}\)
\(\Rightarrow \frac{4}{\sin^2{\theta}}-\frac{7\cos{\theta}}{\sin^2{\theta}}-2=0\) ➜ \(\because cocos \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow 4-7\cos{\theta}-2\sin^2{\theta}=0\) ➜ উভয় পার্শে \(\sin^2{\theta}\) গুণ করে,
\(\Rightarrow 4-7\cos{\theta}-2(1-\cos^2{\theta})=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-7\cos{\theta}-2+2\cos^2{\theta}=0\)
\(\Rightarrow 2\cos^2{\theta}-7\cos{\theta}+2=0\)
\(\Rightarrow \cos{\theta}=\frac{7\pm\sqrt{(-7)^2-4\times2\times2}}{2\times2}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{\theta}=\frac{7\pm\sqrt{49-16}}{4}\)
\(\Rightarrow \cos{\theta}=\frac{7\pm\sqrt{33}}{4}\)
\(\Rightarrow \cos{\theta}=\frac{7-\sqrt{33}}{4}, \ -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{\theta}=0.314\)
\(\Rightarrow \cos{\theta}=\cos{\{\cos^{-1}{(0.314)}\}}\)
\(\therefore \theta=2n\pi\pm\cos^{-1}{(0.314)}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\cos^{-1}{(0.314)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\cos^{-1}{(0.314)}\)
যখন, \(n=1,\) \(\theta=2\pi+\cos^{-1}{(0.314)}, \ 2\pi-\cos^{-1}{(0.314)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\pi+\cos^{-1}{(0.314)}, \ -2\pi-\cos^{-1}{(0.314)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\cos^{-1}{(0.314)}\)
\(4cosec^2{\theta}-7\cot{\theta} cosec \ {\theta}-2=0, \ 0\lt{\theta}\lt{\frac{\pi}{2}}\)
\(\Rightarrow \frac{4}{\sin^2{\theta}}-\frac{7\cos{\theta}}{\sin^2{\theta}}-2=0\) ➜ \(\because cocos \ {A}=\frac{1}{\sin{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow 4-7\cos{\theta}-2\sin^2{\theta}=0\) ➜ উভয় পার্শে \(\sin^2{\theta}\) গুণ করে,
\(\Rightarrow 4-7\cos{\theta}-2(1-\cos^2{\theta})=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-7\cos{\theta}-2+2\cos^2{\theta}=0\)
\(\Rightarrow 2\cos^2{\theta}-7\cos{\theta}+2=0\)
\(\Rightarrow \cos{\theta}=\frac{7\pm\sqrt{(-7)^2-4\times2\times2}}{2\times2}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{\theta}=\frac{7\pm\sqrt{49-16}}{4}\)
\(\Rightarrow \cos{\theta}=\frac{7\pm\sqrt{33}}{4}\)
\(\Rightarrow \cos{\theta}=\frac{7-\sqrt{33}}{4}, \ -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{\theta}=0.314\)
\(\Rightarrow \cos{\theta}=\cos{\{\cos^{-1}{(0.314)}\}}\)
\(\therefore \theta=2n\pi\pm\cos^{-1}{(0.314)}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\cos^{-1}{(0.314)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{\theta}\lt{\frac{\pi}{2}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\cos^{-1}{(0.314)}, \ -\cos^{-1}{(0.314)}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=\cos^{-1}{(0.314)}\)
যখন, \(n=1,\) \(\theta=2\pi+\cos^{-1}{(0.314)}, \ 2\pi-\cos^{-1}{(0.314)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\pi+\cos^{-1}{(0.314)}, \ -2\pi-\cos^{-1}{(0.314)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\cos^{-1}{(0.314)}\)
অধ্যায় \(7H\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
সমাধান করঃ
\(Q.3.(i)\) \(\tan{x}+\tan{2x}+\tan{x}\tan{2x}=1\)উত্তরঃ \((4n+1)\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
মাঃ২০১২; রাঃ ২০০২; বঃ২০০৬; চঃ২০০৩; ঢাঃ২০০৯ ।
\(Q.3.(ii)\) \(\sqrt{3}(\tan{x}+\tan{2x})+\tan{x}\tan{2x}=1\)
উত্তরঃ \((6n+1)\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১১; রাঃ ২০১৪,২০১১,২০০৪,২০০০; দিঃ,চঃ২০১৩; সিঃ২০০৭ ।
\(Q.3.(iii)\) \(\tan{\theta}+\tan{2\theta}+\sqrt{3}\tan{\theta}\tan{2\theta}=\sqrt{3}\)
উত্তরঃ \((3n+1)\frac{\pi}{9}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ, রাঃ ২০০৬; চঃ২০১১,২০০৯; সিঃ২০১৪; বঃ২০০৯ ।
\(Q.3.(iv)\) \(\tan{x}+\tan{2x}+\tan{3x}=\tan{x}\tan{2x}\tan{3x}\)
উত্তরঃ \(\frac{n\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৫; মাঃ২০০৭ ।
\(Q.3.(v)\) \(\cot{x}+\cot{2x}+\cot{3x}=\cot{x}\cot{2x}\cot{3x}\)
উত্তরঃ \(\frac{n\pi}{3}\pm\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
বুয়েটঃ২০১৪-২০১৫ ।
\(Q.3.(vi)\) \(\sin{\theta}+\cos{\theta}=1\)
উত্তরঃ \(2n\pi, \ (4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
দিঃ২০১৬; চঃ২০১০; সিঃ২০০৯; ঢাঃ২০০৫; মাঃ২০১৪ ।
\(Q.3.(vii)\) \(\cos{x}-\sin{x}=1\)
উত্তরঃ \(2n\pi, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০০১ ।
\(Q.3.(viii)\) \(\sin{x}+\sqrt{3}\cos{x}=\sqrt{3}\)
উত্তরঃ \(2n\pi, \ (6n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৫; বঃ২০১২,২০০৫; বিআইটিঃ২০০১-২০০২ ।
\(Q.3.(ix)\) \(\sin{x}+\sqrt{3}\cos{x}=\sqrt{2}\)
উত্তরঃ \(2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{5\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
ঢাঃ২০০৭; রাঃ২০০৫ ।
\(Q.3.(x)\) \(\cos{\theta}+\sqrt{3}\sin{\theta}=2\)
উত্তরঃ \(2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xi)\) \(\sin{x}+\cos{x}=\frac{1}{\sqrt{2}}\)
উত্তরঃ \(2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০০২ ।
\(Q.3.(xii)\) \(\cos{\theta}+2\sin{\theta}=1\)
উত্তরঃ \(2n\pi, \ 2n\pi+2\alpha\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
চঃ২০০৪; বুয়েটঃ২০০৫-২০০৬ ।
\(Q.3.(xiii)\) \(\cos{\theta}-\cos{9\theta}=\sin{5\theta}\)
উত্তরঃ \(\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০১৭ ।
\(Q.3.(xiv)\) \(\cos{x}+\cos{2x}+\cos{3x}=0\)
উত্তরঃ \((2n+1)\frac{\pi}{4}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৩; ঢাঃ২০০৫ ।
\(Q.3.(xv)\) \(\sin{x}+\sin{2x}+\sin{3x}=0\)
উত্তরঃ \(\frac{n\pi}{2}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০১৩; রাঃ২০০৭ ।
\(Q.3.(xvi)\) \(\sqrt{2}\cos{3\theta}-\cos{\theta}=\cos{5\theta}\)
উত্তরঃ \((2n+1)\frac{\pi}{6}, \ n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৩; সিঃ২০১০; কুঃ,চঃ২০০৮ ।
\(Q.3.(xvii)\) \(\sin{7\theta}-\sqrt{3}\cos{4\theta}=\sin{\theta}\)
উত্তরঃ \((2n+1)\frac{\pi}{8}, \ \{3n+(-1)^n\}\frac{\pi}{9}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০০৫; দিঃ২০১০ ।
\(Q.3.(xviii)\) \(\cos{7\theta}=\cos{3\theta}+\sin{5\theta}\)
উত্তরঃ \(\frac{n\pi}{5}, \ \frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০০৫; বঃ২০০৭ ।
\(Q.3.(xix)\) \(\cos{\theta}-\cos{7\theta}=\sin{4\theta}\)
উত্তরঃ \(\frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
ঢাঃ২০১৪,২০১০; চঃ২০১৪; বঃ,মাঃ২০১১; দিঃ২০০৯; যঃ২০০৮; কুঃ২০০৬ ।
\(Q.3.(xx)\) \(\sin{3\theta}+\sin{5\theta}+\sin{7\theta}+\sin{9\theta}=0\)
উত্তরঃ \(\theta=\frac{n\pi}{6}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০০৩ ।
\(Q.3.(xxi)\) \(\sin^2{2x}-3\cos^2{x}=0\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১১; কুঃ২০১০ ।
\(Q.3.(xxii)\) \(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)
উত্তরঃ \(2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
রাঃ২০১৪; কুঃ২০০৫; চুয়েটঃ২০১১-২০১২ ।
\(Q.3.(xxiii)\) \(\tan^2{x}+\cot^2{x}=2\)
উত্তরঃ \(n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০০৭ ।
সমাধান করঃ
\(Q.3.(xxiv)\) \(\cos{2x}+\sin{x}=1\)উত্তরঃ \(n\pi, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(\ n\in{\mathbb{Z}}\)
ঢাঃ২০০৪ ।
\(Q.3.(xxv)\) \(\sec^2{\theta}+\tan^2{\theta}=3\tan{\theta}\)
উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{2}, \ n\in{\mathbb{Z}}\)
কুঃ২০১২; দিঃ২০০৯ ।
\(Q.3.(xxvi)\) \(2\sin{\theta}\tan{\theta}+1=\tan{\theta}+2\sin{\theta}\)
উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(\ n\in{\mathbb{Z}}\)
চুয়েটঃ২০০৫-২০০৬ ।
\(Q.3.(xxvii)\) \(\tan{x}+\tan{2x}+\tan{3x}=0\)
উত্তরঃ \(\frac{n\pi}{3}, \ n\pi\pm\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{\sqrt{2}}, \ n\in{\mathbb{Z}}\)
বঃ২০১১; রাঃ২০০৫ ।
\(Q.3.(xxviii)\) \(\tan^2{\theta}-3cosec^2{\theta}+1=0\)
উত্তরঃ \(n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুয়েটঃ২০০৩-২০০৪; টেক্সটাইলঃ২০০৫-২০০৬ ।
\(Q.3.(xxix)\) \(\cos{6x}+\cos{4x}=\sin{3x}+\sin{x}\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ (4n+1)\frac{\pi}{14}, \ (4n-1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৬; রাঃ২০১৪; মাঃ২০১০ ।
\(Q.3.(xxx)\) \(\cos^3{x}\sin{3x}+\sin^3{x}\cos{3x}=\frac{3}{4}\)
উত্তরঃ \((4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০০২ ।
\(Q.3.(xxxi)\) \(\sin{\theta}+\cos{\theta}=\sqrt{2\sin{2\theta}}\)
উত্তরঃ \(n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
ঢাঃ২০০৪ ।
\(Q.3.(xxxii)\) \(\cot{2x}=\cos{x}+\sin{x}\)
উত্তরঃ \((4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
কুঃ২০১১ ।
\(Q.3.(xxxiii)\) \(\frac{\sqrt{3}}{\sin{2x}}-\frac{1}{\cos{2x}}=4\)
উত্তরঃ \((6n+1)\frac{\pi}{18}, \ (3n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
বুয়েটঃ২০০৬-২০০৭, ২০০৭-২০০৮ ।
\(Q.3.(xxxiv)\) \(\tan{3\theta}\tan{\theta}=1\)
উত্তরঃ \((2n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
মাঃ২০১৫ ।
\(Q.3.(xxxv)\) \(\tan{\theta}+\tan{3\theta}=0\)
উত্তরঃ \(\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০০৫ ।
\(Q.3.(xxxvi)\) \(\cos{x}+\sin{x}=\cos{2x}+\sin{2x}\)
উত্তরঃ \(2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
দিঃ২০১৪; ঢাঃ২০১৩,২০০৬; চঃ২০১২,২০০৬; সিঃ২০১৫,২০০৫; কুঃ২০১৪,২০০৫; যঃ২০১৪,২০১১,২০০৭; রাঃ২০১২,২০০৮; মাঃ২০১৩,২০০৯ ।
\(Q.3.(xxxvii)\) \(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}\)
উত্তরঃ \((2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০৮; বুয়েটঃ২০০৮-২০০৯ ।
\(Q.3.(xxxviii)\) \(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0\)
উত্তরঃ \((2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xxxix)\) \(\sin{3x}\sin{x}=\cos{2x}+\frac{1}{2}\)
উত্তরঃ \((2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xL)\) \(\sin{x}+2\cos{x}=1\)
উত্তরঃ \(2n\pi+\frac{\pi}{4}, \ 2n\pi-\alpha\) যেখানে, \(\sin{\alpha}=\frac{3}{5}, \ n\in{\mathbb{Z}}\)
\(Q.3.(xLi)\) \(\sin{x}+\cos{x}+\sqrt{2}=0\)
উত্তরঃ \((8n-3)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xLii)\) \(\cot{\theta}+\cot{\left(\frac{\pi}{4}+\theta\right)}=2\)
উত্তরঃ \(n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xLiii)\) \(2\cot{\frac{\theta}{2}}=(1+\cot{\theta})^2\)
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xLiv)\) \(\frac{\sin{\alpha}}{\sin{2x}}-\frac{\cos{\alpha}}{\cos{2x}}=2\)
উত্তরঃ \(\frac{1}{6}(2n\pi+\alpha), \ \frac{1}{2}\{(2n+1)\pi-\alpha\}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xLv)\) \(\sin{\left(\frac{\pi}{4}\tan{\theta}\right)}=\cos{\left(\frac{\pi}{4}\cot{\theta}\right)}\)
উত্তরঃ \((4n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xLvi)\) \(1-2\sin{\theta}=\cos{\theta}\)
উত্তরঃ \(2n\pi, \ 2n\pi+2\alpha\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
বুয়েটঃ২০০৫-২০০৬ ।
\(Q.3.(xLvii)\) \(\sqrt{3}\cot^2{\theta}+4\cot{\theta}+\sqrt{3}=0\)
উত্তরঃ \(n\pi+\frac{2\pi}{3}, \ n\pi+\frac{5\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.3.(xLviii)\) \(4\sin^2{\theta}+\sqrt{3}=2(1+\sqrt{3})\sin{\theta}\)
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(i)\) \(\tan{x}+\tan{2x}+\tan{x}\tan{2x}=1\)উত্তরঃ \((4n+1)\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
মাঃ২০১২; রাঃ ২০০২; বঃ২০০৬; চঃ২০০৩; ঢাঃ২০০৯ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{x}+\tan{2x}+\tan{x}\tan{2x}=1\)
\(\Rightarrow \tan{x}+\tan{2x}=1-\tan{x}\tan{2x}\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=1\) ➜ উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=1\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow 3x=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=(4n+1)\frac{\pi}{4}\)
\(\therefore x=(4n+1)\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \((4n+1)\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{x}+\tan{2x}+\tan{x}\tan{2x}=1\)
\(\Rightarrow \tan{x}+\tan{2x}=1-\tan{x}\tan{2x}\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=1\) ➜ উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=1\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow 3x=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=(4n+1)\frac{\pi}{4}\)
\(\therefore x=(4n+1)\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \((4n+1)\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(ii)\) \(\sqrt{3}(\tan{x}+\tan{2x})+\tan{x}\tan{2x}=1\)উত্তরঃ \((6n+1)\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১১; রাঃ ২০১৪,২০১১,২০০৪,২০০০; দিঃ,চঃ২০১৩; সিঃ২০০৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{3}(\tan{x}+\tan{2x})+\tan{x}\tan{2x}=1\)
\(\Rightarrow \sqrt{3}(\tan{x}+\tan{2x})=1-\tan{x}\tan{2x}\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=\frac{1}{\sqrt{3}}\) ➜ উভয় পার্শে \(\sqrt{3}(1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=\frac{1}{\sqrt{3}}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}=\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow 3x=n\pi+\frac{\pi}{6}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=(6n+1)\frac{\pi}{6}\)
\(\therefore x=(6n+1)\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \((6n+1)\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sqrt{3}(\tan{x}+\tan{2x})+\tan{x}\tan{2x}=1\)
\(\Rightarrow \sqrt{3}(\tan{x}+\tan{2x})=1-\tan{x}\tan{2x}\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=\frac{1}{\sqrt{3}}\) ➜ উভয় পার্শে \(\sqrt{3}(1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=\frac{1}{\sqrt{3}}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}=\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow 3x=n\pi+\frac{\pi}{6}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=(6n+1)\frac{\pi}{6}\)
\(\therefore x=(6n+1)\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \((6n+1)\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(iii)\) \(\tan{\theta}+\tan{2\theta}+\sqrt{3}\tan{\theta}\tan{2\theta}=\sqrt{3}\)উত্তরঃ \((3n+1)\frac{\pi}{9}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ, রাঃ ২০০৬; চঃ২০১১,২০০৯; সিঃ২০১৪; বঃ২০০৯ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{\theta}+\tan{2\theta}+\sqrt{3}\tan{\theta}\tan{2\theta}=\sqrt{3}\)
\(\Rightarrow \tan{\theta}+\tan{2\theta}=\sqrt{3}-\sqrt{3}\tan{\theta}\tan{2\theta}\)
\(\Rightarrow \tan{\theta}+\tan{2\theta}=\sqrt{3}(1-\tan{\theta}\tan{2\theta})\)
\(\Rightarrow \frac{\tan{\theta}+\tan{2\theta}}{1-\tan{\theta}\tan{2\theta}}=\sqrt{3}\) ➜ উভয় পার্শে \((1-\tan{\theta}\tan{2\theta})\) ভাগ করে,
\(\Rightarrow \tan{(\theta+2\theta)}=\sqrt{3}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3\theta}=\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow 3\theta=n\pi+\frac{\pi}{3}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=(3n+1)\frac{\pi}{3}\)
\(\therefore x=(3n+1)\frac{\pi}{9}\)
\(\therefore\) সাধারণ সমাধান, \((3n+1)\frac{\pi}{9}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{\theta}+\tan{2\theta}+\sqrt{3}\tan{\theta}\tan{2\theta}=\sqrt{3}\)
\(\Rightarrow \tan{\theta}+\tan{2\theta}=\sqrt{3}-\sqrt{3}\tan{\theta}\tan{2\theta}\)
\(\Rightarrow \tan{\theta}+\tan{2\theta}=\sqrt{3}(1-\tan{\theta}\tan{2\theta})\)
\(\Rightarrow \frac{\tan{\theta}+\tan{2\theta}}{1-\tan{\theta}\tan{2\theta}}=\sqrt{3}\) ➜ উভয় পার্শে \((1-\tan{\theta}\tan{2\theta})\) ভাগ করে,
\(\Rightarrow \tan{(\theta+2\theta)}=\sqrt{3}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3\theta}=\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow 3\theta=n\pi+\frac{\pi}{3}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=(3n+1)\frac{\pi}{3}\)
\(\therefore x=(3n+1)\frac{\pi}{9}\)
\(\therefore\) সাধারণ সমাধান, \((3n+1)\frac{\pi}{9}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(iv)\) \(\tan{x}+\tan{2x}+\tan{3x}=\tan{x}\tan{2x}\tan{3x}\)উত্তরঃ \(\frac{n\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৫; মাঃ২০০৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{x}+\tan{2x}+\tan{3x}=\tan{x}\tan{2x}\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=\tan{x}\tan{2x}\tan{3x}-\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=-\tan{3x}\) ➜ উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=-\tan{3x}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}+\tan{3x}=0\)
\(\Rightarrow 2\tan{3x}=0\)
\(\Rightarrow \tan{3x}=0\)
\(\Rightarrow 3x=n\pi\) ➜ \(\because \tan{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\frac{n\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{x}+\tan{2x}+\tan{3x}=\tan{x}\tan{2x}\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=\tan{x}\tan{2x}\tan{3x}-\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=-\tan{3x}\) ➜ উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=-\tan{3x}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}+\tan{3x}=0\)
\(\Rightarrow 2\tan{3x}=0\)
\(\Rightarrow \tan{3x}=0\)
\(\Rightarrow 3x=n\pi\) ➜ \(\because \tan{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\frac{n\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(v)\) \(\cot{x}+\cot{2x}+\cot{3x}=\cot{x}\cot{2x}\cot{3x}\)উত্তরঃ \(\frac{n\pi}{3}\pm\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
বুয়েটঃ২০১৪-২০১৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{x}+\cot{2x}+\cot{3x}=\cot{x}\cot{2x}\cot{3x}\)
\(\Rightarrow \cot{x}\cot{2x}\cot{3x}=\cot{x}+\cot{2x}+\cot{3x}\)
\(\Rightarrow \cot{x}\cot{2x}\cot{3x}-\cot{3x}=\cot{x}+\cot{2x}\)
\(\Rightarrow \cot{3x}(\cot{x}\cot{2x}-1)=\cot{x}+\cot{2x}\)
\(\Rightarrow \frac{\cot{x}\cot{2x}-1}{\cot{x}+\cot{2x}}=\frac{1}{\cot{3x}}\) ➜ উভয় পার্শে \(\cot{3x}(\cot{x}+\cot{2x})\) ভাগ করে,
\(\Rightarrow \frac{\cot{x}\cot{2x}-1}{\cot{2x}+\cot{x}}=\tan{3x}\) ➜ \(\because \frac{1}{\cot{3x}}=\tan{3x}\)
\(\Rightarrow \cot{(x+2x)}=\tan{3x}\) ➜ \(\because \frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}=\cot{(A+B)}\)
\(\Rightarrow \cot{3x}=\tan{3x}\)
\(\Rightarrow \frac{1}{\tan{3x}}=\tan{3x}\) ➜ \(\because \cot{3x}=\frac{1}{\tan{3x}}\)
\(\Rightarrow \tan^2{3x}=1\)
\(\Rightarrow \tan{3x}=\pm\sqrt{1}\)
\(\Rightarrow \tan{3x}=\pm1\)
\(\Rightarrow \tan{3x}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{3x}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow 3x=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=n\pi\pm\frac{\pi}{4}\)
\(\therefore x=\frac{n\pi}{3}\pm\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\frac{n\pi}{3}\pm\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cot{x}+\cot{2x}+\cot{3x}=\cot{x}\cot{2x}\cot{3x}\)
\(\Rightarrow \cot{x}\cot{2x}\cot{3x}=\cot{x}+\cot{2x}+\cot{3x}\)
\(\Rightarrow \cot{x}\cot{2x}\cot{3x}-\cot{3x}=\cot{x}+\cot{2x}\)
\(\Rightarrow \cot{3x}(\cot{x}\cot{2x}-1)=\cot{x}+\cot{2x}\)
\(\Rightarrow \frac{\cot{x}\cot{2x}-1}{\cot{x}+\cot{2x}}=\frac{1}{\cot{3x}}\) ➜ উভয় পার্শে \(\cot{3x}(\cot{x}+\cot{2x})\) ভাগ করে,
\(\Rightarrow \frac{\cot{x}\cot{2x}-1}{\cot{2x}+\cot{x}}=\tan{3x}\) ➜ \(\because \frac{1}{\cot{3x}}=\tan{3x}\)
\(\Rightarrow \cot{(x+2x)}=\tan{3x}\) ➜ \(\because \frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}=\cot{(A+B)}\)
\(\Rightarrow \cot{3x}=\tan{3x}\)
\(\Rightarrow \frac{1}{\tan{3x}}=\tan{3x}\) ➜ \(\because \cot{3x}=\frac{1}{\tan{3x}}\)
\(\Rightarrow \tan^2{3x}=1\)
\(\Rightarrow \tan{3x}=\pm\sqrt{1}\)
\(\Rightarrow \tan{3x}=\pm1\)
\(\Rightarrow \tan{3x}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{3x}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow 3x=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=n\pi\pm\frac{\pi}{4}\)
\(\therefore x=\frac{n\pi}{3}\pm\frac{\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\frac{n\pi}{3}\pm\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(vi)\) \(\sin{\theta}+\cos{\theta}=1\)উত্তরঃ \(2n\pi, \ (4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
দিঃ২০১৬; চঃ২০১০; সিঃ২০০৯; ঢাঃ২০০৫; মাঃ২০১৪ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{\theta}+\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}+\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}+\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}+\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta-\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \theta-\frac{\pi}{4}=2n\pi\pm{\frac{\pi}{4}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{4}}+\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi-\frac{\pi}{4}+\frac{\pi}{4}, \ 2n\pi+\frac{\pi}{4}+\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi+\frac{2\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi+\frac{\pi}{2}\)
\(\therefore \theta=2n\pi, \ (4n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ (4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{\theta}+\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}+\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}+\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}+\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta-\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \theta-\frac{\pi}{4}=2n\pi\pm{\frac{\pi}{4}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{4}}+\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi-\frac{\pi}{4}+\frac{\pi}{4}, \ 2n\pi+\frac{\pi}{4}+\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi+\frac{2\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi+\frac{\pi}{2}\)
\(\therefore \theta=2n\pi, \ (4n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ (4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(vii)\) \(\cos{x}-\sin{x}=1\)উত্তরঃ \(2n\pi, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০০১ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{x}-\sin{x}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \theta+\frac{\pi}{4}=2n\pi\pm{\frac{\pi}{4}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{4}}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{4}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi-\frac{2\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi-\frac{\pi}{2}\)
\(\therefore \theta=2n\pi, \ (4n-1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{x}-\sin{x}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{4}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \theta+\frac{\pi}{4}=2n\pi\pm{\frac{\pi}{4}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{4}}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}-\frac{\pi}{4}, \ 2n\pi-\frac{\pi}{4}-\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi-\frac{2\pi}{4}\)
\(\Rightarrow \theta=2n\pi, \ 2n\pi-\frac{\pi}{2}\)
\(\therefore \theta=2n\pi, \ (4n-1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ (4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(viii)\) \(\sin{x}+\sqrt{3}\cos{x}=\sqrt{3}\)উত্তরঃ \(2n\pi, \ (6n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৫; বঃ২০১২,২০০৫; বিআইটিঃ২০০১-২০০২ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}+\sqrt{3}\cos{x}=\sqrt{3}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{3}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{3}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{\theta}\sin{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{6}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm{\frac{\pi}{6}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm{\frac{\pi}{6}}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi-\frac{\pi}{6}+\frac{\pi}{6}, \ 2n\pi+\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi, \ 2n\pi+\frac{2\pi}{6}\)
\(\Rightarrow x=2n\pi, \ 2n\pi+\frac{\pi}{3}\)
\(\therefore x=2n\pi, \ (6n+1)\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ (6n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{x}+\sqrt{3}\cos{x}=\sqrt{3}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{3}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{3}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{\theta}\sin{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{6}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm{\frac{\pi}{6}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm{\frac{\pi}{6}}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi-\frac{\pi}{6}+\frac{\pi}{6}, \ 2n\pi+\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi, \ 2n\pi+\frac{2\pi}{6}\)
\(\Rightarrow x=2n\pi, \ 2n\pi+\frac{\pi}{3}\)
\(\therefore x=2n\pi, \ (6n+1)\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ (6n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(ix)\) \(\sin{x}+\sqrt{3}\cos{x}=\sqrt{2}\)উত্তরঃ \(2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{5\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
ঢাঃ২০০৭; রাঃ২০০৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}+\sqrt{3}\cos{x}=\sqrt{2}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{2}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{\theta}\sin{\frac{\pi}{6}}=\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{\theta}\sin{\frac{\pi}{6}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm{\frac{\pi}{4}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm{\frac{\pi}{4}}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi-\frac{\pi}{4}+\frac{\pi}{6}, \ 2n\pi+\frac{\pi}{4}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi-\left(\frac{\pi}{4}-\frac{\pi}{6}\right), \ 2n\pi+\frac{3\pi+2\pi}{12}\)
\(\Rightarrow x=2n\pi-\frac{3\pi-2\pi}{12}, \ 2n\pi+\frac{3\pi+2\pi}{12}\)
\(\therefore x=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{5\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{5\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{x}+\sqrt{3}\cos{x}=\sqrt{2}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{2}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{2}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{\theta}\sin{\frac{\pi}{6}}=\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{\theta}\sin{\frac{\pi}{6}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm{\frac{\pi}{4}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm{\frac{\pi}{4}}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi-\frac{\pi}{4}+\frac{\pi}{6}, \ 2n\pi+\frac{\pi}{4}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi-\left(\frac{\pi}{4}-\frac{\pi}{6}\right), \ 2n\pi+\frac{3\pi+2\pi}{12}\)
\(\Rightarrow x=2n\pi-\frac{3\pi-2\pi}{12}, \ 2n\pi+\frac{3\pi+2\pi}{12}\)
\(\therefore x=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{5\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{5\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(x)\) \(\cos{\theta}+\sqrt{3}\sin{\theta}=2\)উত্তরঃ \(2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}+\sqrt{3}\sin{\theta}=2\)
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}+\sin{\theta}\sin{\frac{\pi}{3}}=1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{3}\right)}=1\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \theta-\frac{\pi}{3}=2n\pi\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi+\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{\theta}+\sqrt{3}\sin{\theta}=2\)
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}+\sin{\theta}\sin{\frac{\pi}{3}}=1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{3}\right)}=1\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \theta-\frac{\pi}{3}=2n\pi\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi+\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xi)\) \(\sin{x}+\cos{x}=\frac{1}{\sqrt{2}}\)উত্তরঃ \(2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০০২ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}+\cos{x}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}+\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}\times\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}+\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta-\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta-\frac{\pi}{4}=2n\pi\pm{\frac{\pi}{3}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{3}}+\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{4\pi+3\pi}{12}, \ 2n\pi-\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\)
\(\Rightarrow \theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi-\frac{4\pi-3\pi}{12}\)
\(\Rightarrow \theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi-\frac{\pi}{12}\)
\(\therefore \theta=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{x}+\cos{x}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}+\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}\times\sqrt{2}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}+\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\theta-\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta-\frac{\pi}{4}=2n\pi\pm{\frac{\pi}{3}}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\frac{\pi}{3}}+\frac{\pi}{4}\)
\(\Rightarrow \theta=2n\pi+\frac{4\pi+3\pi}{12}, \ 2n\pi-\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\)
\(\Rightarrow \theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi-\frac{4\pi-3\pi}{12}\)
\(\Rightarrow \theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi-\frac{\pi}{12}\)
\(\therefore \theta=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi-\frac{\pi}{12}, \ 2n\pi+\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xii)\) \(\cos{\theta}+2\sin{\theta}=1\)উত্তরঃ \(2n\pi, \ 2n\pi+2\alpha\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
চঃ২০০৪; বুয়েটঃ২০০৫-২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}+2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{5}}+\sin{\theta}\frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+2^2}=\sqrt{5}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha}=\cos{\alpha}\) ➜ \(\because \cos{\alpha}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow \sin{\alpha}=\frac{2}{\sqrt{5}}\)
\(\Rightarrow \cos{\left(\theta-\alpha\right)}=\cos{\alpha}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \theta-\alpha=2n\pi\pm{\alpha}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\alpha}+\alpha\)
\(\Rightarrow \theta=2n\pi-{\alpha}+\alpha, \ 2n\pi+{\alpha}+\alpha\)
\(\therefore \theta=2n\pi, \ 2n\pi+2{\alpha}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ 2n\pi+2{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
\(\cos{\theta}+2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{5}}+\sin{\theta}\frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+2^2}=\sqrt{5}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha}=\cos{\alpha}\) ➜ \(\because \cos{\alpha}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow \sin{\alpha}=\frac{2}{\sqrt{5}}\)
\(\Rightarrow \cos{\left(\theta-\alpha\right)}=\cos{\alpha}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \theta-\alpha=2n\pi\pm{\alpha}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\alpha}+\alpha\)
\(\Rightarrow \theta=2n\pi-{\alpha}+\alpha, \ 2n\pi+{\alpha}+\alpha\)
\(\therefore \theta=2n\pi, \ 2n\pi+2{\alpha}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ 2n\pi+2{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xiii)\) \(\cos{\theta}-\cos{9\theta}=\sin{5\theta}\)উত্তরঃ \(\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০১৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+9\theta}{2}}\sin{\frac{9\theta-\theta}{2}}=\sin{5\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{10\theta}{2}}\sin{\frac{8\theta}{2}}=\sin{5\theta}\)
\(\Rightarrow 2\sin{5\theta}\sin{4\theta}-\sin{5\theta}=0\)
\(\Rightarrow \sin{5\theta}(2\sin{4\theta}-1)=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}-1=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}=1\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 5\theta=n\pi, \ 4\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{5}, \ \theta=\frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore \theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+9\theta}{2}}\sin{\frac{9\theta-\theta}{2}}=\sin{5\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{10\theta}{2}}\sin{\frac{8\theta}{2}}=\sin{5\theta}\)
\(\Rightarrow 2\sin{5\theta}\sin{4\theta}-\sin{5\theta}=0\)
\(\Rightarrow \sin{5\theta}(2\sin{4\theta}-1)=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}-1=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}=1\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 5\theta=n\pi, \ 4\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{5}, \ \theta=\frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore \theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xiv)\) \(\cos{x}+\cos{2x}+\cos{3x}=0\)উত্তরঃ \((2n+1)\frac{\pi}{4}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৩; ঢাঃ২০০৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{x}+\cos{2x}+\cos{3x}=0\)
\(\Rightarrow \cos{3x}+\cos{x}+\cos{2x}=0\)
\(\Rightarrow 2\cos{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}+\cos{2x}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{4x}{2}}\cos{\frac{2x}{2}}+\cos{2x}=0\)
\(\Rightarrow 2\cos{2x}\cos{x}+\cos{2x}=0\)
\(\Rightarrow \cos{2x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{x}+\cos{2x}+\cos{3x}=0\)
\(\Rightarrow \cos{3x}+\cos{x}+\cos{2x}=0\)
\(\Rightarrow 2\cos{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}+\cos{2x}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{4x}{2}}\cos{\frac{2x}{2}}+\cos{2x}=0\)
\(\Rightarrow 2\cos{2x}\cos{x}+\cos{2x}=0\)
\(\Rightarrow \cos{2x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xv)\) \(\sin{x}+\sin{2x}+\sin{3x}=0\)উত্তরঃ \(\frac{n\pi}{2}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০১৩; রাঃ২০০৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}+\sin{2x}+\sin{3x}=0\)
\(\Rightarrow \sin{3x}+\sin{x}+\sin{2x}=0\)
\(\Rightarrow 2\sin{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}+\sin{2x}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\sin{\frac{4x}{2}}\cos{\frac{2x}{2}}+\sin{2x}=0\)
\(\Rightarrow 2\sin{2x}\cos{x}+\sin{2x}=0\)
\(\Rightarrow \sin{2x}(2\cos{x}+1)=0\)
\(\Rightarrow \sin{2x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \sin{2x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \sin{2x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \sin{2x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=n\pi, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{x}+\sin{2x}+\sin{3x}=0\)
\(\Rightarrow \sin{3x}+\sin{x}+\sin{2x}=0\)
\(\Rightarrow 2\sin{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}+\sin{2x}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\sin{\frac{4x}{2}}\cos{\frac{2x}{2}}+\sin{2x}=0\)
\(\Rightarrow 2\sin{2x}\cos{x}+\sin{2x}=0\)
\(\Rightarrow \sin{2x}(2\cos{x}+1)=0\)
\(\Rightarrow \sin{2x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \sin{2x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \sin{2x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \sin{2x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=n\pi, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xvi)\) \(\sqrt{2}\cos{3\theta}-\cos{\theta}=\cos{5\theta}\)উত্তরঃ \((2n+1)\frac{\pi}{6}, \ n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৩; সিঃ২০১০; কুঃ,চঃ২০০৮ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{2}\cos{3\theta}-\cos{\theta}=\cos{5\theta}\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-\cos{\theta}-\cos{5\theta}=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-(\cos{5\theta}+\cos{\theta})=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-2\cos{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-2\cos{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-2\cos{3\theta}\cos{2\theta}=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}(1-\sqrt{2}\cos{2\theta})=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}=0, \ 1-\sqrt{2}\cos{2\theta}=0\)
\(\Rightarrow \cos{3\theta}=0, \ -\sqrt{2}\cos{2\theta}=-1\)
\(\Rightarrow \cos{3\theta}=0, \ \sqrt{2}\cos{2\theta}=1\)
\(\Rightarrow \cos{3\theta}=0, \ \cos{2\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{3\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow 3\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{6}, \ \theta=n\pi\pm\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{6}, \ \theta=n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sqrt{2}\cos{3\theta}-\cos{\theta}=\cos{5\theta}\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-\cos{\theta}-\cos{5\theta}=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-(\cos{5\theta}+\cos{\theta})=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-2\cos{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-2\cos{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}-2\cos{3\theta}\cos{2\theta}=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}(1-\sqrt{2}\cos{2\theta})=0\)
\(\Rightarrow \sqrt{2}\cos{3\theta}=0, \ 1-\sqrt{2}\cos{2\theta}=0\)
\(\Rightarrow \cos{3\theta}=0, \ -\sqrt{2}\cos{2\theta}=-1\)
\(\Rightarrow \cos{3\theta}=0, \ \sqrt{2}\cos{2\theta}=1\)
\(\Rightarrow \cos{3\theta}=0, \ \cos{2\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{3\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow 3\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{6}, \ \theta=n\pi\pm\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{6}, \ \theta=n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xvii)\) \(\sin{7\theta}-\sqrt{3}\cos{4\theta}=\sin{\theta}\)উত্তরঃ \((2n+1)\frac{\pi}{8}, \ \{3n+(-1)^n\}\frac{\pi}{9}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০০৫; দিঃ২০১০ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{7\theta}-\sqrt{3}\cos{4\theta}=\sin{\theta}\)
\(\Rightarrow \sin{7\theta}-\sin{\theta}-\sqrt{3}\cos{4\theta}=0\)
\(\Rightarrow 2\cos{\frac{7\theta+\theta}{2}}\sin{\frac{7\theta-\theta}{2}}-\sqrt{3}\cos{4\theta}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{8\theta}{2}}\sin{\frac{6\theta}{2}}-\sqrt{3}\cos{4\theta}=0\)
\(\Rightarrow 2\cos{4\theta}\sin{3\theta}-\sqrt{3}\cos{4\theta}=0\)
\(\Rightarrow \cos{4\theta}(2\sin{3\theta}-\sqrt{3})=0\)
\(\Rightarrow \cos{4\theta}=0, \ 2\sin{3\theta}-\sqrt{3}=0\)
\(\Rightarrow \cos{4\theta}=0, \ 2\sin{3\theta}=\sqrt{3}\)
\(\Rightarrow \cos{4\theta}=0, \ \sin{3\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{4\theta}=0, \ \sin{3\theta}=\sin{\frac{\pi}{3}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow 4\theta=(2n+1)\frac{\pi}{2}, \ 3\theta=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{8}, \ 3\theta=\{3n+(-1)^n\}\frac{\pi}{3}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{8}, \ \theta=\{3n+(-1)^n\}\frac{\pi}{9}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}, \ \{3n+(-1)^n\}\frac{\pi}{9}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{6}, \ \theta=n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{7\theta}-\sqrt{3}\cos{4\theta}=\sin{\theta}\)
\(\Rightarrow \sin{7\theta}-\sin{\theta}-\sqrt{3}\cos{4\theta}=0\)
\(\Rightarrow 2\cos{\frac{7\theta+\theta}{2}}\sin{\frac{7\theta-\theta}{2}}-\sqrt{3}\cos{4\theta}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{8\theta}{2}}\sin{\frac{6\theta}{2}}-\sqrt{3}\cos{4\theta}=0\)
\(\Rightarrow 2\cos{4\theta}\sin{3\theta}-\sqrt{3}\cos{4\theta}=0\)
\(\Rightarrow \cos{4\theta}(2\sin{3\theta}-\sqrt{3})=0\)
\(\Rightarrow \cos{4\theta}=0, \ 2\sin{3\theta}-\sqrt{3}=0\)
\(\Rightarrow \cos{4\theta}=0, \ 2\sin{3\theta}=\sqrt{3}\)
\(\Rightarrow \cos{4\theta}=0, \ \sin{3\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{4\theta}=0, \ \sin{3\theta}=\sin{\frac{\pi}{3}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow 4\theta=(2n+1)\frac{\pi}{2}, \ 3\theta=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{8}, \ 3\theta=\{3n+(-1)^n\}\frac{\pi}{3}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{8}, \ \theta=\{3n+(-1)^n\}\frac{\pi}{9}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}, \ \{3n+(-1)^n\}\frac{\pi}{9}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{6}, \ \theta=n\pi\pm\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xviii)\) \(\cos{7\theta}=\cos{3\theta}+\sin{5\theta}\)উত্তরঃ \(\frac{n\pi}{5}, \ \frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০০৫; বঃ২০০৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{7\theta}=\cos{3\theta}+\sin{5\theta}\)
\(\Rightarrow \cos{7\theta}-\cos{3\theta}=\sin{5\theta}\)
\(\Rightarrow -(\cos{3\theta}-\cos{7\theta})=\sin{5\theta}\)
\(\Rightarrow -2\sin{\frac{3\theta+7\theta}{2}}\sin{\frac{7\theta-3\theta}{2}}=\sin{5\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow -2\sin{\frac{10\theta}{2}}\sin{\frac{4\theta}{2}}=\sin{5\theta}\)
\(\Rightarrow -2\sin{5\theta}\sin{2\theta}-\sin{5\theta}=0\)
\(\Rightarrow -\sin{5\theta}(2\sin{2\theta}+1)=0\)
\(\Rightarrow -\sin{5\theta}=0, \ 2\sin{2\theta}+1=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{2\theta}=-1\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{2\theta}=-\frac{1}{2}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{2\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{2\theta}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\sin{\frac{\pi}{6}}=\sin{\left(\pi+\frac{\pi}{6}\right)}\)
\(\Rightarrow 5\theta=n\pi, \ 2\theta=n\pi+(-1)^n\left(\pi+\frac{\pi}{6}\right)\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{5}, \ 2\theta=n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore \theta=\frac{n\pi}{5}, \ \theta=\frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{5}, \ \frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{7\theta}=\cos{3\theta}+\sin{5\theta}\)
\(\Rightarrow \cos{7\theta}-\cos{3\theta}=\sin{5\theta}\)
\(\Rightarrow -(\cos{3\theta}-\cos{7\theta})=\sin{5\theta}\)
\(\Rightarrow -2\sin{\frac{3\theta+7\theta}{2}}\sin{\frac{7\theta-3\theta}{2}}=\sin{5\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow -2\sin{\frac{10\theta}{2}}\sin{\frac{4\theta}{2}}=\sin{5\theta}\)
\(\Rightarrow -2\sin{5\theta}\sin{2\theta}-\sin{5\theta}=0\)
\(\Rightarrow -\sin{5\theta}(2\sin{2\theta}+1)=0\)
\(\Rightarrow -\sin{5\theta}=0, \ 2\sin{2\theta}+1=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{2\theta}=-1\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{2\theta}=-\frac{1}{2}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{2\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{2\theta}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\sin{\frac{\pi}{6}}=\sin{\left(\pi+\frac{\pi}{6}\right)}\)
\(\Rightarrow 5\theta=n\pi, \ 2\theta=n\pi+(-1)^n\left(\pi+\frac{\pi}{6}\right)\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{5}, \ 2\theta=n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore \theta=\frac{n\pi}{5}, \ \theta=\frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{5}, \ \frac{n\pi}{2}+(-1)^n\frac{7\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xix)\) \(\cos{\theta}-\cos{7\theta}=\sin{4\theta}\)উত্তরঃ \(\frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
ঢাঃ২০১৪,২০১০; চঃ২০১৪; বঃ,মাঃ২০১১; দিঃ২০০৯; যঃ২০০৮; কুঃ২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}-\cos{7\theta}=\sin{4\theta}\)
\(\Rightarrow \cos{\theta}-\cos{7\theta}=\sin{4\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+7\theta}{2}}\sin{\frac{7\theta-\theta}{2}}=\sin{4\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{8\theta}{2}}\sin{\frac{6\theta}{2}}=\sin{4\theta}\)
\(\Rightarrow 2\sin{4\theta}\sin{3\theta}-\sin{4\theta}=0\)
\(\Rightarrow \sin{4\theta}(2\sin{3\theta}-1)=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}-1=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}=1\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 4\theta=n\pi, \ 3\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}, \ \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{\theta}-\cos{7\theta}=\sin{4\theta}\)
\(\Rightarrow \cos{\theta}-\cos{7\theta}=\sin{4\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+7\theta}{2}}\sin{\frac{7\theta-\theta}{2}}=\sin{4\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{8\theta}{2}}\sin{\frac{6\theta}{2}}=\sin{4\theta}\)
\(\Rightarrow 2\sin{4\theta}\sin{3\theta}-\sin{4\theta}=0\)
\(\Rightarrow \sin{4\theta}(2\sin{3\theta}-1)=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}-1=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}=1\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 4\theta=n\pi, \ 3\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}, \ \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xx)\) \(\sin{3\theta}+\sin{5\theta}+\sin{7\theta}+\sin{9\theta}=0\)উত্তরঃ \(\theta=\frac{n\pi}{6}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০০৩ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{3\theta}+\sin{5\theta}+\sin{7\theta}+\sin{9\theta}=0\)
\(\Rightarrow \sin{9\theta}+\sin{3\theta}+\sin{7\theta}+\sin{5\theta}=0\)
\(\Rightarrow 2\sin{\frac{9\theta+3\theta}{2}}\cos{\frac{9\theta-3\theta}{2}}+2\sin{\frac{7\theta+5\theta}{2}}\cos{\frac{7\theta-5\theta}{2}}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{12\theta}{2}}\cos{\frac{6\theta}{2}}+2\sin{\frac{12\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\sin{6\theta}\cos{3\theta}+2\sin{6\theta}\cos{\theta}=0\)
\(\Rightarrow 2\sin{6\theta}(\cos{3\theta}+\cos{\theta})=0\)
\(\Rightarrow 2\sin{6\theta}=0, \ \cos{3\theta}+\cos{\theta}=0\)
\(\Rightarrow 6\theta=n\pi, \ 2\cos{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(\sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ 2\cos{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ \cos{2\theta}\cos{\theta}=0\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ \cos{2\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ 2\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=\frac{n\pi}{6}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{6}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{3\theta}+\sin{5\theta}+\sin{7\theta}+\sin{9\theta}=0\)
\(\Rightarrow \sin{9\theta}+\sin{3\theta}+\sin{7\theta}+\sin{5\theta}=0\)
\(\Rightarrow 2\sin{\frac{9\theta+3\theta}{2}}\cos{\frac{9\theta-3\theta}{2}}+2\sin{\frac{7\theta+5\theta}{2}}\cos{\frac{7\theta-5\theta}{2}}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{12\theta}{2}}\cos{\frac{6\theta}{2}}+2\sin{\frac{12\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\sin{6\theta}\cos{3\theta}+2\sin{6\theta}\cos{\theta}=0\)
\(\Rightarrow 2\sin{6\theta}(\cos{3\theta}+\cos{\theta})=0\)
\(\Rightarrow 2\sin{6\theta}=0, \ \cos{3\theta}+\cos{\theta}=0\)
\(\Rightarrow 6\theta=n\pi, \ 2\cos{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(\sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ 2\cos{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ \cos{2\theta}\cos{\theta}=0\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ \cos{2\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ 2\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{6}, \ \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=\frac{n\pi}{6}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{6}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxi)\) \(\sin^2{2x}-3\cos^2{x}=0\)উত্তরঃ \((2n+1)\frac{\pi}{2}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সিঃ২০১১; কুঃ২০১০ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin^2{2x}-3\cos^2{x}=0\)
\(\Rightarrow \sin^2{2x}-\frac{3}{2}\times2\cos^2{x}=0\)
\(\Rightarrow 1-\cos^2{2x}-\frac{3}{2}(1+\cos{2x})=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
এবং \(2\cos^2{A}=1+\cos{2A}\)
\(\Rightarrow -2+2\cos^2{2x}+3(1+\cos{2x})=0\) ➜ উভয় পার্শে \(-2\) গুণ করে,
\(\Rightarrow 2\cos^2{2x}+3+3\cos{2x}-2=0\)
\(\Rightarrow 2\cos^2{2x}+3\cos{2x}+1=0\)
\(\Rightarrow 2\cos^2{2x}+2\cos{2x}+\cos{2x}+1=0\)
\(\Rightarrow 2\cos{2x}(\cos{2x}+1)+1(\cos{2x}+1)=0\)
\(\Rightarrow (\cos{2x}+1)(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{2x}+1=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{2x}=-1, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{2x}=-1, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=-1, \ \cos{2x}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\pi, \ 2x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{2}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{2}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin^2{2x}-3\cos^2{x}=0\)
\(\Rightarrow \sin^2{2x}-\frac{3}{2}\times2\cos^2{x}=0\)
\(\Rightarrow 1-\cos^2{2x}-\frac{3}{2}(1+\cos{2x})=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
এবং \(2\cos^2{A}=1+\cos{2A}\)
\(\Rightarrow -2+2\cos^2{2x}+3(1+\cos{2x})=0\) ➜ উভয় পার্শে \(-2\) গুণ করে,
\(\Rightarrow 2\cos^2{2x}+3+3\cos{2x}-2=0\)
\(\Rightarrow 2\cos^2{2x}+3\cos{2x}+1=0\)
\(\Rightarrow 2\cos^2{2x}+2\cos{2x}+\cos{2x}+1=0\)
\(\Rightarrow 2\cos{2x}(\cos{2x}+1)+1(\cos{2x}+1)=0\)
\(\Rightarrow (\cos{2x}+1)(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{2x}+1=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{2x}=-1, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{2x}=-1, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=-1, \ \cos{2x}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\pi, \ 2x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{2}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{2}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxii)\) \(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)উত্তরঃ \(2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
রাঃ২০১৪; কুঃ২০০৫; চুয়েটঃ২০১১-২০১২ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)
\(\Rightarrow \sec^2{\theta}-1-2\sqrt{3}\sec{\theta}+4=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow \sec^2{\theta}-2\sqrt{3}\sec{\theta}+3=0\)
\(\Rightarrow (\sec{\theta}-\sqrt{3})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sec{\theta}-\sqrt{3}=0\)
\(\Rightarrow \sec{\theta}=\sqrt{3}\)
\(\Rightarrow \sec{\theta}=\sec{\alpha}\) যেখানে, \(\sec{\alpha}=\sqrt{3}\)
\(\therefore \theta=2n\pi\pm\alpha\) ➜ \(\because \sec{A}=\sec{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
\(\tan^2{\theta}-2\sqrt{3}\sec{\theta}+4=0\)
\(\Rightarrow \sec^2{\theta}-1-2\sqrt{3}\sec{\theta}+4=0\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow \sec^2{\theta}-2\sqrt{3}\sec{\theta}+3=0\)
\(\Rightarrow (\sec{\theta}-\sqrt{3})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sec{\theta}-\sqrt{3}=0\)
\(\Rightarrow \sec{\theta}=\sqrt{3}\)
\(\Rightarrow \sec{\theta}=\sec{\alpha}\) যেখানে, \(\sec{\alpha}=\sqrt{3}\)
\(\therefore \theta=2n\pi\pm\alpha\) ➜ \(\because \sec{A}=\sec{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\alpha\) যেখানে, \(\sec{\alpha}=\sqrt{3}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxiii)\) \(\tan^2{x}+\cot^2{x}=2\)উত্তরঃ \(n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০০৭ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^2{x}+\cot^2{x}=2\)
\(\Rightarrow \tan^2{x}+\frac{1}{\tan^2{x}}=2\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(\Rightarrow \frac{\tan^4{x}+1}{\tan^2{x}}=2\)
\(\Rightarrow \tan^4{x}+1=2\tan^2{x}\)
\(\Rightarrow \tan^4{x}-2\tan^2{x}+1=0\)
\(\Rightarrow (\tan^2{x}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \tan^2{x}=1\)
\(\Rightarrow \tan{x}=\pm\sqrt{1}\)
\(\Rightarrow \tan{x}=\pm1\)
\(\Rightarrow \tan{x}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{x}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow x=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan^2{x}+\cot^2{x}=2\)
\(\Rightarrow \tan^2{x}+\frac{1}{\tan^2{x}}=2\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(\Rightarrow \frac{\tan^4{x}+1}{\tan^2{x}}=2\)
\(\Rightarrow \tan^4{x}+1=2\tan^2{x}\)
\(\Rightarrow \tan^4{x}-2\tan^2{x}+1=0\)
\(\Rightarrow (\tan^2{x}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \tan^2{x}=1\)
\(\Rightarrow \tan{x}=\pm\sqrt{1}\)
\(\Rightarrow \tan{x}=\pm1\)
\(\Rightarrow \tan{x}=\pm\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \tan{x}=\tan{\left(\pm\frac{\pi}{4}\right)}\)
\(\Rightarrow x=n\pi+\left(\pm\frac{\pi}{4}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxiv)\) \(\cos{2x}+\sin{x}=1\)উত্তরঃ \(n\pi, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(\ n\in{\mathbb{Z}}\)
ঢাঃ২০০৪ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{2x}+\sin{x}=1\)
\(\Rightarrow 1-2\sin^2{x}+\sin{x}=1\) ➜ \(\because \cos{2A}=1-2\sin^2{A}\)
\(\Rightarrow 1-2\sin^2{x}+\sin{x}-1=0\)
\(\Rightarrow -2\sin^2{x}+\sin{x}=0\)
\(\Rightarrow -\sin{x}(2\sin{x}-1)=0\)
\(\Rightarrow \sin{x}=0, \ 2\sin{x}-1=0\)
\(\Rightarrow \sin{x}=0, \ 2\sin{x}=1\)
\(\Rightarrow \sin{x}=0, \ \sin{x}=\frac{1}{2}\)
\(\Rightarrow \sin{x}=0, \ \sin{x}=\sin{\frac{\pi}{6}}\)
\(\therefore x=n\pi, \ x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{2x}+\sin{x}=1\)
\(\Rightarrow 1-2\sin^2{x}+\sin{x}=1\) ➜ \(\because \cos{2A}=1-2\sin^2{A}\)
\(\Rightarrow 1-2\sin^2{x}+\sin{x}-1=0\)
\(\Rightarrow -2\sin^2{x}+\sin{x}=0\)
\(\Rightarrow -\sin{x}(2\sin{x}-1)=0\)
\(\Rightarrow \sin{x}=0, \ 2\sin{x}-1=0\)
\(\Rightarrow \sin{x}=0, \ 2\sin{x}=1\)
\(\Rightarrow \sin{x}=0, \ \sin{x}=\frac{1}{2}\)
\(\Rightarrow \sin{x}=0, \ \sin{x}=\sin{\frac{\pi}{6}}\)
\(\therefore x=n\pi, \ x=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxv)\) \(\sec^2{\theta}+\tan^2{\theta}=3\tan{\theta}\)উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{2}, \ n\in{\mathbb{Z}}\)
কুঃ২০১২; দিঃ২০০৯ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sec^2{\theta}+\tan^2{\theta}=3\tan{\theta}\)
\(\Rightarrow 1+\tan^2{\theta}+\tan^2{\theta}=3\tan{\theta}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{\theta}-3\tan{\theta}+1=0\)
\(\Rightarrow 2\tan^2{\theta}-2\tan{\theta}-\tan{\theta}+1=0\)
\(\Rightarrow 2\tan{\theta}(\tan{\theta}-1)-1(\tan{\theta}-1)=0\)
\(\Rightarrow (\tan{\theta}-1)(2\tan{\theta}-1)=0\)
\(\Rightarrow \tan{\theta}-1=0, \ 2\tan{\theta}-1=0\)
\(\Rightarrow \tan{\theta}=1, \ 2\tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=1, \ \tan{\theta}=\frac{1}{2}\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}, \ \tan{\theta}=\tan{\alpha}\) ➜ যেখানে, \(\frac{1}{2}=\tan{\alpha}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{4}, \ \theta=n\pi+\alpha\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+\frac{\pi}{4}, \ n\pi+\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}, \ n\pi+\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{2}, \ n\in{\mathbb{Z}}\)
\(\sec^2{\theta}+\tan^2{\theta}=3\tan{\theta}\)
\(\Rightarrow 1+\tan^2{\theta}+\tan^2{\theta}=3\tan{\theta}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{\theta}-3\tan{\theta}+1=0\)
\(\Rightarrow 2\tan^2{\theta}-2\tan{\theta}-\tan{\theta}+1=0\)
\(\Rightarrow 2\tan{\theta}(\tan{\theta}-1)-1(\tan{\theta}-1)=0\)
\(\Rightarrow (\tan{\theta}-1)(2\tan{\theta}-1)=0\)
\(\Rightarrow \tan{\theta}-1=0, \ 2\tan{\theta}-1=0\)
\(\Rightarrow \tan{\theta}=1, \ 2\tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=1, \ \tan{\theta}=\frac{1}{2}\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}, \ \tan{\theta}=\tan{\alpha}\) ➜ যেখানে, \(\frac{1}{2}=\tan{\alpha}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{4}, \ \theta=n\pi+\alpha\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+\frac{\pi}{4}, \ n\pi+\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}, \ n\pi+\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{2}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxvi)\) \(2\sin{\theta}\tan{\theta}+1=\tan{\theta}+2\sin{\theta}\)উত্তরঃ \(n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(\ n\in{\mathbb{Z}}\)
চুয়েটঃ২০০৫-২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin{\theta}\tan{\theta}+1=\tan{\theta}+2\sin{\theta}\)
\(\Rightarrow 2\sin{\theta}\tan{\theta}-\tan{\theta}-2\sin{\theta}+1=0\)
\(\Rightarrow \tan{\theta}(2\sin{\theta}-1)-1(2\sin{\theta}-1)=0\)
\(\Rightarrow (2\sin{\theta}-1)(\tan{\theta}-1)=0\)
\(\Rightarrow 2\sin{\theta}-1=0, \ \tan{\theta}-1=0\)
\(\Rightarrow 2\sin{\theta}=1, \ \tan{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}, \ \tan{\theta}=1\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(2\sin{\theta}\tan{\theta}+1=\tan{\theta}+2\sin{\theta}\)
\(\Rightarrow 2\sin{\theta}\tan{\theta}-\tan{\theta}-2\sin{\theta}+1=0\)
\(\Rightarrow \tan{\theta}(2\sin{\theta}-1)-1(2\sin{\theta}-1)=0\)
\(\Rightarrow (2\sin{\theta}-1)(\tan{\theta}-1)=0\)
\(\Rightarrow 2\sin{\theta}-1=0, \ \tan{\theta}-1=0\)
\(\Rightarrow 2\sin{\theta}=1, \ \tan{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}, \ \tan{\theta}=1\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}, \ n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxvii)\) \(\tan{x}+\tan{2x}+\tan{3x}=0\)উত্তরঃ \(\frac{n\pi}{3}, \ n\pi\pm\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{\sqrt{2}}, \ n\in{\mathbb{Z}}\)
বঃ২০১১; রাঃ২০০৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{x}+\tan{2x}+\tan{3x}=0\)
\(\Rightarrow \tan{x}+\tan{2x}+\tan{3x}=0\)
\(\Rightarrow \tan{x}+\tan{2x}+\tan{(x+2x)}=0\)
\(\Rightarrow \tan{x}+\tan{2x}+\frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=0\) ➜ \(\because \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
\(\Rightarrow (\tan{x}+\tan{2x})\left(1+\frac{1}{1-\tan{x}\tan{2x}}\right)=0\)
\(\Rightarrow \tan{x}+\tan{2x}=0, \ 1+\frac{1}{1-\tan{x}\tan{2x}}=0\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=0, \ \frac{1}{1-\tan{x}\tan{2x}}=-1\) ➜ প্রথম সমীকরণের উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=0, \ -1+\tan{x}\tan{2x}=1\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}=0, \ \tan{x}\tan{2x}=2\)
\(\Rightarrow 3x=n\pi\) ➜ \(\because \tan{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{3}\)
আবার,
\(\tan{x}\tan{2x}=2\)
\(\Rightarrow \tan{x}\frac{2\tan{x}}{1-\tan^2{x}}=2\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\Rightarrow \frac{\tan^2{x}}{1-\tan^2{x}}=1\)
\(\Rightarrow \tan^2{x}=1-\tan^2{x}\)
\(\Rightarrow \tan^2{x}+\tan^2{x}=1\)
\(\Rightarrow 2\tan^2{x}=1\)
\(\Rightarrow \tan^2{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\pm\sqrt{\frac{1}{2}}\)
\(\Rightarrow \tan{x}=\pm\frac{1}{\sqrt{2}}\)
\(\Rightarrow \tan{x}=\pm\tan{\alpha}\) ➜ যেখানে, \(\frac{1}{\sqrt{2}}=\tan{\alpha}\)
\(\Rightarrow \tan{x}=\tan{(\pm\alpha)}\)
\(\Rightarrow x=n\pi+(\pm\alpha)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{3}, \ n\pi\pm\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{\sqrt{2}}, \ n\in{\mathbb{Z}}\)
\(\tan{x}+\tan{2x}+\tan{3x}=0\)
\(\Rightarrow \tan{x}+\tan{2x}+\tan{3x}=0\)
\(\Rightarrow \tan{x}+\tan{2x}+\tan{(x+2x)}=0\)
\(\Rightarrow \tan{x}+\tan{2x}+\frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=0\) ➜ \(\because \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
\(\Rightarrow (\tan{x}+\tan{2x})\left(1+\frac{1}{1-\tan{x}\tan{2x}}\right)=0\)
\(\Rightarrow \tan{x}+\tan{2x}=0, \ 1+\frac{1}{1-\tan{x}\tan{2x}}=0\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=0, \ \frac{1}{1-\tan{x}\tan{2x}}=-1\) ➜ প্রথম সমীকরণের উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=0, \ -1+\tan{x}\tan{2x}=1\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}=0, \ \tan{x}\tan{2x}=2\)
\(\Rightarrow 3x=n\pi\) ➜ \(\because \tan{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{3}\)
আবার,
\(\tan{x}\tan{2x}=2\)
\(\Rightarrow \tan{x}\frac{2\tan{x}}{1-\tan^2{x}}=2\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\Rightarrow \frac{\tan^2{x}}{1-\tan^2{x}}=1\)
\(\Rightarrow \tan^2{x}=1-\tan^2{x}\)
\(\Rightarrow \tan^2{x}+\tan^2{x}=1\)
\(\Rightarrow 2\tan^2{x}=1\)
\(\Rightarrow \tan^2{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\pm\sqrt{\frac{1}{2}}\)
\(\Rightarrow \tan{x}=\pm\frac{1}{\sqrt{2}}\)
\(\Rightarrow \tan{x}=\pm\tan{\alpha}\) ➜ যেখানে, \(\frac{1}{\sqrt{2}}=\tan{\alpha}\)
\(\Rightarrow \tan{x}=\tan{(\pm\alpha)}\)
\(\Rightarrow x=n\pi+(\pm\alpha)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\alpha\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{3}, \ n\pi\pm\alpha\) যেখানে, \(\tan{\alpha}=\frac{1}{\sqrt{2}}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxviii)\) \(\tan^2{\theta}-3cosec^2{\theta}+1=0\)উত্তরঃ \(n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুয়েটঃ২০০৩-২০০৪; টেক্সটাইলঃ২০০৫-২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan^2{\theta}-3cosec^2{\theta}+1=0\)
\(\Rightarrow \tan^2{\theta}-3(1+\cot^2{\theta})+1=0\) ➜ \(\because cosec^2{A}=1+cot^2{A}\)
\(\Rightarrow \tan^2{\theta}-3-3\cot^2{\theta}+1=0\)
\(\Rightarrow \tan^2{\theta}-\frac{3}{\tan^2{\theta}}-2=0\) ➜ \(\because cot^2{A}=\frac{1}{\tan^2{A}}\)
\(\Rightarrow \tan^4{\theta}-3-2\tan^2{\theta}=0\) ➜ উভয় পার্শে \(\tan^2{\theta}\) গুণ করে,
\(\Rightarrow \tan^4{\theta}-2\tan^2{\theta}-3=0\)
\(\Rightarrow \tan^4{\theta}-3\tan^2{\theta}+\tan^2{\theta}-3=0\)
\(\Rightarrow \tan^2{\theta}(\tan^2{\theta}-3)+1(\tan^2{\theta}-3)=0\)
\(\Rightarrow (\tan^2{\theta}-3)(\tan^2{\theta}+1)=0\)
\(\Rightarrow \tan^2{\theta}-3=0, \ (\tan^2{\theta}+1)\ne{0}\)
\(\Rightarrow \tan^2{\theta}=3\)
\(\Rightarrow \tan{\theta}=\pm\sqrt{3}\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{3}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right)\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan^2{\theta}-3cosec^2{\theta}+1=0\)
\(\Rightarrow \tan^2{\theta}-3(1+\cot^2{\theta})+1=0\) ➜ \(\because cosec^2{A}=1+cot^2{A}\)
\(\Rightarrow \tan^2{\theta}-3-3\cot^2{\theta}+1=0\)
\(\Rightarrow \tan^2{\theta}-\frac{3}{\tan^2{\theta}}-2=0\) ➜ \(\because cot^2{A}=\frac{1}{\tan^2{A}}\)
\(\Rightarrow \tan^4{\theta}-3-2\tan^2{\theta}=0\) ➜ উভয় পার্শে \(\tan^2{\theta}\) গুণ করে,
\(\Rightarrow \tan^4{\theta}-2\tan^2{\theta}-3=0\)
\(\Rightarrow \tan^4{\theta}-3\tan^2{\theta}+\tan^2{\theta}-3=0\)
\(\Rightarrow \tan^2{\theta}(\tan^2{\theta}-3)+1(\tan^2{\theta}-3)=0\)
\(\Rightarrow (\tan^2{\theta}-3)(\tan^2{\theta}+1)=0\)
\(\Rightarrow \tan^2{\theta}-3=0, \ (\tan^2{\theta}+1)\ne{0}\)
\(\Rightarrow \tan^2{\theta}=3\)
\(\Rightarrow \tan{\theta}=\pm\sqrt{3}\)
\(\Rightarrow \tan{\theta}=\pm\tan{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\tan{\frac{\pi}{3}}\)
\(\Rightarrow \tan{\theta}=\tan{\left(\pm\frac{\pi}{3}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right)\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{3}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxix)\) \(\cos{6x}+\cos{4x}=\sin{3x}+\sin{x}\)উত্তরঃ \((2n+1)\frac{\pi}{2}, \ (4n+1)\frac{\pi}{14}, \ (4n-1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৬; রাঃ২০১৪; মাঃ২০১০ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{6x}+\cos{4x}=\sin{3x}+\sin{x}\)
\(\Rightarrow 2\cos{\frac{6x+4x}{2}}\cos{\frac{6x-4x}{2}}=2\sin{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(\sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10x}{2}}\cos{\frac{2x}{2}}=2\sin{\frac{4x}{2}}\cos{\frac{2x}{2}}\)
\(\Rightarrow 2\cos{5x}\cos{x}=2\sin{2x}\cos{x}\)
\(\Rightarrow \cos{5x}\cos{x}=\sin{2x}\cos{x}\)
\(\Rightarrow \cos{x}(\cos{5x}-\sin{2x})=0\)
\(\Rightarrow \cos{x}=0, \ \cos{5x}-\sin{2x}=0\)
\(\Rightarrow \cos{x}=0, \ \cos{5x}=\sin{2x}\)
\(\Rightarrow \cos{x}=0, \ \cos{5x}=\cos{\left(\frac{\pi}{2}-2x\right)}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ 5x=2n\pi\pm\left(\frac{\pi}{2}-2x\right)\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{2}\)
আবার,
\(5x=2n\pi+\left(\frac{\pi}{2}-2x\right), \ +ve \text{ চিহ্ন ব্যবহার করে, } \)
\(\Rightarrow 5x=2n\pi+\frac{\pi}{2}-2x\)
\(\Rightarrow 5x+2x=2n\pi+\frac{\pi}{2}\)
\(\Rightarrow 7x=(4n+1)\frac{\pi}{2}\)
\(\therefore x=(4n+1)\frac{\pi}{14}\)
আবার,
\(5x=2n\pi-\left(\frac{\pi}{2}-2x\right), \ -ve \text{ চিহ্ন ব্যবহার করে, } \)
\(\Rightarrow 5x=2n\pi-\frac{\pi}{2}+2x\)
\(\Rightarrow 5x-2x=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 3x=(4n-1)\frac{\pi}{2}\)
\(\therefore x=(4n-1)\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{2}, \ (4n+1)\frac{\pi}{14}, \ (4n-1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{6x}+\cos{4x}=\sin{3x}+\sin{x}\)
\(\Rightarrow 2\cos{\frac{6x+4x}{2}}\cos{\frac{6x-4x}{2}}=2\sin{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(\sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10x}{2}}\cos{\frac{2x}{2}}=2\sin{\frac{4x}{2}}\cos{\frac{2x}{2}}\)
\(\Rightarrow 2\cos{5x}\cos{x}=2\sin{2x}\cos{x}\)
\(\Rightarrow \cos{5x}\cos{x}=\sin{2x}\cos{x}\)
\(\Rightarrow \cos{x}(\cos{5x}-\sin{2x})=0\)
\(\Rightarrow \cos{x}=0, \ \cos{5x}-\sin{2x}=0\)
\(\Rightarrow \cos{x}=0, \ \cos{5x}=\sin{2x}\)
\(\Rightarrow \cos{x}=0, \ \cos{5x}=\cos{\left(\frac{\pi}{2}-2x\right)}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{2}, \ 5x=2n\pi\pm\left(\frac{\pi}{2}-2x\right)\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{2}\)
আবার,
\(5x=2n\pi+\left(\frac{\pi}{2}-2x\right), \ +ve \text{ চিহ্ন ব্যবহার করে, } \)
\(\Rightarrow 5x=2n\pi+\frac{\pi}{2}-2x\)
\(\Rightarrow 5x+2x=2n\pi+\frac{\pi}{2}\)
\(\Rightarrow 7x=(4n+1)\frac{\pi}{2}\)
\(\therefore x=(4n+1)\frac{\pi}{14}\)
আবার,
\(5x=2n\pi-\left(\frac{\pi}{2}-2x\right), \ -ve \text{ চিহ্ন ব্যবহার করে, } \)
\(\Rightarrow 5x=2n\pi-\frac{\pi}{2}+2x\)
\(\Rightarrow 5x-2x=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 3x=(4n-1)\frac{\pi}{2}\)
\(\therefore x=(4n-1)\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{2}, \ (4n+1)\frac{\pi}{14}, \ (4n-1)\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxx)\) \(\cos^3{x}\sin{3x}+\sin^3{x}\cos{3x}=\frac{3}{4}\)উত্তরঃ \((4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০০২ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos^3{x}\sin{3x}+\sin^3{x}\cos{3x}=\frac{3}{4}\)
\(\Rightarrow 4\cos^3{x}\sin{3x}+4\sin^3{x}\cos{3x}=3\) ➜ উভয় পার্শে \(4\) গুণ করে,
\(\Rightarrow (\cos{3x}+3\cos{x})\sin{3x}+(3\sin{x}-\sin{3x})\cos{3x}=3\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
এবং \(4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(\Rightarrow \sin{3x}\cos{3x}+3\sin{3x}\cos{x}+3\sin{x}\cos{3x}-\sin{3x}\cos{3x}=3\)
\(\Rightarrow 3\sin{3x}\cos{x}+3\sin{x}\cos{3x}=3\)
\(\Rightarrow 3(\sin{3x}\cos{x}+\sin{x}\cos{3x})=3\)
\(\Rightarrow \sin{3x}\cos{x}+\sin{x}\cos{3x}=1\)
\(\Rightarrow \sin{(3x+x)}=1\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \sin{4x}=1\)
\(\Rightarrow 4x=(4n+1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=1\)
\(\Rightarrow A=(4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(x=(4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos^3{x}\sin{3x}+\sin^3{x}\cos{3x}=\frac{3}{4}\)
\(\Rightarrow 4\cos^3{x}\sin{3x}+4\sin^3{x}\cos{3x}=3\) ➜ উভয় পার্শে \(4\) গুণ করে,
\(\Rightarrow (\cos{3x}+3\cos{x})\sin{3x}+(3\sin{x}-\sin{3x})\cos{3x}=3\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
এবং \(4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(\Rightarrow \sin{3x}\cos{3x}+3\sin{3x}\cos{x}+3\sin{x}\cos{3x}-\sin{3x}\cos{3x}=3\)
\(\Rightarrow 3\sin{3x}\cos{x}+3\sin{x}\cos{3x}=3\)
\(\Rightarrow 3(\sin{3x}\cos{x}+\sin{x}\cos{3x})=3\)
\(\Rightarrow \sin{3x}\cos{x}+\sin{x}\cos{3x}=1\)
\(\Rightarrow \sin{(3x+x)}=1\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \sin{4x}=1\)
\(\Rightarrow 4x=(4n+1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=1\)
\(\Rightarrow A=(4n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(x=(4n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxi)\) \(\sin{\theta}+\cos{\theta}=\sqrt{2\sin{2\theta}}\)উত্তরঃ \(n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
ঢাঃ২০০৪ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{\theta}+\cos{\theta}=\sqrt{2\sin{2\theta}}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{2\times2\sin{\theta}\cos{\theta}}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{4\sin{\theta}\cos{\theta}}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=2\sqrt{\sin{\theta}}\sqrt{\cos{\theta}}\)
\(\Rightarrow (\sqrt{\sin{\theta}})^2-2\sqrt{\sin{\theta}}\sqrt{\cos{\theta}}+(\sqrt{\cos{\theta}})^2=0\)
\(\Rightarrow (\sqrt{\sin{\theta}}-\sqrt{\cos{\theta}})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{\sin{\theta}}-\sqrt{\cos{\theta}}=0\)
\(\Rightarrow \sqrt{\sin{\theta}}=\sqrt{\cos{\theta}}\)
\(\Rightarrow \sin{\theta}=\cos{\theta}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}=1\) ➜ উভয় পার্শে \(\cos{\theta}\) ভাগ করে,
\(\Rightarrow \tan{\theta}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{\theta}+\cos{\theta}=\sqrt{2\sin{2\theta}}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{2\times2\sin{\theta}\cos{\theta}}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=\sqrt{4\sin{\theta}\cos{\theta}}\)
\(\Rightarrow \sin{\theta}+\cos{\theta}=2\sqrt{\sin{\theta}}\sqrt{\cos{\theta}}\)
\(\Rightarrow (\sqrt{\sin{\theta}})^2-2\sqrt{\sin{\theta}}\sqrt{\cos{\theta}}+(\sqrt{\cos{\theta}})^2=0\)
\(\Rightarrow (\sqrt{\sin{\theta}}-\sqrt{\cos{\theta}})^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \sqrt{\sin{\theta}}-\sqrt{\cos{\theta}}=0\)
\(\Rightarrow \sqrt{\sin{\theta}}=\sqrt{\cos{\theta}}\)
\(\Rightarrow \sin{\theta}=\cos{\theta}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}=1\) ➜ উভয় পার্শে \(\cos{\theta}\) ভাগ করে,
\(\Rightarrow \tan{\theta}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxii)\) \(\cot{2x}=\cos{x}+\sin{x}\)উত্তরঃ \((4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
কুঃ২০১১ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{2x}=\cos{x}+\sin{x}\)
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\cos{x}+\sin{x}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos^2{2x}}{\sin^2{2x}}=(\cos{x}+\sin{x})^2\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=\cos^2{x}+\sin^2{x}+2\sin{x}\cos{x}\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
এবং \((a+b)^2=a^2+b^2+2ab\)
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=1+\sin{2x}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})=1-\sin^2{2x}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1-\sin^2{2x})=0\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1+\sin{2x})(1-\sin{2x})=0\) ➜ \(\because a^2-b^2=(a+b)(a-b)\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}-1+\sin{2x})=0\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}+\sin{2x}-1)=0\)
\(\Rightarrow 1+\sin{2x}=0, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=-1, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow 2x=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n-1)\frac{\pi}{4}\)
আবার,
\(\sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1+\sqrt{5}}{2}, \ \sin{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\sin{2x}}\le{1}\)
\(\Rightarrow \sin{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(\cot{2x}=\cos{x}+\sin{x}\)
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\cos{x}+\sin{x}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos^2{2x}}{\sin^2{2x}}=(\cos{x}+\sin{x})^2\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=\cos^2{x}+\sin^2{x}+2\sin{x}\cos{x}\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
এবং \((a+b)^2=a^2+b^2+2ab\)
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=1+\sin{2x}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})=1-\sin^2{2x}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1-\sin^2{2x})=0\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1+\sin{2x})(1-\sin{2x})=0\) ➜ \(\because a^2-b^2=(a+b)(a-b)\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}-1+\sin{2x})=0\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}+\sin{2x}-1)=0\)
\(\Rightarrow 1+\sin{2x}=0, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=-1, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow 2x=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n-1)\frac{\pi}{4}\)
আবার,
\(\sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1+\sqrt{5}}{2}, \ \sin{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\sin{2x}}\le{1}\)
\(\Rightarrow \sin{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxiii)\) \(\frac{\sqrt{3}}{\sin{2x}}-\frac{1}{\cos{2x}}=4\)উত্তরঃ \((6n+1)\frac{\pi}{18}, \ (3n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
বুয়েটঃ২০০৬-২০০৭, ২০০৭-২০০৮ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\frac{\sqrt{3}}{\sin{2x}}-\frac{1}{\cos{2x}}=4\)
\(\Rightarrow \frac{\sqrt{3}\cos{2x}-\sin{2x}}{\sin{2x}\cos{2x}}=4\)
\(\Rightarrow \sqrt{3}\cos{2x}-\sin{2x}=4\sin{2x}\cos{2x}\)
\(\Rightarrow \sqrt{3}\cos{2x}-\sin{2x}=2\times2\sin{2x}\cos{2x}\)
\(\Rightarrow \sqrt{3}\cos{2x}-\sin{2x}=2\sin{4x}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \frac{\sqrt{3}}{2}\cos{2x}-\frac{1}{2}\sin{2x}=\sin{4x}\) ➜ উভয় পার্শে \(2\) ভাগ করে,
\(\Rightarrow \sin{\frac{\pi}{3}}\cos{2x}-\cos{\frac{\pi}{3}}\sin{2x}=\sin{4x}\) ➜ \(\because \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \sin{\left(\frac{\pi}{3}-2x\right)}=\sin{4x}\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \sin{\left(\frac{\pi}{3}-2x\right)}-\sin{4x}=0\)
\(\Rightarrow -\sin{\left(2x-\frac{\pi}{3}\right)}-\sin{4x}=0\)
\(\Rightarrow \sin{\left(2x-\frac{\pi}{3}\right)}+\sin{4x}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\sin{\frac{2x-\frac{\pi}{3}+4x}{2}}\cos{\frac{2x-\frac{\pi}{3}-4x}{2}}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\left(\frac{6x-\frac{\pi}{3}}{2}\right)}\cos{\left(\frac{-\frac{\pi}{3}-2x}{2}\right)}=0\)
\(\Rightarrow \sin{\left(\frac{6x-\frac{\pi}{3}}{2}\right)}=0, \ \cos{\left(\frac{\frac{\pi}{3}+2x}{2}\right)}=0, \ \because \cos{(-\theta)}=\cos{\theta}\)
\(\Rightarrow \sin{\left(3x-\frac{\pi}{6}\right)}=0, \ \cos{\left(\frac{\pi}{6}+x\right)}=0\)
\(\Rightarrow 3x-\frac{\pi}{6}=n\pi, \ \frac{\pi}{6}+x=(2n+1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=n\pi+\frac{\pi}{6}, \ x=(2n+1)\frac{\pi}{2}-\frac{\pi}{6}\)
\(\Rightarrow 3x=(6n+1)\frac{\pi}{6}, \ x=(6n+3-1)\frac{\pi}{6}\)
\(\Rightarrow x=(6n+1)\frac{\pi}{18}, \ x=(6n+2)\frac{\pi}{6}\)
\(\Rightarrow x=(6n+1)\frac{\pi}{18}, \ x=2(3n+1)\frac{\pi}{6}\)
\(\Rightarrow x=(6n+1)\frac{\pi}{18}, \ x=(3n+1)\frac{\pi}{3}\)
\(\therefore x=(6n+1)\frac{\pi}{18}, \ (3n+1)\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(6n+1)\frac{\pi}{18}, \ (3n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\frac{\sqrt{3}}{\sin{2x}}-\frac{1}{\cos{2x}}=4\)
\(\Rightarrow \frac{\sqrt{3}\cos{2x}-\sin{2x}}{\sin{2x}\cos{2x}}=4\)
\(\Rightarrow \sqrt{3}\cos{2x}-\sin{2x}=4\sin{2x}\cos{2x}\)
\(\Rightarrow \sqrt{3}\cos{2x}-\sin{2x}=2\times2\sin{2x}\cos{2x}\)
\(\Rightarrow \sqrt{3}\cos{2x}-\sin{2x}=2\sin{4x}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \frac{\sqrt{3}}{2}\cos{2x}-\frac{1}{2}\sin{2x}=\sin{4x}\) ➜ উভয় পার্শে \(2\) ভাগ করে,
\(\Rightarrow \sin{\frac{\pi}{3}}\cos{2x}-\cos{\frac{\pi}{3}}\sin{2x}=\sin{4x}\) ➜ \(\because \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \sin{\left(\frac{\pi}{3}-2x\right)}=\sin{4x}\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \sin{\left(\frac{\pi}{3}-2x\right)}-\sin{4x}=0\)
\(\Rightarrow -\sin{\left(2x-\frac{\pi}{3}\right)}-\sin{4x}=0\)
\(\Rightarrow \sin{\left(2x-\frac{\pi}{3}\right)}+\sin{4x}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\sin{\frac{2x-\frac{\pi}{3}+4x}{2}}\cos{\frac{2x-\frac{\pi}{3}-4x}{2}}=0\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\left(\frac{6x-\frac{\pi}{3}}{2}\right)}\cos{\left(\frac{-\frac{\pi}{3}-2x}{2}\right)}=0\)
\(\Rightarrow \sin{\left(\frac{6x-\frac{\pi}{3}}{2}\right)}=0, \ \cos{\left(\frac{\frac{\pi}{3}+2x}{2}\right)}=0, \ \because \cos{(-\theta)}=\cos{\theta}\)
\(\Rightarrow \sin{\left(3x-\frac{\pi}{6}\right)}=0, \ \cos{\left(\frac{\pi}{6}+x\right)}=0\)
\(\Rightarrow 3x-\frac{\pi}{6}=n\pi, \ \frac{\pi}{6}+x=(2n+1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3x=n\pi+\frac{\pi}{6}, \ x=(2n+1)\frac{\pi}{2}-\frac{\pi}{6}\)
\(\Rightarrow 3x=(6n+1)\frac{\pi}{6}, \ x=(6n+3-1)\frac{\pi}{6}\)
\(\Rightarrow x=(6n+1)\frac{\pi}{18}, \ x=(6n+2)\frac{\pi}{6}\)
\(\Rightarrow x=(6n+1)\frac{\pi}{18}, \ x=2(3n+1)\frac{\pi}{6}\)
\(\Rightarrow x=(6n+1)\frac{\pi}{18}, \ x=(3n+1)\frac{\pi}{3}\)
\(\therefore x=(6n+1)\frac{\pi}{18}, \ (3n+1)\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(6n+1)\frac{\pi}{18}, \ (3n+1)\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxiv)\) \(\tan{3\theta}\tan{\theta}=1\)উত্তরঃ \((2n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
মাঃ২০১৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{3\theta}\tan{\theta}=1\)
\(\Rightarrow \frac{\sin{3\theta}\sin{\theta}}{\cos{3\theta}\cos{\theta}}=1\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \cos{3\theta}\cos{\theta}=\sin{3\theta}\sin{\theta}\)
\(\Rightarrow \cos{3\theta}\cos{\theta}-\sin{3\theta}\sin{\theta}=0\)
\(\Rightarrow \cos{(3\theta+\theta)}=0\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \cos{4\theta}=0\)
\(\Rightarrow 4\theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{3\theta}\tan{\theta}=1\)
\(\Rightarrow \frac{\sin{3\theta}\sin{\theta}}{\cos{3\theta}\cos{\theta}}=1\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \cos{3\theta}\cos{\theta}=\sin{3\theta}\sin{\theta}\)
\(\Rightarrow \cos{3\theta}\cos{\theta}-\sin{3\theta}\sin{\theta}=0\)
\(\Rightarrow \cos{(3\theta+\theta)}=0\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \cos{4\theta}=0\)
\(\Rightarrow 4\theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{8}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxv)\) \(\tan{\theta}+\tan{3\theta}=0\)উত্তরঃ \(\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০০৫ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\tan{\theta}+\tan{3\theta}=0\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}+\frac{\sin{3\theta}}{\cos{3\theta}}=0\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\sin{\theta}\cos{3\theta}+\cos{\theta}\sin{3\theta}}{\cos{\theta}\cos{3\theta}}=0\)
\(\Rightarrow \frac{\sin{(\theta+3\theta)}}{\cos{\theta}\cos{3\theta}}=0\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \frac{\sin{4\theta}}{\cos{\theta}\cos{3\theta}}=0\)
\(\Rightarrow \sin{4\theta}=0\)
\(\Rightarrow 4\theta=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\tan{\theta}+\tan{3\theta}=0\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}+\frac{\sin{3\theta}}{\cos{3\theta}}=0\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\sin{\theta}\cos{3\theta}+\cos{\theta}\sin{3\theta}}{\cos{\theta}\cos{3\theta}}=0\)
\(\Rightarrow \frac{\sin{(\theta+3\theta)}}{\cos{\theta}\cos{3\theta}}=0\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \frac{\sin{4\theta}}{\cos{\theta}\cos{3\theta}}=0\)
\(\Rightarrow \sin{4\theta}=0\)
\(\Rightarrow 4\theta=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxvi)\) \(\cos{x}+\sin{x}=\cos{2x}+\sin{2x}\)উত্তরঃ \(2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
দিঃ২০১৪; ঢাঃ২০১৩,২০০৬; চঃ২০১২,২০০৬; সিঃ২০১৫,২০০৫; কুঃ২০১৪,২০০৫; যঃ২০১৪,২০১১,২০০৭; রাঃ২০১২,২০০৮; মাঃ২০১৩,২০০৯ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\Rightarrow \cos{x}+\sin{x}=\cos{2x}+\sin{2x}\)
\(\Rightarrow \cos{x}-\cos{2x}=\sin{2x}-\sin{x}\)
\(\Rightarrow 2\sin{\frac{x+2x}{2}}\sin{\frac{2x-x}{2}}=2\cos{\frac{2x+x}{2}}\sin{\frac{2x-x}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}=\cos{\frac{3x}{2}}\sin{\frac{x}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}-\cos{\frac{3x}{2}}\sin{\frac{x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}\left(\sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}=\cos{\frac{3x}{2}}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \frac{\sin{\frac{3x}{2}}}{\cos{\frac{3x}{2}}}=1\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{x}{2}=n\pi, \ \frac{3x}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
\(\therefore\) নির্ণেয় সাধারণ সমাধান, \(x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \cos{x}+\sin{x}=\cos{2x}+\sin{2x}\)
\(\Rightarrow \cos{x}-\cos{2x}=\sin{2x}-\sin{x}\)
\(\Rightarrow 2\sin{\frac{x+2x}{2}}\sin{\frac{2x-x}{2}}=2\cos{\frac{2x+x}{2}}\sin{\frac{2x-x}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}=\cos{\frac{3x}{2}}\sin{\frac{x}{2}}\)
\(\Rightarrow \sin{\frac{3x}{2}}\sin{\frac{x}{2}}-\cos{\frac{3x}{2}}\sin{\frac{x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}\left(\sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}-\cos{\frac{3x}{2}}=0\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \sin{\frac{3x}{2}}=\cos{\frac{3x}{2}}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \frac{\sin{\frac{3x}{2}}}{\cos{\frac{3x}{2}}}=1\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=1\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(\Rightarrow \sin{\frac{x}{2}}=0, \ \tan{\frac{3x}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{x}{2}=n\pi, \ \frac{3x}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\)
\(\therefore\) নির্ণেয় সাধারণ সমাধান, \(x=2n\pi, \ \frac{2}{3}\left(n\pi+\frac{\pi}{4}\right)\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxvii)\) \(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}\)উত্তরঃ \((2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০৮; বুয়েটঃ২০০৮-২০০৯ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}\)
\(\Rightarrow \sin{3\theta}+\sin{\theta}+\sin{2\theta}=1+\cos{2\theta}+\cos{\theta}\)
\(\Rightarrow 2\sin{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}+\sin{2\theta}=2\cos^2{\theta}+\cos{\theta}\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow 2\sin{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\sin{2\theta}\cos{\theta}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow \sin{2\theta}(2\cos{\theta}+1)-\cos{\theta}(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(\sin{2\theta}-\cos{\theta})=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \sin{2\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ \cos{\theta}=0, \ 2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \cos{\theta}=0, \ \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\cos{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ (2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{\theta}+\sin{2\theta}+\sin{3\theta}=1+\cos{\theta}+\cos{2\theta}\)
\(\Rightarrow \sin{3\theta}+\sin{\theta}+\sin{2\theta}=1+\cos{2\theta}+\cos{\theta}\)
\(\Rightarrow 2\sin{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}+\sin{2\theta}=2\cos^2{\theta}+\cos{\theta}\) ➜ \(\because \sin{C}+\sin{D}=2\sin{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
এবং \(1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow 2\sin{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\sin{2\theta}\cos{\theta}+\sin{2\theta}-2\cos^2{\theta}-\cos{\theta}=0\)
\(\Rightarrow \sin{2\theta}(2\cos{\theta}+1)-\cos{\theta}(2\cos{\theta}+1)=0\)
\(\Rightarrow (2\cos{\theta}+1)(\sin{2\theta}-\cos{\theta})=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \sin{2\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}\cos{\theta}-\cos{\theta}=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ \cos{\theta}=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ \cos{\theta}=0, \ 2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \cos{\theta}=0, \ \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=-\cos{\frac{\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\left(\pi-\frac{\pi}{3}\right)}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \cos{\theta}=0, \ \sin{\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=(2n+1)\frac{\pi}{2}, \ \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\cos{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ (2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{2}, \ n\pi+(-1)^n\frac{\pi}{6}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxviii)\) \(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0\)উত্তরঃ \((2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0\)
\(\Rightarrow \cos{5\theta}+\cos{\theta}+\cos{7\theta}+\cos{3\theta}=0\)
\(\Rightarrow 2\cos{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}+2\cos{\frac{7\theta+3\theta}{2}}\cos{\frac{7\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}+2\cos{\frac{10\theta}{2}}\cos{\frac{4\theta}{2}}=0\)
\(\Rightarrow 2\cos{3\theta}\cos{2\theta}+2\cos{5\theta}\cos{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}(\cos{3\theta}+\cos{5\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}+\cos{5\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{5\theta+3\theta}{2}}\cos{\frac{5\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{8\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 4\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{8}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0\)
\(\Rightarrow \cos{5\theta}+\cos{\theta}+\cos{7\theta}+\cos{3\theta}=0\)
\(\Rightarrow 2\cos{\frac{5\theta+\theta}{2}}\cos{\frac{5\theta-\theta}{2}}+2\cos{\frac{7\theta+3\theta}{2}}\cos{\frac{7\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{6\theta}{2}}\cos{\frac{4\theta}{2}}+2\cos{\frac{10\theta}{2}}\cos{\frac{4\theta}{2}}=0\)
\(\Rightarrow 2\cos{3\theta}\cos{2\theta}+2\cos{5\theta}\cos{2\theta}=0\)
\(\Rightarrow 2\cos{2\theta}(\cos{3\theta}+\cos{5\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{3\theta}+\cos{5\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{5\theta+3\theta}{2}}\cos{\frac{5\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{\frac{8\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{4\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 4\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=(2n+1)\frac{\pi}{8}, \ \theta=(2n+1)\frac{\pi}{2}\)
\(\therefore \theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{8}, \ (2n+1)\frac{\pi}{4}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xxxix)\) \(\sin{3x}\sin{x}=\cos{2x}+\frac{1}{2}\)উত্তরঃ \((2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{3x}\sin{x}=\cos{2x}+\frac{1}{2}\)
\(\Rightarrow 2\sin{3x}\sin{x}=2\cos{2x}+1\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \cos{(3x-x)}-\cos{(3x+x)}=2\cos{2x}+1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2x}-\cos{4x}-2\cos{2x}-1=0\)
\(\Rightarrow -\cos{4x}-\cos{2x}-1=0\)
\(\Rightarrow \cos{4x}+\cos{2x}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos^2{2x}-1+\cos{2x}+1=0\) ➜ \(\because \cos{2A}=2\cos^2{A}-1\)
\(\Rightarrow 2\cos^2{2x}+\cos{2x}=0\)
\(\Rightarrow \cos{2x}(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=-\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=\cos{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\left(\pi-\frac{\pi}{3}\right)\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ 2x=2n\pi\pm\frac{2\pi}{3}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{3x}\sin{x}=\cos{2x}+\frac{1}{2}\)
\(\Rightarrow 2\sin{3x}\sin{x}=2\cos{2x}+1\) ➜ উভয় পার্শে \(2\) গুণ করে,
\(\Rightarrow \cos{(3x-x)}-\cos{(3x+x)}=2\cos{2x}+1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2x}-\cos{4x}-2\cos{2x}-1=0\)
\(\Rightarrow -\cos{4x}-\cos{2x}-1=0\)
\(\Rightarrow \cos{4x}+\cos{2x}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos^2{2x}-1+\cos{2x}+1=0\) ➜ \(\because \cos{2A}=2\cos^2{A}-1\)
\(\Rightarrow 2\cos^2{2x}+\cos{2x}=0\)
\(\Rightarrow \cos{2x}(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=-\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{2x}=0, \ \cos{2x}=\cos{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cos{A}=\cos{\left(\pi-A\right)}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\left(\pi-\frac{\pi}{3}\right)\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ 2x=2n\pi\pm\frac{2\pi}{3}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{4}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xL)\) \(\sin{x}+2\cos{x}=1\)উত্তরঃ \(2n\pi+\frac{\pi}{4}, \ 2n\pi-\alpha\) যেখানে, \(\sin{\alpha}=\frac{3}{5}, \ n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}+2\cos{x}=1\)
\(\Rightarrow \cos{x}\times2+\sin{x}=1\)
\(\Rightarrow \cos{x}\frac{2}{\sqrt{5}}+\sin{x}\frac{1}{\sqrt{5}}=\frac{1}{\sqrt{5}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+2^2}=\sqrt{5}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\theta}+\sin{x}\sin{\theta}=\sin{\theta}\) ➜ \(\because \cos{\theta}=\frac{2}{\sqrt{5}}\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow \cos{(x-\theta)}=\cos{\left(\frac{\pi}{2}-\theta\right)}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\sin{A}=\cos{\left(\frac{\pi}{2}-A\right)}\)
\(\Rightarrow x-\theta=2n\pi\pm\left(\frac{\pi}{2}-\theta\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\left(\frac{\pi}{2}-\theta\right)+\theta\)
\(\Rightarrow x=2n\pi+\left(\frac{\pi}{2}-\theta\right)+\theta, \ 2n\pi-\left(\frac{\pi}{2}-\theta\right)+\theta\)
\(\Rightarrow x=2n\pi+\frac{\pi}{2}-\theta+\theta, \ 2n\pi-\frac{\pi}{2}+\theta+\theta\)
\(\Rightarrow x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\frac{\pi}{2}+2\theta\)
\(\Rightarrow x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\left(\frac{\pi}{2}-2\theta\right)\)
\(\therefore x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\alpha\)
যেখানে, \(\alpha=\frac{\pi}{2}-2\theta\)
\(\Rightarrow \sin{\alpha}=\sin{\left(\frac{\pi}{2}-2\theta\right)}\)
\(\Rightarrow \sin{\alpha}=\cos{2\theta}\)
\(\Rightarrow \sin{\alpha}=\cos^2{\theta}-\sin^2{\theta}\)
\(\Rightarrow \sin{\alpha}=\left(\frac{2}{\sqrt{5}}\right)^2-\left(\frac{1}{\sqrt{5}}\right)^2\)
\(\Rightarrow \sin{\alpha}=\frac{4}{5}-\frac{1}{5}\)
\(\Rightarrow \sin{\alpha}=\frac{4-1}{5}\)
\(\therefore \sin{\alpha}=\frac{3}{5}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\alpha\) যেখানে, \(\sin{\alpha}=\frac{3}{5}, \ n\in{\mathbb{Z}}\)
\(\sin{x}+2\cos{x}=1\)
\(\Rightarrow \cos{x}\times2+\sin{x}=1\)
\(\Rightarrow \cos{x}\frac{2}{\sqrt{5}}+\sin{x}\frac{1}{\sqrt{5}}=\frac{1}{\sqrt{5}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+2^2}=\sqrt{5}\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\theta}+\sin{x}\sin{\theta}=\sin{\theta}\) ➜ \(\because \cos{\theta}=\frac{2}{\sqrt{5}}\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow \cos{(x-\theta)}=\cos{\left(\frac{\pi}{2}-\theta\right)}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\sin{A}=\cos{\left(\frac{\pi}{2}-A\right)}\)
\(\Rightarrow x-\theta=2n\pi\pm\left(\frac{\pi}{2}-\theta\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\left(\frac{\pi}{2}-\theta\right)+\theta\)
\(\Rightarrow x=2n\pi+\left(\frac{\pi}{2}-\theta\right)+\theta, \ 2n\pi-\left(\frac{\pi}{2}-\theta\right)+\theta\)
\(\Rightarrow x=2n\pi+\frac{\pi}{2}-\theta+\theta, \ 2n\pi-\frac{\pi}{2}+\theta+\theta\)
\(\Rightarrow x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\frac{\pi}{2}+2\theta\)
\(\Rightarrow x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\left(\frac{\pi}{2}-2\theta\right)\)
\(\therefore x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\alpha\)
যেখানে, \(\alpha=\frac{\pi}{2}-2\theta\)
\(\Rightarrow \sin{\alpha}=\sin{\left(\frac{\pi}{2}-2\theta\right)}\)
\(\Rightarrow \sin{\alpha}=\cos{2\theta}\)
\(\Rightarrow \sin{\alpha}=\cos^2{\theta}-\sin^2{\theta}\)
\(\Rightarrow \sin{\alpha}=\left(\frac{2}{\sqrt{5}}\right)^2-\left(\frac{1}{\sqrt{5}}\right)^2\)
\(\Rightarrow \sin{\alpha}=\frac{4}{5}-\frac{1}{5}\)
\(\Rightarrow \sin{\alpha}=\frac{4-1}{5}\)
\(\therefore \sin{\alpha}=\frac{3}{5}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{2}, \ 2n\pi-\alpha\) যেখানে, \(\sin{\alpha}=\frac{3}{5}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLi)\) \(\sin{x}+\cos{x}+\sqrt{2}=0\)উত্তরঃ \((8n-3)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{x}+\cos{x}+\sqrt{2}=0\)
\(\Rightarrow \sin{x}+\cos{x}=-\sqrt{2}\)
\(\Rightarrow \sin{x}\frac{1}{\sqrt{2}}+\cos{x}\frac{1}{\sqrt{2}}=-1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \sin{x}\cos{\frac{\pi}{4}}+\cos{x}\sin{\frac{\pi}{4}}=-1\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \sin{\left(x+\frac{\pi}{4}\right)}=-1\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{4}=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(4n-1)\frac{\pi}{2}-\frac{\pi}{4}\)
\(\Rightarrow x=\{2(4n-1)-1\}\frac{\pi}{4}\)
\(\Rightarrow x=\{8n-2-1\}\frac{\pi}{4}\)
\(\therefore x=(8n-3)\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=(8n-3)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{x}+\cos{x}+\sqrt{2}=0\)
\(\Rightarrow \sin{x}+\cos{x}=-\sqrt{2}\)
\(\Rightarrow \sin{x}\frac{1}{\sqrt{2}}+\cos{x}\frac{1}{\sqrt{2}}=-1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \sin{x}\cos{\frac{\pi}{4}}+\cos{x}\sin{\frac{\pi}{4}}=-1\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \sin{\left(x+\frac{\pi}{4}\right)}=-1\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{4}=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(4n-1)\frac{\pi}{2}-\frac{\pi}{4}\)
\(\Rightarrow x=\{2(4n-1)-1\}\frac{\pi}{4}\)
\(\Rightarrow x=\{8n-2-1\}\frac{\pi}{4}\)
\(\therefore x=(8n-3)\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(x=(8n-3)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLii)\) \(\cot{\theta}+\cot{\left(\frac{\pi}{4}+\theta\right)}=2\)উত্তরঃ \(n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cot{\theta}+\cot{\left(\frac{\pi}{4}+\theta\right)}=2\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}+\frac{\cos{\left(\frac{\pi}{4}+\theta\right)}}{\sin{\left(\frac{\pi}{4}+\theta\right)}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin{\left(\frac{\pi}{4}+\theta\right)}\cos{\theta}+\cos{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}}{\sin{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}}=2\)
\(\Rightarrow \frac{\sin{\left(\frac{\pi}{4}+\theta+\theta\right)}}{\sin{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}}=2\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \sin{\left(\frac{\pi}{4}+2\theta\right)}=2\sin{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}\)
\(\Rightarrow \sin{\left(\frac{\pi}{4}+2\theta\right)}=\cos{\left(\frac{\pi}{4}+\theta-\theta\right)}-\cos{\left(\frac{\pi}{4}+\theta+\theta\right)}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \sin{\left(\frac{\pi}{4}+2\theta\right)}=\cos{\frac{\pi}{4}}-\cos{\left(\frac{\pi}{4}+2\theta\right)}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta\right)}+\sin{\left(\frac{\pi}{4}+2\theta\right)}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta\right)}\frac{1}{\sqrt{2}}+\sin{\left(\frac{\pi}{4}+2\theta\right)}\frac{1}{\sqrt{2}}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta\right)}\cos{\frac{\pi}{4}}+\sin{\left(\frac{\pi}{4}+2\theta\right)}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta-\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{2\theta}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cot{\theta}+\cot{\left(\frac{\pi}{4}+\theta\right)}=2\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}+\frac{\cos{\left(\frac{\pi}{4}+\theta\right)}}{\sin{\left(\frac{\pi}{4}+\theta\right)}}=2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin{\left(\frac{\pi}{4}+\theta\right)}\cos{\theta}+\cos{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}}{\sin{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}}=2\)
\(\Rightarrow \frac{\sin{\left(\frac{\pi}{4}+\theta+\theta\right)}}{\sin{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}}=2\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\Rightarrow \sin{\left(\frac{\pi}{4}+2\theta\right)}=2\sin{\left(\frac{\pi}{4}+\theta\right)}\sin{\theta}\)
\(\Rightarrow \sin{\left(\frac{\pi}{4}+2\theta\right)}=\cos{\left(\frac{\pi}{4}+\theta-\theta\right)}-\cos{\left(\frac{\pi}{4}+\theta+\theta\right)}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \sin{\left(\frac{\pi}{4}+2\theta\right)}=\cos{\frac{\pi}{4}}-\cos{\left(\frac{\pi}{4}+2\theta\right)}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta\right)}+\sin{\left(\frac{\pi}{4}+2\theta\right)}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta\right)}\frac{1}{\sqrt{2}}+\sin{\left(\frac{\pi}{4}+2\theta\right)}\frac{1}{\sqrt{2}}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+1^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta\right)}\cos{\frac{\pi}{4}}+\sin{\left(\frac{\pi}{4}+2\theta\right)}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}+2\theta-\frac{\pi}{4}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{2\theta}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLiii)\) \(2\cot{\frac{\theta}{2}}=(1+\cot{\theta})^2\)উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\cot{\frac{\theta}{2}}=(1+\cot{\theta})^2\)
\(\Rightarrow \frac{2\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\left(1+\frac{\cos{\theta}}{\sin{\theta}}\right)^2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{2\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\left(\frac{\sin{\theta}+\cos{\theta}}{\sin{\theta}}\right)^2\)
\(\Rightarrow \frac{2\times2\cos^2{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}=\frac{(\sin{\theta}+\cos{\theta})^2}{\sin^2{\theta}}\) ➜ বাম পাশের লব ও হরকে \(2\cos{\frac{\theta}{2}}\) দ্বারা গুণ করে,
\(\Rightarrow \frac{2(1+\cos{\theta})}{\sin{\theta}}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\sin^2{\theta}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow 2+2\cos{\theta}=\frac{1+\sin{2\theta}}{\sin{\theta}}, \ \because \sin{\theta}\ne{0}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow 2\sin{\theta}+2\sin{\theta}\cos{\theta}=1+\sin{2\theta}\)
\(\Rightarrow 2\sin{\theta}+\sin{2\theta}=1+\sin{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow 2\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(2\cot{\frac{\theta}{2}}=(1+\cot{\theta})^2\)
\(\Rightarrow \frac{2\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\left(1+\frac{\cos{\theta}}{\sin{\theta}}\right)^2\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{2\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\left(\frac{\sin{\theta}+\cos{\theta}}{\sin{\theta}}\right)^2\)
\(\Rightarrow \frac{2\times2\cos^2{\frac{\theta}{2}}}{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}=\frac{(\sin{\theta}+\cos{\theta})^2}{\sin^2{\theta}}\) ➜ বাম পাশের লব ও হরকে \(2\cos{\frac{\theta}{2}}\) দ্বারা গুণ করে,
\(\Rightarrow \frac{2(1+\cos{\theta})}{\sin{\theta}}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\sin^2{\theta}}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow 2+2\cos{\theta}=\frac{1+\sin{2\theta}}{\sin{\theta}}, \ \because \sin{\theta}\ne{0}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow 2\sin{\theta}+2\sin{\theta}\cos{\theta}=1+\sin{2\theta}\)
\(\Rightarrow 2\sin{\theta}+\sin{2\theta}=1+\sin{2\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow 2\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLiv)\) \(\frac{\sin{\alpha}}{\sin{2x}}-\frac{\cos{\alpha}}{\cos{2x}}=2\)উত্তরঃ \(\frac{1}{6}(2n\pi+\alpha), \ \frac{1}{2}\{(2n+1)\pi-\alpha\}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\frac{\sin{\alpha}}{\sin{2x}}-\frac{\cos{\alpha}}{\cos{2x}}=2\)
\(\Rightarrow \frac{\cos{\alpha}}{\cos{2x}}-\frac{\sin{\alpha}}{\sin{2x}}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \frac{\sin{2x}\cos{\alpha}-\cos{2x}\sin{\alpha}}{\sin{2x}\cos{2x}}=-2\)
\(\Rightarrow \frac{\sin{(2x-\alpha)}}{\sin{2x}\cos{2x}}=-2\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \sin{(2x-\alpha)}=-2\sin{2x}\cos{2x}\)
\(\Rightarrow \sin{(2x-\alpha)}=-\sin{4x}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{4x}=-\sin{(2x-\alpha)}\)
\(\Rightarrow \sin{4x}=\sin{(\alpha-2x)}\)
\(\Rightarrow 4x=m\pi+(-1)^m(\alpha-2x)\) ➜ \(\because \sin{A}=\sin{\theta}\)
\(\Rightarrow A=m\pi+(-1)^m\theta\) যেখানে, \(m\in{\mathbb{Z}}\)
\(\Rightarrow 4x=2n\pi+(-1)^(2n)(\alpha-2x)\) যখন, \(m\) জোড় সংখ্যা অর্থাৎ \(m=2n\)
\(\Rightarrow 4x=2n\pi+\alpha-2x\)
\(\Rightarrow 4x+2x=2n\pi+\alpha\)
\(\Rightarrow 6x=2n\pi+\alpha\)
\(\therefore x=\frac{1}{6}(2n\pi+\alpha)\)
আবার,
\(\Rightarrow 4x=2n\pi+(-1)^(2n+1)(\alpha-2x)\) যখন, \(m\) বিজোড় সংখ্যা অর্থাৎ \(m=2n+1\)
\(\Rightarrow 4x=(2n+1)\pi-(\alpha-2x)\)
\(\Rightarrow 4x=(2n+1)\pi-\alpha+2x\)
\(\Rightarrow 4x-2x=(2n+1)\pi-\alpha\)
\(\Rightarrow 2x=(2n+1)\pi-\alpha\)
\(\therefore x=\frac{1}{2}\{(2n+1)\pi-\alpha\}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{1}{6}(2n\pi+\alpha), \ \frac{1}{2}\{(2n+1)\pi-\alpha\}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\frac{\sin{\alpha}}{\sin{2x}}-\frac{\cos{\alpha}}{\cos{2x}}=2\)
\(\Rightarrow \frac{\cos{\alpha}}{\cos{2x}}-\frac{\sin{\alpha}}{\sin{2x}}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \frac{\sin{2x}\cos{\alpha}-\cos{2x}\sin{\alpha}}{\sin{2x}\cos{2x}}=-2\)
\(\Rightarrow \frac{\sin{(2x-\alpha)}}{\sin{2x}\cos{2x}}=-2\) ➜ \(\because \sin{A}\cos{B}-\cos{A}\sin{B}=\sin{(A-B)}\)
\(\Rightarrow \sin{(2x-\alpha)}=-2\sin{2x}\cos{2x}\)
\(\Rightarrow \sin{(2x-\alpha)}=-\sin{4x}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin{4x}=-\sin{(2x-\alpha)}\)
\(\Rightarrow \sin{4x}=\sin{(\alpha-2x)}\)
\(\Rightarrow 4x=m\pi+(-1)^m(\alpha-2x)\) ➜ \(\because \sin{A}=\sin{\theta}\)
\(\Rightarrow A=m\pi+(-1)^m\theta\) যেখানে, \(m\in{\mathbb{Z}}\)
\(\Rightarrow 4x=2n\pi+(-1)^(2n)(\alpha-2x)\) যখন, \(m\) জোড় সংখ্যা অর্থাৎ \(m=2n\)
\(\Rightarrow 4x=2n\pi+\alpha-2x\)
\(\Rightarrow 4x+2x=2n\pi+\alpha\)
\(\Rightarrow 6x=2n\pi+\alpha\)
\(\therefore x=\frac{1}{6}(2n\pi+\alpha)\)
আবার,
\(\Rightarrow 4x=2n\pi+(-1)^(2n+1)(\alpha-2x)\) যখন, \(m\) বিজোড় সংখ্যা অর্থাৎ \(m=2n+1\)
\(\Rightarrow 4x=(2n+1)\pi-(\alpha-2x)\)
\(\Rightarrow 4x=(2n+1)\pi-\alpha+2x\)
\(\Rightarrow 4x-2x=(2n+1)\pi-\alpha\)
\(\Rightarrow 2x=(2n+1)\pi-\alpha\)
\(\therefore x=\frac{1}{2}\{(2n+1)\pi-\alpha\}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{1}{6}(2n\pi+\alpha), \ \frac{1}{2}\{(2n+1)\pi-\alpha\}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLv)\) \(\sin{\left(\frac{\pi}{4}\tan{\theta}\right)}=\cos{\left(\frac{\pi}{4}\cot{\theta}\right)}\)উত্তরঃ \((4n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{\left(\frac{\pi}{4}\tan{\theta}\right)}=\cos{\left(\frac{\pi}{4}\cot{\theta}\right)}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}\cot{\theta}\right)}=\sin{\left(\frac{\pi}{4}\tan{\theta}\right)}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}\cot{\theta}\right)}=\cos{\left(\frac{\pi}{2}-\frac{\pi}{4}\tan{\theta}\right)}\)
\(\Rightarrow \frac{\pi}{4}\cot{\theta}=\frac{\pi}{2}-\frac{\pi}{4}\tan{\theta}\)
\(\Rightarrow \cot{\theta}=2-\tan{\theta}\) ➜ উভয় পার্শে \(\frac{4}{\pi}\) গুণ করে,
\(\Rightarrow \tan{\theta}+\cot{\theta}=2\)
\(\Rightarrow \tan{\theta}+\frac{1}{\tan{\theta}}=2\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(\Rightarrow \frac{\tan^2{\theta}+1}{\tan{\theta}}=2\)
\(\Rightarrow \tan^2{\theta}+1=2\tan{\theta}\)
\(\Rightarrow \tan^2{\theta}-2\tan{\theta}+1=0\)
\(\Rightarrow (\tan{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \tan{\theta}-1=0\)
\(\Rightarrow \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(4n+1)\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sin{\left(\frac{\pi}{4}\tan{\theta}\right)}=\cos{\left(\frac{\pi}{4}\cot{\theta}\right)}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}\cot{\theta}\right)}=\sin{\left(\frac{\pi}{4}\tan{\theta}\right)}\)
\(\Rightarrow \cos{\left(\frac{\pi}{4}\cot{\theta}\right)}=\cos{\left(\frac{\pi}{2}-\frac{\pi}{4}\tan{\theta}\right)}\)
\(\Rightarrow \frac{\pi}{4}\cot{\theta}=\frac{\pi}{2}-\frac{\pi}{4}\tan{\theta}\)
\(\Rightarrow \cot{\theta}=2-\tan{\theta}\) ➜ উভয় পার্শে \(\frac{4}{\pi}\) গুণ করে,
\(\Rightarrow \tan{\theta}+\cot{\theta}=2\)
\(\Rightarrow \tan{\theta}+\frac{1}{\tan{\theta}}=2\) ➜ \(\because \cot{A}=\frac{1}{\tan{A}}\)
\(\Rightarrow \frac{\tan^2{\theta}+1}{\tan{\theta}}=2\)
\(\Rightarrow \tan^2{\theta}+1=2\tan{\theta}\)
\(\Rightarrow \tan^2{\theta}-2\tan{\theta}+1=0\)
\(\Rightarrow (\tan{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow \tan{\theta}-1=0\)
\(\Rightarrow \tan{\theta}=1\)
\(\Rightarrow \tan{\theta}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \theta=n\pi+\frac{\pi}{4}\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(4n+1)\frac{\pi}{4}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLvi)\) \(1-2\sin{\theta}=\cos{\theta}\)উত্তরঃ \(2n\pi, \ 2n\pi+2\alpha\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
বুয়েটঃ২০০৫-২০০৬ ।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(1-2\sin{\theta}=\cos{\theta}\)
\(\Rightarrow \cos{\theta}=1-2\sin{\theta}\)
\(\Rightarrow \cos{\theta}+2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{5}}+\sin{\theta}\frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+2^2}=\sqrt{5}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha}=\cos{\alpha}\) ➜ \(\because \cos{\alpha}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow \sin{\alpha}=\frac{2}{\sqrt{5}}\)
\(\Rightarrow \cos{\left(\theta-\alpha\right)}=\cos{\alpha}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \theta-\alpha=2n\pi\pm{\alpha}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\alpha}+\alpha\)
\(\Rightarrow \theta=2n\pi-{\alpha}+\alpha, \ 2n\pi+{\alpha}+\alpha\)
\(\therefore \theta=2n\pi, \ 2n\pi+2{\alpha}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ 2n\pi+2{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
\(1-2\sin{\theta}=\cos{\theta}\)
\(\Rightarrow \cos{\theta}=1-2\sin{\theta}\)
\(\Rightarrow \cos{\theta}+2\sin{\theta}=1\)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{5}}+\sin{\theta}\frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+2^2}=\sqrt{5}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha}=\cos{\alpha}\) ➜ \(\because \cos{\alpha}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow \sin{\alpha}=\frac{2}{\sqrt{5}}\)
\(\Rightarrow \cos{\left(\theta-\alpha\right)}=\cos{\alpha}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow \theta-\alpha=2n\pi\pm{\alpha}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm{\alpha}+\alpha\)
\(\Rightarrow \theta=2n\pi-{\alpha}+\alpha, \ 2n\pi+{\alpha}+\alpha\)
\(\therefore \theta=2n\pi, \ 2n\pi+2{\alpha}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ 2n\pi+2{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{1}{\sqrt{5}}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLvii)\) \(\sqrt{3}\cot^2{\theta}+4\cot{\theta}+\sqrt{3}=0\)উত্তরঃ \(n\pi+\frac{2\pi}{3}, \ n\pi+\frac{5\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sqrt{3}\cot^2{\theta}+4\cot{\theta}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot^2{\theta}+3\cot{\theta}+\cot{\theta}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot{\theta}(\cot{\theta}+\sqrt{3})+1(\cot{\theta}+\sqrt{3})=0\)
\(\Rightarrow (\cot{\theta}+\sqrt{3})(\sqrt{3}\cot{\theta}+1)=0\)
\(\Rightarrow \cot{\theta}+\sqrt{3}=0, \ \sqrt{3}\cot{\theta}+1=0\)
\(\Rightarrow \cot{\theta}=-\sqrt{3}, \ \sqrt{3}\cot{\theta}=-1\)
\(\Rightarrow \cot{\theta}=-\sqrt{3}, \ \cot{\theta}=-\frac{1}{\sqrt{3}}\)
\(\Rightarrow \cot{\theta}=-\cot{\frac{\pi}{6}}, \ \cot{\theta}=-\cot{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\cot{\frac{\pi}{6}}\)
এবং \(\frac{1}{\sqrt{3}}=\cot{\frac{\pi}{3}}\)
\(\Rightarrow \cot{\theta}=\cot{\left(\pi-\frac{\pi}{6}\right)}, \ \cot{\theta}=\cot{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cot{A}=\cot{(\pi-A)}\)
\(\Rightarrow \cot{\theta}=\cot{\frac{5\pi}{6}}, \ \cot{\theta}=\cot{\frac{2\pi}{3}}\)
\(\Rightarrow \theta=n\pi+\frac{5\pi}{6}, \ \theta=n\pi+\frac{2\pi}{3}\) ➜ \(\because \cot{A}=\cot{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{2\pi}{3}, \ n\pi+\frac{5\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sqrt{3}\cot^2{\theta}+4\cot{\theta}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot^2{\theta}+3\cot{\theta}+\cot{\theta}+\sqrt{3}=0\)
\(\Rightarrow \sqrt{3}\cot{\theta}(\cot{\theta}+\sqrt{3})+1(\cot{\theta}+\sqrt{3})=0\)
\(\Rightarrow (\cot{\theta}+\sqrt{3})(\sqrt{3}\cot{\theta}+1)=0\)
\(\Rightarrow \cot{\theta}+\sqrt{3}=0, \ \sqrt{3}\cot{\theta}+1=0\)
\(\Rightarrow \cot{\theta}=-\sqrt{3}, \ \sqrt{3}\cot{\theta}=-1\)
\(\Rightarrow \cot{\theta}=-\sqrt{3}, \ \cot{\theta}=-\frac{1}{\sqrt{3}}\)
\(\Rightarrow \cot{\theta}=-\cot{\frac{\pi}{6}}, \ \cot{\theta}=-\cot{\frac{\pi}{3}}\) ➜ \(\because \sqrt{3}=\cot{\frac{\pi}{6}}\)
এবং \(\frac{1}{\sqrt{3}}=\cot{\frac{\pi}{3}}\)
\(\Rightarrow \cot{\theta}=\cot{\left(\pi-\frac{\pi}{6}\right)}, \ \cot{\theta}=\cot{\left(\pi-\frac{\pi}{3}\right)}\) ➜ \(\because -\cot{A}=\cot{(\pi-A)}\)
\(\Rightarrow \cot{\theta}=\cot{\frac{5\pi}{6}}, \ \cot{\theta}=\cot{\frac{2\pi}{3}}\)
\(\Rightarrow \theta=n\pi+\frac{5\pi}{6}, \ \theta=n\pi+\frac{2\pi}{3}\) ➜ \(\because \cot{A}=\cot{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+\frac{5\pi}{6}, \ n\pi+\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+\frac{2\pi}{3}, \ n\pi+\frac{5\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.3.(xLviii)\) \(4\sin^2{\theta}+\sqrt{3}=2(1+\sqrt{3})\sin{\theta}\)উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
প্রদত্ত সমীকরণ,
\(4\sin^2{\theta}+\sqrt{3}=2(1+\sqrt{3})\sin{\theta}\)
\(\Rightarrow 4\sin^2{\theta}+\sqrt{3}=2\sin{\theta}+2\sqrt{3}\sin{\theta}\)
\(\Rightarrow 4\sin^2{\theta}-2\sin{\theta}-2\sqrt{3}\sin{\theta}+\sqrt{3}=0\)
\(\Rightarrow 2\sin{\theta}(2\sin{\theta}-1)-\sqrt{3}(2\sin{\theta}-1)=0\)
\(\Rightarrow (2\sin{\theta}-1)(2\sin{\theta}-\sqrt{3})=0\)
\(\Rightarrow 2\sin{\theta}-1=0, \ 2\sin{\theta}-\sqrt{3}=0\)
\(\Rightarrow 2\sin{\theta}=1, \ 2\sin{\theta}=\sqrt{3}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}, \ \sin{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \sin{\theta}=\sin{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(4\sin^2{\theta}+\sqrt{3}=2(1+\sqrt{3})\sin{\theta}\)
\(\Rightarrow 4\sin^2{\theta}+\sqrt{3}=2\sin{\theta}+2\sqrt{3}\sin{\theta}\)
\(\Rightarrow 4\sin^2{\theta}-2\sin{\theta}-2\sqrt{3}\sin{\theta}+\sqrt{3}=0\)
\(\Rightarrow 2\sin{\theta}(2\sin{\theta}-1)-\sqrt{3}(2\sin{\theta}-1)=0\)
\(\Rightarrow (2\sin{\theta}-1)(2\sin{\theta}-\sqrt{3})=0\)
\(\Rightarrow 2\sin{\theta}-1=0, \ 2\sin{\theta}-\sqrt{3}=0\)
\(\Rightarrow 2\sin{\theta}=1, \ 2\sin{\theta}=\sqrt{3}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}, \ \sin{\theta}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}, \ \sin{\theta}=\sin{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}, \ \theta=n\pi+(-1)^n\frac{\pi}{3}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}, \ n\pi+(-1)^n\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
অধ্যায় \(7H\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
\(Q.4.(i)\) \(1+\sin^2{x}-2\cos^2{x}+3\cos{x}=3-\cos^2{x}\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \(2n\pi, \ 2n\pi\pm\frac{\pi}{3}\)
\(Q.4.(ii)\) \(f(x)=\tan{x}. \ \{f(x)\}^2+f^{\prime}(x)=3f(x)\) হলে বিশেষ সমাধান নির্ণয় কর যখন \(0\le{x}\le{2\pi}\)
উত্তরঃ \(\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(iii)\) \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \(2n\pi\pm\frac{2\pi}{3}, \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(iv)\) \(h(x)=\cos{x}. \ 2\{h(x)\}^2+\{h(2x)\}^2=2\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \(n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(Q.4.(v)\) \(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\) হলে, \(f(\sin{\theta})=0.\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(vi)\) \(\sqrt{2}x=\sin^{-1}{A}, \ \frac{-x}{2}=\cos^{-1}{B}\) এবং \(A-B=0.\) হলে, \(x\) এর সমাধানের জন্য সাধারণ রাশিমালা নির্ণয় কর।
উত্তরঃ \(\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(vii)\) \(g(x)=\cot{x}\) হলে, সমাধান করঃ \(g\left(\frac{\pi}{2}-\theta\right)g\left(\frac{3\pi}{2}-2\theta\right)=1, \ 0\le{\theta}\le{\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{5\pi}{6}\)
\(Q.4.(viii)\) \(f(x)=\sin{x}, \ g(x)=\cos{x}\) হলে, \(\sqrt{3}g(x)+f(x)=\sqrt{3}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(ix)\) \(g(x)=\cos{x}\) হলে, সমাধান করঃ \(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1, \ -2\pi\lt{x}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ -\frac{11\pi}{6}, \ -\frac{\pi}{2}, \ \frac{3\pi}{2}\)
\(Q.4.(x)\) \(f(x)=\sin{x}\) হলে, সমাধান করঃ \(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2, \ -2\pi\lt{x}\lt{2\pi}\)
উত্তরঃ \(-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(Q.4.(xi)\) \(\sin{2\theta}=\cos{3\theta}\) সমীকরণের যে সকল সমাধান \(0^{o}\) হতে \(360^{o}\) এর মধ্যে অবস্থিত, তাদের মান নির্ণয় কর।
উত্তরঃ \(18^{o}, \ 90^{o}, \ 162^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\)
\(Q.4.(xii)\) \(0\le{\theta}\le{2\pi}\) ব্যাবধিতে \(2\sin{\theta}\sin{3\theta}=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
উত্তরঃ \(2n\pi, \ 2n\pi\pm\frac{\pi}{3}\)
সকল বোঃ২০১৮।
\(Q.4.(ii)\) \(f(x)=\tan{x}. \ \{f(x)\}^2+f^{\prime}(x)=3f(x)\) হলে বিশেষ সমাধান নির্ণয় কর যখন \(0\le{x}\le{2\pi}\)
উত্তরঃ \(\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০১৭।
\(Q.4.(iii)\) \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \(2n\pi\pm\frac{2\pi}{3}, \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৭।
\(Q.4.(iv)\) \(h(x)=\cos{x}. \ 2\{h(x)\}^2+\{h(2x)\}^2=2\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \(n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
বঃ২০১৭।
\(Q.4.(v)\) \(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\) হলে, \(f(\sin{\theta})=0.\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০১৯।
\(Q.4.(vi)\) \(\sqrt{2}x=\sin^{-1}{A}, \ \frac{-x}{2}=\cos^{-1}{B}\) এবং \(A-B=0.\) হলে, \(x\) এর সমাধানের জন্য সাধারণ রাশিমালা নির্ণয় কর।
উত্তরঃ \(\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৯।
\(Q.4.(vii)\) \(g(x)=\cot{x}\) হলে, সমাধান করঃ \(g\left(\frac{\pi}{2}-\theta\right)g\left(\frac{3\pi}{2}-2\theta\right)=1, \ 0\le{\theta}\le{\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{5\pi}{6}\)
দিঃ২০১৯।
\(Q.4.(viii)\) \(f(x)=\sin{x}, \ g(x)=\cos{x}\) হলে, \(\sqrt{3}g(x)+f(x)=\sqrt{3}\) এর সমাধান নির্ণয় কর।
উত্তরঃ \(2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৭।
\(Q.4.(ix)\) \(g(x)=\cos{x}\) হলে, সমাধান করঃ \(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1, \ -2\pi\lt{x}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ -\frac{11\pi}{6}, \ -\frac{\pi}{2}, \ \frac{3\pi}{2}\)
যঃ২০১৯।
\(Q.4.(x)\) \(f(x)=\sin{x}\) হলে, সমাধান করঃ \(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2, \ -2\pi\lt{x}\lt{2\pi}\)
উত্তরঃ \(-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
রাঃ২০১৯।
\(Q.4.(xi)\) \(\sin{2\theta}=\cos{3\theta}\) সমীকরণের যে সকল সমাধান \(0^{o}\) হতে \(360^{o}\) এর মধ্যে অবস্থিত, তাদের মান নির্ণয় কর।
উত্তরঃ \(18^{o}, \ 90^{o}, \ 162^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\)
বুয়েটঃ২০০৪-২০০৫।
\(Q.4.(xii)\) \(0\le{\theta}\le{2\pi}\) ব্যাবধিতে \(2\sin{\theta}\sin{3\theta}=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
ঢাঃ২০১৯।
\(Q.4.(xiii)\) \(f(x)=\tan{x}\) হলে, সমাধান করঃ \(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
উত্তরঃ \((4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(Q.4.(xiv)\) \(\cos{\theta}+\sin{\theta}=\cos{2\theta}+\sin{2\theta}\) হলে, সমীকরণটির \(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে সমাধান আছে কিনা যাচাই কর।
উত্তরঃ \(0, \ \frac{\pi}{6}\)
\(Q.4.(xv)\) \(\sin{\sqrt{2}x}-\cos{\frac{x}{2}}=0\) হলে, \(x\) এর সমাধানের জন্য সাধারণ রাশিমালা নির্ণয় কর।
উত্তরঃ \(\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(xvi)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\sin^2{x}=\frac{1}{2}\sin{2x}, \ 0\le{x}\lt{2\pi} \)
উত্তরঃ \(0, \ \frac{\pi}{4}, \ \pi, \ \frac{5\pi}{4}\)
\(Q.4.(xvii)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi} \)
উত্তরঃ \(\frac{\pi}{8}, \ \frac{3\pi}{8}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\) এবং \(\frac{\pi}{2}\)
\(Q.4.(xviii)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -\pi\lt{\theta}\lt{\pi} \)
উত্তরঃ \(-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
\(Q.4.(xix)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\sin{4\theta}=\cos{3\theta}+\sin{2\theta}, \ 0\lt{\theta}\lt{\pi} \)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{5\pi}{6}\)
\(Q.4.(xx)\) \(h(x)=\sin{x}\) হলে, \(0\le{\theta}\le{2\pi}\) ব্যবধিতে \(2h(\theta).h(3\theta)=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}\) এবং \(\frac{11\pi}{6}\)
\(Q.4.(xxi)\) যদি \(\sin{(\pi\cot{\theta})}=\cos{(\pi\tan{\theta})}\) হয়, তবে প্রমাণ কর যে, \(cosec \ {2\theta}\) বা \(\cot{2\theta}\) এর মান \(\left(n+\frac{1}{4}\right)\) এর সমান, যখন \(n\) এর মান একটি পূর্ণ সংখ্যা হয়।
\(Q.4.(xxii)\) যদি \(\sin{A}=\sin{B}\) এবং \(\cos{A}=\cos{B}\) হয়, তবে প্রমাণ কর যে, \(A=B\) অথবা এদের পার্থক্য চার সমকোণের যেকোনো গুণিতকের সমান।
\(Q.4.(xxiii)\) যদি \(a\cos{\theta}+b\sin{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\) হয়, তবে প্রমাণ কর যে, \(\sin{(\alpha+\beta)}=\frac{2ab}{a^2+b^2}\)
\(Q.4.(xxiv)\) \(1+\cos{(y-z)}+\cos{(z-x)}+\cos{(x-y)}=0\) হলে, দেখাও যে, \((y-z)\) অথবা, \((z-x)\) অথবা, \((x-y)\) হবে \(\pi\) এর বিজোড় গুণিতকের সমান।
উত্তরঃ \((4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সিঃ২০১৯।
\(Q.4.(xiv)\) \(\cos{\theta}+\sin{\theta}=\cos{2\theta}+\sin{2\theta}\) হলে, সমীকরণটির \(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে সমাধান আছে কিনা যাচাই কর।
উত্তরঃ \(0, \ \frac{\pi}{6}\)
দিঃ২০১৭।
\(Q.4.(xv)\) \(\sin{\sqrt{2}x}-\cos{\frac{x}{2}}=0\) হলে, \(x\) এর সমাধানের জন্য সাধারণ রাশিমালা নির্ণয় কর।
উত্তরঃ \(\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৯।
\(Q.4.(xvi)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\sin^2{x}=\frac{1}{2}\sin{2x}, \ 0\le{x}\lt{2\pi} \)
উত্তরঃ \(0, \ \frac{\pi}{4}, \ \pi, \ \frac{5\pi}{4}\)
\(Q.4.(xvii)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi} \)
উত্তরঃ \(\frac{\pi}{8}, \ \frac{3\pi}{8}, \ \frac{5\pi}{8}, \ \frac{7\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\) এবং \(\frac{\pi}{2}\)
\(Q.4.(xviii)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -\pi\lt{\theta}\lt{\pi} \)
উত্তরঃ \(-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
\(Q.4.(xix)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\sin{4\theta}=\cos{3\theta}+\sin{2\theta}, \ 0\lt{\theta}\lt{\pi} \)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{5\pi}{6}\)
\(Q.4.(xx)\) \(h(x)=\sin{x}\) হলে, \(0\le{\theta}\le{2\pi}\) ব্যবধিতে \(2h(\theta).h(3\theta)=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}\) এবং \(\frac{11\pi}{6}\)
ঢাঃ২০১৯।
\(Q.4.(xxi)\) যদি \(\sin{(\pi\cot{\theta})}=\cos{(\pi\tan{\theta})}\) হয়, তবে প্রমাণ কর যে, \(cosec \ {2\theta}\) বা \(\cot{2\theta}\) এর মান \(\left(n+\frac{1}{4}\right)\) এর সমান, যখন \(n\) এর মান একটি পূর্ণ সংখ্যা হয়।
\(Q.4.(xxii)\) যদি \(\sin{A}=\sin{B}\) এবং \(\cos{A}=\cos{B}\) হয়, তবে প্রমাণ কর যে, \(A=B\) অথবা এদের পার্থক্য চার সমকোণের যেকোনো গুণিতকের সমান।
\(Q.4.(xxiii)\) যদি \(a\cos{\theta}+b\sin{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\) হয়, তবে প্রমাণ কর যে, \(\sin{(\alpha+\beta)}=\frac{2ab}{a^2+b^2}\)
\(Q.4.(xxiv)\) \(1+\cos{(y-z)}+\cos{(z-x)}+\cos{(x-y)}=0\) হলে, দেখাও যে, \((y-z)\) অথবা, \((z-x)\) অথবা, \((x-y)\) হবে \(\pi\) এর বিজোড় গুণিতকের সমান।
\(Q.4.(i)\) \(1+\sin^2{x}-2\cos^2{x}+3\cos{x}=3-\cos^2{x}\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \(2n\pi, \ 2n\pi\pm\frac{\pi}{3}\)
উত্তরঃ \(2n\pi, \ 2n\pi\pm\frac{\pi}{3}\)
সকল বোঃ২০১৮।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(1+\sin^2{x}-2\cos^2{x}+3\cos{x}=3-\cos^2{x}\)
\(\Rightarrow 1+1-\cos^2{x}-2\cos^2{x}+3\cos{x}-3+\cos^2{x}=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow -2\cos^2{x}+3\cos{x}-1=0\)
\(\Rightarrow 2\cos^2{x}-3\cos{x}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{x}(\cos{x}-1)-1(\cos{x}-1)=0\)
\(\Rightarrow (\cos{x}-1)(2\cos{x}-1)=0\)
\(\Rightarrow \cos{x}-1=0, \ 2\cos{x}-1=0\)
\(\Rightarrow \cos{x}=1, \ 2\cos{x}=1\)
\(\Rightarrow \cos{x}=1, \ \cos{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=1, \ \cos{x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x=2n\pi, \ x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi, \ 2n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ 2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(1+\sin^2{x}-2\cos^2{x}+3\cos{x}=3-\cos^2{x}\)
\(\Rightarrow 1+1-\cos^2{x}-2\cos^2{x}+3\cos{x}-3+\cos^2{x}=0\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow -2\cos^2{x}+3\cos{x}-1=0\)
\(\Rightarrow 2\cos^2{x}-3\cos{x}+1=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{x}(\cos{x}-1)-1(\cos{x}-1)=0\)
\(\Rightarrow (\cos{x}-1)(2\cos{x}-1)=0\)
\(\Rightarrow \cos{x}-1=0, \ 2\cos{x}-1=0\)
\(\Rightarrow \cos{x}=1, \ 2\cos{x}=1\)
\(\Rightarrow \cos{x}=1, \ \cos{x}=\frac{1}{2}\)
\(\Rightarrow \cos{x}=1, \ \cos{x}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow x=2n\pi, \ x=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=1\)
\(\Rightarrow A=2n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=2n\pi, \ 2n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ 2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.4.(ii)\) \(f(x)=\tan{x}. \ \{f(x)\}^2+f^{\prime}(x)=3f(x)\) হলে বিশেষ সমাধান নির্ণয় কর যখন \(0\le{x}\le{2\pi}\)উত্তরঃ \(\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
রাঃ২০১৭।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(f(x)=\tan{x}. \ \{f(x)\}^2+f^{\prime}(x)=3f(x)\) যখন \(0\le{x}\le{2\pi}\)
\(\Rightarrow \{f(x)\}^2+f^{\prime}(x)=3f(x)\)
\(\Rightarrow \{f(x)\}^2+\frac{d}{dx}\{f(x)\}=3f(x)\)
\(\Rightarrow \{\tan{x}\}^2+\frac{d}{dx}(\tan{x})=3\tan{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan^2{x}+\sec^2{x}=3\tan{x}\) ➜ \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(f(x)=\tan{x}. \ \{f(x)\}^2+f^{\prime}(x)=3f(x)\) যখন \(0\le{x}\le{2\pi}\)
\(\Rightarrow \{f(x)\}^2+f^{\prime}(x)=3f(x)\)
\(\Rightarrow \{f(x)\}^2+\frac{d}{dx}\{f(x)\}=3f(x)\)
\(\Rightarrow \{\tan{x}\}^2+\frac{d}{dx}(\tan{x})=3\tan{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan^2{x}+\sec^2{x}=3\tan{x}\) ➜ \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{x}\le{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
সমাধান করঃ
\(Q.4.(iii)\) \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর। উত্তরঃ \(2n\pi\pm\frac{2\pi}{3}, \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৭।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\(\Rightarrow 2\times2\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}+1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}+1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=-1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=-\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=\sin{\left(-\frac{\pi}{6}\right)}\) ➜ \(\because \sin{(-\theta)}=-\sin{\theta}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=n\pi+(-1)^n\left(-\frac{\pi}{6}\right)\) ➜ \(\because \cos{A}=cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=n\pi-(-1)^n\frac{\pi}{6}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi-(-1)^n\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\(\Rightarrow 2\times2\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}+1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}+1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=-1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=-\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=\sin{\left(-\frac{\pi}{6}\right)}\) ➜ \(\because \sin{(-\theta)}=-\sin{\theta}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=n\pi+(-1)^n\left(-\frac{\pi}{6}\right)\) ➜ \(\because \cos{A}=cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=n\pi-(-1)^n\frac{\pi}{6}\)
\(\therefore \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi-(-1)^n\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi-(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.4.(iv)\) \(h(x)=\cos{x}. \ 2\{h(x)\}^2+\{h(2x)\}^2=2\) সমীকরণটির সাধারণ সমাধান নির্ণয় কর। উত্তরঃ \(n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
বঃ২০১৭।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(h(x)=\cos{x}. \ 2\{h(x)\}^2+\{h(2x)\}^2=2\)
\(\Rightarrow 2\{\cos{x}\}^2+\{\cos{2x}\}^2=2\) ➜ \(\because h(x)=\cos{x}\)
\(\Rightarrow 2\cos^2{x}+\cos^2{2x}=2\)
\(\Rightarrow \cos{2x}+1+\cos^2{2x}=2\) ➜ \(\because 2\cos^2{A}=\cos{2A}+1\)
\(\Rightarrow \cos^2{2x}+\cos{2x}+1-2=0\)
\(\Rightarrow \cos^2{2x}+\cos{2x}-1=0\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1+\sqrt{5}}{2}, \ \cos{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \cos{2x}=\cos{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\theta}\)
\(\Rightarrow A=2n\pi\pm\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(h(x)=\cos{x}. \ 2\{h(x)\}^2+\{h(2x)\}^2=2\)
\(\Rightarrow 2\{\cos{x}\}^2+\{\cos{2x}\}^2=2\) ➜ \(\because h(x)=\cos{x}\)
\(\Rightarrow 2\cos^2{x}+\cos^2{2x}=2\)
\(\Rightarrow \cos{2x}+1+\cos^2{2x}=2\) ➜ \(\because 2\cos^2{A}=\cos{2A}+1\)
\(\Rightarrow \cos^2{2x}+\cos{2x}+1-2=0\)
\(\Rightarrow \cos^2{2x}+\cos{2x}-1=0\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1+\sqrt{5}}{2}, \ \cos{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \cos{2x}=\cos{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\theta}\)
\(\Rightarrow A=2n\pi\pm\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi\pm\frac{\alpha}{2}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.4.(v)\) \(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\) হলে, \(f(\sin{\theta})=0.\) এর সমাধান নির্ণয় কর। উত্তরঃ \(n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
বঃ২০১৯।
সমাধানঃ
দেওয়া আছে,
\(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
প্রদত্ত সমীকরণ,
\(f(\sin{\theta})=0\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\) ➜ \(\because f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
\(\Rightarrow f(\sin{\theta})=\sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}(\sin{\theta}-\sqrt{2})-1(\sin{\theta}-\sqrt{2})=0\)
\(\Rightarrow (\sin{\theta}-\sqrt{2})(\sqrt{2}\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}-\sqrt{2}\ne{0}, \ \sqrt{2}\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sqrt{2}\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
প্রদত্ত সমীকরণ,
\(f(\sin{\theta})=0\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\) ➜ \(\because f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
\(\Rightarrow f(\sin{\theta})=\sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}(\sin{\theta}-\sqrt{2})-1(\sin{\theta}-\sqrt{2})=0\)
\(\Rightarrow (\sin{\theta}-\sqrt{2})(\sqrt{2}\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}-\sqrt{2}\ne{0}, \ \sqrt{2}\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sqrt{2}\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.4.(vi)\) \(\sqrt{2}x=\sin^{-1}{A}, \ \frac{-x}{2}=\cos^{-1}{B}\) এবং \(A-B=0.\) হলে, \(x\) এর সমাধানের জন্য সাধারণ রাশিমালা নির্ণয় কর। উত্তরঃ \(\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৯।
সমাধানঃ
দেওয়া আছে,
\(\sqrt{2}x=\sin^{-1}{A}, \ \frac{-x}{2}=\cos^{-1}{B}\)
\(\therefore \sin{\sqrt{2}x}=A, \ \cos{\frac{-x}{2}}=B\)
প্রদত্ত সমীকরণ,
\(A-B=0\)
\(\Rightarrow \sin{\sqrt{2}x}-\cos{\frac{-x}{2}}=0\) ➜ \(\because \sin{\sqrt{2}x}=A\)
এবং \(\cos{\frac{-x}{2}}=B\)
\(\Rightarrow \sin{\sqrt{2}x}=\cos{\frac{-x}{2}}\)
\(\Rightarrow \sin{\sqrt{2}x}=\cos{\frac{x}{2}}\) ➜ \(\because \cos{(-\theta)}=\cos{\theta}\)
\(\Rightarrow \cos{\frac{x}{2}}=\sin{\sqrt{2}x}\)
\(\Rightarrow \cos{\frac{x}{2}}=\cos{\left(\frac{\pi}{2}-\sqrt{2}x\right)}\)
\(\Rightarrow \frac{x}{2}=2n\pi\pm\left(\frac{\pi}{2}-\sqrt{2}x\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \frac{x}{2}=2n\pi+\left(\frac{\pi}{2}-\sqrt{2}x\right), \ \frac{x}{2}=2n\pi-\left(\frac{\pi}{2}-\sqrt{2}x\right)\)
\(\Rightarrow \frac{x}{2}=2n\pi+\frac{\pi}{2}-\sqrt{2}x, \ \frac{x}{2}=2n\pi-\frac{\pi}{2}+\sqrt{2}x\)
\(\Rightarrow \frac{x}{2}+\sqrt{2}x=2n\pi+\frac{\pi}{2}, \ \frac{x}{2}-\sqrt{2}x=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow \left(\frac{1}{2}+\sqrt{2}\right)x=(4n+1)\frac{\pi}{2}, \ \left(\frac{1}{2}-\sqrt{2}\right)x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \frac{1+2\sqrt{2}}{2}x=(4n+1)\frac{\pi}{2}, \ \frac{1-2\sqrt{2}}{2}x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow x=(4n+1)\frac{\pi}{2}\times\frac{2}{1+2\sqrt{2}}, \ x=(4n-1)\frac{\pi}{2}\times\frac{2}{1-2\sqrt{2}}\)
\(\Rightarrow x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ x=\frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\sqrt{2}x=\sin^{-1}{A}, \ \frac{-x}{2}=\cos^{-1}{B}\)
\(\therefore \sin{\sqrt{2}x}=A, \ \cos{\frac{-x}{2}}=B\)
প্রদত্ত সমীকরণ,
\(A-B=0\)
\(\Rightarrow \sin{\sqrt{2}x}-\cos{\frac{-x}{2}}=0\) ➜ \(\because \sin{\sqrt{2}x}=A\)
এবং \(\cos{\frac{-x}{2}}=B\)
\(\Rightarrow \sin{\sqrt{2}x}=\cos{\frac{-x}{2}}\)
\(\Rightarrow \sin{\sqrt{2}x}=\cos{\frac{x}{2}}\) ➜ \(\because \cos{(-\theta)}=\cos{\theta}\)
\(\Rightarrow \cos{\frac{x}{2}}=\sin{\sqrt{2}x}\)
\(\Rightarrow \cos{\frac{x}{2}}=\cos{\left(\frac{\pi}{2}-\sqrt{2}x\right)}\)
\(\Rightarrow \frac{x}{2}=2n\pi\pm\left(\frac{\pi}{2}-\sqrt{2}x\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \frac{x}{2}=2n\pi+\left(\frac{\pi}{2}-\sqrt{2}x\right), \ \frac{x}{2}=2n\pi-\left(\frac{\pi}{2}-\sqrt{2}x\right)\)
\(\Rightarrow \frac{x}{2}=2n\pi+\frac{\pi}{2}-\sqrt{2}x, \ \frac{x}{2}=2n\pi-\frac{\pi}{2}+\sqrt{2}x\)
\(\Rightarrow \frac{x}{2}+\sqrt{2}x=2n\pi+\frac{\pi}{2}, \ \frac{x}{2}-\sqrt{2}x=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow \left(\frac{1}{2}+\sqrt{2}\right)x=(4n+1)\frac{\pi}{2}, \ \left(\frac{1}{2}-\sqrt{2}\right)x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \frac{1+2\sqrt{2}}{2}x=(4n+1)\frac{\pi}{2}, \ \frac{1-2\sqrt{2}}{2}x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow x=(4n+1)\frac{\pi}{2}\times\frac{2}{1+2\sqrt{2}}, \ x=(4n-1)\frac{\pi}{2}\times\frac{2}{1-2\sqrt{2}}\)
\(\Rightarrow x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ x=\frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধান করঃ
\(Q.4.(vii)\) \(g(x)=\cot{x}\) হলে, সমাধান করঃ \(g\left(\frac{\pi}{2}-\theta\right)g\left(\frac{3\pi}{2}-2\theta\right)=1, \ 0\le{\theta}\le{\pi}\) উত্তরঃ \(\frac{\pi}{6}, \ \frac{5\pi}{6}\)
দিঃ২০১৯।
সমাধানঃ
দেওয়া আছে,
\(g(x)=\cot{x}\)
প্রদত্ত সমীকরণ,
\(g\left(\frac{\pi}{2}-\theta\right)g\left(\frac{3\pi}{2}-2\theta\right)=1\)
\(\Rightarrow \cot{\left(\frac{\pi}{2}-\theta\right)}\cot{\left(\frac{3\pi}{2}-2\theta\right)}=1\) ➜ \(\because g(x)=\cot{x}\)
\(\Rightarrow \cot{\left(\frac{\pi}{2}-\theta\right)}\cot{\left(\frac{\pi}{2}\times3-2\theta\right)}=1\)
\(\Rightarrow \tan{\theta}\tan{2\theta}=1\) ➜
প্রথম পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং কোট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে ট্যানজেন্ট হয়েছে।

দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(3\) একটি বিজোড় সংখ্যা,
সুতরাং কোট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে ট্যানজেন্ট হয়েছে।
\(\Rightarrow \tan{\theta}\frac{2\tan{\theta}}{1-tan^2{\theta}}=1\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-tan^2{A}}\)
\(\Rightarrow \frac{2\tan^2{\theta}}{1-tan^2{\theta}}=1\)
\(\Rightarrow 2\tan^2{\theta}=1-tan^2{\theta}\)
\(\Rightarrow 2\tan^2{\theta}+tan^2{\theta}=1\)
\(\Rightarrow 3tan^2{\theta}=1\)
\(\Rightarrow tan^2{\theta}=\frac{1}{3}\)
\(\Rightarrow tan{\theta}=\pm\frac{1}{\sqrt{3}}\)
\(\Rightarrow tan{\theta}=\pm\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow tan{\theta}=\tan{\left(\pm\frac{\pi}{6}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{6}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{6}\) \(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{6}, \ \frac{5\pi}{6}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{5\pi}{6}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{6}, \ \frac{5\pi}{6}\)
\(g(x)=\cot{x}\)
প্রদত্ত সমীকরণ,
\(g\left(\frac{\pi}{2}-\theta\right)g\left(\frac{3\pi}{2}-2\theta\right)=1\)
\(\Rightarrow \cot{\left(\frac{\pi}{2}-\theta\right)}\cot{\left(\frac{3\pi}{2}-2\theta\right)}=1\) ➜ \(\because g(x)=\cot{x}\)
\(\Rightarrow \cot{\left(\frac{\pi}{2}-\theta\right)}\cot{\left(\frac{\pi}{2}\times3-2\theta\right)}=1\)
\(\Rightarrow \tan{\theta}\tan{2\theta}=1\) ➜

প্রথম পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং কোট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে ট্যানজেন্ট হয়েছে।

দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি তৃতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(3\) একটি বিজোড় সংখ্যা,
সুতরাং কোট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে ট্যানজেন্ট হয়েছে।
\(\Rightarrow \tan{\theta}\frac{2\tan{\theta}}{1-tan^2{\theta}}=1\) ➜ \(\because \tan{2A}=\frac{2\tan{A}}{1-tan^2{A}}\)
\(\Rightarrow \frac{2\tan^2{\theta}}{1-tan^2{\theta}}=1\)
\(\Rightarrow 2\tan^2{\theta}=1-tan^2{\theta}\)
\(\Rightarrow 2\tan^2{\theta}+tan^2{\theta}=1\)
\(\Rightarrow 3tan^2{\theta}=1\)
\(\Rightarrow tan^2{\theta}=\frac{1}{3}\)
\(\Rightarrow tan{\theta}=\pm\frac{1}{\sqrt{3}}\)
\(\Rightarrow tan{\theta}=\pm\tan{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow tan{\theta}=\tan{\left(\pm\frac{\pi}{6}\right)}\)
\(\Rightarrow \theta=n\pi+\left(\pm\frac{\pi}{6}\right)\) ➜ \(\because \tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=n\pi\pm\frac{\pi}{6}\) \(\therefore\) সাধারণ সমাধান, \(\theta=n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{6}, \ -\frac{\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{6}, \ \frac{5\pi}{6}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow x=-\frac{5\pi}{6}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{6}, \ \frac{5\pi}{6}\)
সমাধান করঃ
\(Q.4.(viii)\) \(f(x)=\sin{x}, \ g(x)=\cos{x}\) হলে, \(\sqrt{3}g(x)+f(x)=\sqrt{3}\) এর সমাধান নির্ণয় কর। উত্তরঃ \(2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৭।
সমাধানঃ
দেওয়া আছে,
\(f(x)=\sin{x}, \ g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{3}\) ➜ \(\because g(x)=\cos{x}\)
এবং \(f(x)=\sin{x}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{3}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+1^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{6}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi}{6}, \ x=2n\pi\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}, \ x=2n\pi\)
\(\therefore x=2n\pi+\frac{\pi}{3}, \ 2n\pi\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(f(x)=\sin{x}, \ g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{3}\) ➜ \(\because g(x)=\cos{x}\)
এবং \(f(x)=\sin{x}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{3}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+1^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{6}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi}{6}, \ x=2n\pi\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}, \ x=2n\pi\)
\(\therefore x=2n\pi+\frac{\pi}{3}, \ 2n\pi\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(ix)\) \(g(x)=\cos{x}\) হলে, সমাধান করঃ \(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1, \ -2\pi\lt{x}\lt{2\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ -\frac{11\pi}{6}, \ -\frac{\pi}{2}, \ \frac{3\pi}{2}\)
উত্তরঃ \(\frac{\pi}{6}, \ -\frac{11\pi}{6}, \ -\frac{\pi}{2}, \ \frac{3\pi}{2}\)
যঃ২০১৯।
সমাধানঃ
দেওয়া আছে,
\(g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}+x\right)}=1\) ➜ \(\because g(x)=\cos{x}\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}\times1+x\right)}=1\)
\(\Rightarrow \sqrt{3}\cos{x}-\sin{x}=1\) ➜
দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}-\sin{x}\frac{1}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+(-1)^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}-\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{6}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{6}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}-\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi-\pi}{6}, \ x=2n\pi-\frac{2\pi+\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{3\pi}{6}\)
\(\therefore x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{6}, \ -\frac{\pi}{2}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=\frac{13\pi}{6}, \ \frac{3\pi}{2}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{2}\)
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=-\frac{11\pi}{6}, \ -\frac{5\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{11\pi}{6}\)
যখন, \(n=2,\) \(x=4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{2}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{6}, \ -\frac{11\pi}{6}, \ -\frac{\pi}{2}, \ \frac{3\pi}{2}\)
\(g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}+x\right)}=1\) ➜ \(\because g(x)=\cos{x}\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}\times1+x\right)}=1\)
\(\Rightarrow \sqrt{3}\cos{x}-\sin{x}=1\) ➜

দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}-\sin{x}\frac{1}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+(-1)^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}-\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{6}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{6}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}-\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi-\pi}{6}, \ x=2n\pi-\frac{2\pi+\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{3\pi}{6}\)
\(\therefore x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{6}, \ -\frac{\pi}{2}\) সবগুলি মান গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{6}, \ -\frac{\pi}{2}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=\frac{13\pi}{6}, \ \frac{3\pi}{2}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{2}\)
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=-\frac{11\pi}{6}, \ -\frac{5\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{11\pi}{6}\)
যখন, \(n=2,\) \(x=4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{2}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{6}, \ -\frac{11\pi}{6}, \ -\frac{\pi}{2}, \ \frac{3\pi}{2}\)
\(Q.4.(x)\) \(f(x)=\sin{x}\) হলে, সমাধান করঃ \(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2, \ -2\pi\lt{x}\lt{2\pi}\)
উত্তরঃ \(-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
উত্তরঃ \(-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
রাঃ২০১৯।
সমাধানঃ
দেওয়া আছে,
\(f(x)=\sin{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}+x\right)}=2\) ➜ \(\because f(x)=\sin{x}\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}\times1+x\right)}=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\cos{x}=2\) ➜
দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow \cos{x}-\sqrt{3}\sin{x}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{x}\frac{1}{2}-\sin{x}\frac{\sqrt{3}}{2}=-\frac{2}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{3}}-\sin{x}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow x=(6n+3-1)\frac{\pi}{3}\)
\(\Rightarrow x=(6n+2)\frac{\pi}{3}\)
\(\Rightarrow x=2(3n+1)\frac{\pi}{3}\)
\(\therefore x=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(x=4\frac{2\pi}{3}\)
\(\Rightarrow x=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\frac{2\pi}{3}\)
\(\Rightarrow x=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{4\pi}{3}\)
যখন, \(n=2,\) \(x=\frac{14\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(x=-\frac{10\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(f(x)=\sin{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}+x\right)}=2\) ➜ \(\because f(x)=\sin{x}\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}\times1+x\right)}=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\cos{x}=2\) ➜

দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow \cos{x}-\sqrt{3}\sin{x}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{x}\frac{1}{2}-\sin{x}\frac{\sqrt{3}}{2}=-\frac{2}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{3}}-\sin{x}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow x=(6n+3-1)\frac{\pi}{3}\)
\(\Rightarrow x=(6n+2)\frac{\pi}{3}\)
\(\Rightarrow x=2(3n+1)\frac{\pi}{3}\)
\(\therefore x=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{2\pi}{3}\) মানটি গ্রহনযোগ্য।\(\therefore x=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(x=4\frac{2\pi}{3}\)
\(\Rightarrow x=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\frac{2\pi}{3}\)
\(\Rightarrow x=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{4\pi}{3}\)
যখন, \(n=2,\) \(x=\frac{14\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(x=-\frac{10\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(Q.4.(xi)\) \(\sin{2\theta}=\cos{3\theta}\) সমীকরণের যে সকল সমাধান \(0^{o}\) হতে \(360^{o}\) এর মধ্যে অবস্থিত, তাদের মান নির্ণয় কর।
উত্তরঃ \(18^{o}, \ 90^{o}, \ 162^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\)
উত্তরঃ \(18^{o}, \ 90^{o}, \ 162^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\)
বুয়েটঃ২০০৪-২০০৫।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\sin{2\theta}=\cos{3\theta}\)
\(\Rightarrow \cos{3\theta}=\sin{2\theta}\)
\(\Rightarrow \cos{3\theta}=\cos{\left(\frac{\pi}{2}-2\theta\right)}\) ➜ \(\because \sin(x)=\cos{\left(\frac{\pi}{2}-x\right)}\)
\(\Rightarrow 3\theta=2n\pi\pm\left(\frac{\pi}{2}-2\theta\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=2n\pi+\left(\frac{\pi}{2}-2\theta\right), \ 3\theta=2n\pi-\left(\frac{\pi}{2}-2\theta\right)\)
\(\Rightarrow 3\theta=2n\pi+\frac{\pi}{2}-2\theta, \ 3\theta=2n\pi-\frac{\pi}{2}+2\theta\)
\(\Rightarrow 3\theta+2\theta=2n\pi+\frac{\pi}{2}, \ 3\theta-2\theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=2n\pi+\frac{\pi}{2}, \ \theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=(4n+1)\frac{\pi}{2}, \ \theta=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n+1)\frac{\pi}{10}, \ \theta=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n+1)18^{o}, \ \theta=(4n-1)90^{o}\)
\(\therefore \theta=(4n+1)18^{o}, \ (4n-1)90^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)18^{o}, \ (4n-1)90^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=18^{o}\)
যখন, \(n=1,\) \(\theta=5\times18^{o}, \ 3\times90^{o}\)
\(\Rightarrow \theta=90^{o}, \ 270^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=90^{o}, \ 270^{o}\)
যখন, \(n=-1,\) \(\theta=-3\times18^{o}, \ -5\times90^{o}\)
\(\Rightarrow \theta=-54^{o}, \ -450^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=9\times18^{o}, \ 7\times90^{o}\)
\(\Rightarrow \theta=162^{o}, \ 630^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=3,\) \(\theta=13\times18^{o}, \ 11\times90^{o}\)
\(\Rightarrow \theta=234^{o}, \ 990^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=234^{o}\)
যখন, \(n=4,\) \(\theta=17\times18^{o}, \ 15\times90^{o}\)
\(\Rightarrow \theta=306^{o}, \ 1350^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=306^{o}\)
যখন, \(n=5,\) \(\theta=21\times18^{o}, \ 19\times90^{o}\)
\(\Rightarrow \theta=378^{o}, \ 1350^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore 0^{o}\) হতে \(360^{o}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=18^{o}, \ 90^{o}, \ 162^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\)
\(\sin{2\theta}=\cos{3\theta}\)
\(\Rightarrow \cos{3\theta}=\sin{2\theta}\)
\(\Rightarrow \cos{3\theta}=\cos{\left(\frac{\pi}{2}-2\theta\right)}\) ➜ \(\because \sin(x)=\cos{\left(\frac{\pi}{2}-x\right)}\)
\(\Rightarrow 3\theta=2n\pi\pm\left(\frac{\pi}{2}-2\theta\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow 3\theta=2n\pi+\left(\frac{\pi}{2}-2\theta\right), \ 3\theta=2n\pi-\left(\frac{\pi}{2}-2\theta\right)\)
\(\Rightarrow 3\theta=2n\pi+\frac{\pi}{2}-2\theta, \ 3\theta=2n\pi-\frac{\pi}{2}+2\theta\)
\(\Rightarrow 3\theta+2\theta=2n\pi+\frac{\pi}{2}, \ 3\theta-2\theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=2n\pi+\frac{\pi}{2}, \ \theta=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow 5\theta=(4n+1)\frac{\pi}{2}, \ \theta=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n+1)\frac{\pi}{10}, \ \theta=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \theta=(4n+1)18^{o}, \ \theta=(4n-1)90^{o}\)
\(\therefore \theta=(4n+1)18^{o}, \ (4n-1)90^{o}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n+1)18^{o}, \ (4n-1)90^{o}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0^{o}\) হতে \(360^{o}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=18^{o}, \ -90^{o}\) প্রথম মানটি গ্রহনযোগ্য।\(\therefore \theta=18^{o}\)
যখন, \(n=1,\) \(\theta=5\times18^{o}, \ 3\times90^{o}\)
\(\Rightarrow \theta=90^{o}, \ 270^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=90^{o}, \ 270^{o}\)
যখন, \(n=-1,\) \(\theta=-3\times18^{o}, \ -5\times90^{o}\)
\(\Rightarrow \theta=-54^{o}, \ -450^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=9\times18^{o}, \ 7\times90^{o}\)
\(\Rightarrow \theta=162^{o}, \ 630^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=3,\) \(\theta=13\times18^{o}, \ 11\times90^{o}\)
\(\Rightarrow \theta=234^{o}, \ 990^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=234^{o}\)
যখন, \(n=4,\) \(\theta=17\times18^{o}, \ 15\times90^{o}\)
\(\Rightarrow \theta=306^{o}, \ 1350^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=306^{o}\)
যখন, \(n=5,\) \(\theta=21\times18^{o}, \ 19\times90^{o}\)
\(\Rightarrow \theta=378^{o}, \ 1350^{o}\) মান দুইটি গ্রহনযোগ্য নয়।
\(\therefore 0^{o}\) হতে \(360^{o}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=18^{o}, \ 90^{o}, \ 162^{o}, \ 234^{o}, \ 270^{o}, \ 306^{o}\)
\(Q.4.(xii)\) \(0\le{\theta}\le{2\pi}\) ব্যাবধিতে \(2\sin{\theta}\sin{3\theta}=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
উত্তরঃ \(\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
ঢাঃ২০১৯।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(2\sin{\theta}\sin{3\theta}=1\)
\(\Rightarrow 2\sin{3\theta}\sin{\theta}=1\)
\(\Rightarrow \cos{(3\theta-\theta)}-\cos{(3\theta+\theta)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=1\)
\(\Rightarrow \cos{2\theta}=1+\cos{4\theta}\)
\(\Rightarrow \cos{2\theta}=2\cos^2{2\theta}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2\theta}-2\cos^2{2\theta}=0\)
\(\Rightarrow -\cos{2\theta}(2\cos{2\theta}-1)=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}-1=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সকল মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
\(2\sin{\theta}\sin{3\theta}=1\)
\(\Rightarrow 2\sin{3\theta}\sin{\theta}=1\)
\(\Rightarrow \cos{(3\theta-\theta)}-\cos{(3\theta+\theta)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=1\)
\(\Rightarrow \cos{2\theta}=1+\cos{4\theta}\)
\(\Rightarrow \cos{2\theta}=2\cos^2{2\theta}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2\theta}-2\cos^2{2\theta}=0\)
\(\Rightarrow -\cos{2\theta}(2\cos{2\theta}-1)=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}-1=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{4}, \ \frac{\pi}{6}, \ -\frac{\pi}{6}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সকল মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
\(Q.4.(xiii)\) \(f(x)=\tan{x}\) হলে, সমাধান করঃ \(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
উত্তরঃ \((4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
উত্তরঃ \((4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সিঃ২০১৯।
সমাধানঃ
দেওয়া আছে,
\(f(x)=\tan{x}\)
প্রদত্ত সমীকরণ,
\(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}-2x\right)}=\cos{x}+\sin{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}\times1-2x\right)}=\cos{x}+\sin{x}\)
\(\Rightarrow \cot{2x}=\cos{x}+\sin{x}\) ➜
প্রথম পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং ট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং ট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে কোট্যানজেন্ট হয়েছে।
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\cos{x}+\sin{x}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos^2{2x}}{\sin^2{2x}}=(\cos{x}+\sin{x})^2\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=\cos^2{x}+\sin^2{x}+2\sin{x}\cos{x}\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
এবং \((a+b)^2=a^2+b^2+2ab\)
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=1+\sin{2x}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})=1-\sin^2{2x}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1-\sin^2{2x})=0\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1+\sin{2x})(1-\sin{2x})=0\) ➜ \(\because a^2-b^2=(a+b)(a-b)\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}-1+\sin{2x})=0\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}+\sin{2x}-1)=0\)
\(\Rightarrow 1+\sin{2x}=0, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=-1, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow 2x=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n-1)\frac{\pi}{4}\)
আবার,
\(\sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1+\sqrt{5}}{2}, \ \sin{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\sin{2x}}\le{1}\)
\(\Rightarrow \sin{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(f(x)=\tan{x}\)
প্রদত্ত সমীকরণ,
\(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}-2x\right)}=\cos{x}+\sin{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}\times1-2x\right)}=\cos{x}+\sin{x}\)
\(\Rightarrow \cot{2x}=\cos{x}+\sin{x}\) ➜

প্রথম পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং ট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং ট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে কোট্যানজেন্ট হয়েছে।
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\cos{x}+\sin{x}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos^2{2x}}{\sin^2{2x}}=(\cos{x}+\sin{x})^2\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=\cos^2{x}+\sin^2{x}+2\sin{x}\cos{x}\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
এবং \((a+b)^2=a^2+b^2+2ab\)
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=1+\sin{2x}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})=1-\sin^2{2x}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1-\sin^2{2x})=0\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1+\sin{2x})(1-\sin{2x})=0\) ➜ \(\because a^2-b^2=(a+b)(a-b)\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}-1+\sin{2x})=0\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}+\sin{2x}-1)=0\)
\(\Rightarrow 1+\sin{2x}=0, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=-1, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow 2x=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n-1)\frac{\pi}{4}\)
আবার,
\(\sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1+\sqrt{5}}{2}, \ \sin{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\sin{2x}}\le{1}\)
\(\Rightarrow \sin{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(4n-1)\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(Q.4.(xiv)\) \(\cos{\theta}+\sin{\theta}=\cos{2\theta}+\sin{2\theta}\) হলে, সমীকরণটির \(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে সমাধান আছে কিনা যাচাই কর।
উত্তরঃ \(0, \ \frac{\pi}{6}\)
উত্তরঃ \(0, \ \frac{\pi}{6}\)
দিঃ২০১৭।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\cos{\theta}+\sin{\theta}=\cos{2\theta}+\sin{2\theta}\)
\(\Rightarrow \cos{\theta}-\cos{2\theta}=\sin{2\theta}-\sin{\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+2\theta}{2}}\sin{\frac{2\theta-\theta}{2}}=2\cos{\frac{2\theta+\theta}{2}}\sin{\frac{2\theta-\theta}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}-\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}\left(\sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}=\cos{\frac{3\theta}{2}}\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \frac{\sin{\frac{3\theta}{2}}}{\cos{\frac{3\theta}{2}}}=1\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\cos{\frac{3\theta}{2}}\) ভাগ করে,
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=1\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi, \ \frac{3\theta}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi, \ \theta=\frac{2n\pi}{3}+\frac{\pi}{6}\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\frac{2}{3}\) গুণ করে,
\(\Rightarrow \theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=0, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=2\pi, \ \frac{2\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow \theta=2\pi, \ \frac{5\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=0, \ \frac{\pi}{6}\)
\(\cos{\theta}+\sin{\theta}=\cos{2\theta}+\sin{2\theta}\)
\(\Rightarrow \cos{\theta}-\cos{2\theta}=\sin{2\theta}-\sin{\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+2\theta}{2}}\sin{\frac{2\theta-\theta}{2}}=2\cos{\frac{2\theta+\theta}{2}}\sin{\frac{2\theta-\theta}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}-\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}\left(\sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}=\cos{\frac{3\theta}{2}}\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \frac{\sin{\frac{3\theta}{2}}}{\cos{\frac{3\theta}{2}}}=1\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\cos{\frac{3\theta}{2}}\) ভাগ করে,
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=1\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi, \ \frac{3\theta}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi, \ \theta=\frac{2n\pi}{3}+\frac{\pi}{6}\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\frac{2}{3}\) গুণ করে,
\(\Rightarrow \theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=0, \ \frac{\pi}{6}\) মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=0, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=2\pi, \ \frac{2\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow \theta=2\pi, \ \frac{5\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=0, \ \frac{\pi}{6}\)
\(Q.4.(xv)\) \(\sin{\sqrt{2}x}-\cos{\frac{x}{2}}=0\) হলে, \(x\) এর সমাধানের জন্য সাধারণ রাশিমালা নির্ণয় কর।
উত্তরঃ \(\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
উত্তরঃ \(\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৯।
সমাধানঃ
প্রদত্ত সমীকরণ,
\(\Rightarrow \sin{\sqrt{2}x}-\cos{\frac{x}{2}}=0\)
\(\Rightarrow \sin{\sqrt{2}x}=\cos{\frac{x}{2}}\)
\(\Rightarrow \cos{\frac{x}{2}}=\sin{\sqrt{2}x}\)
\(\Rightarrow \cos{\frac{x}{2}}=\cos{\left(\frac{\pi}{2}-\sqrt{2}x\right)}\) ➜ \(\because \sin{A}=\cos{\left(\frac{\pi}{2}-A\right)}\)
\(\Rightarrow \frac{x}{2}=2n\pi\pm\left(\frac{\pi}{2}-\sqrt{2}x\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \frac{x}{2}=2n\pi+\left(\frac{\pi}{2}-\sqrt{2}x\right), \ \frac{x}{2}=2n\pi-\left(\frac{\pi}{2}-\sqrt{2}x\right)\)
\(\Rightarrow \frac{x}{2}=2n\pi+\frac{\pi}{2}-\sqrt{2}x, \ \frac{x}{2}=2n\pi-\frac{\pi}{2}+\sqrt{2}x\)
\(\Rightarrow \frac{x}{2}+\sqrt{2}x=2n\pi+\frac{\pi}{2}, \ \frac{x}{2}-\sqrt{2}x=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow \left(\frac{1}{2}+\sqrt{2}\right)x=(4n+1)\frac{\pi}{2}, \ \left(\frac{1}{2}-\sqrt{2}\right)x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \frac{1+2\sqrt{2}}{2}x=(4n+1)\frac{\pi}{2}, \ \frac{1-2\sqrt{2}}{2}x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow x=(4n+1)\frac{\pi}{2}\times\frac{2}{1+2\sqrt{2}}, \ x=(4n-1)\frac{\pi}{2}\times\frac{2}{1-2\sqrt{2}}\)
\(\Rightarrow x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ x=\frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \sin{\sqrt{2}x}-\cos{\frac{x}{2}}=0\)
\(\Rightarrow \sin{\sqrt{2}x}=\cos{\frac{x}{2}}\)
\(\Rightarrow \cos{\frac{x}{2}}=\sin{\sqrt{2}x}\)
\(\Rightarrow \cos{\frac{x}{2}}=\cos{\left(\frac{\pi}{2}-\sqrt{2}x\right)}\) ➜ \(\because \sin{A}=\cos{\left(\frac{\pi}{2}-A\right)}\)
\(\Rightarrow \frac{x}{2}=2n\pi\pm\left(\frac{\pi}{2}-\sqrt{2}x\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \frac{x}{2}=2n\pi+\left(\frac{\pi}{2}-\sqrt{2}x\right), \ \frac{x}{2}=2n\pi-\left(\frac{\pi}{2}-\sqrt{2}x\right)\)
\(\Rightarrow \frac{x}{2}=2n\pi+\frac{\pi}{2}-\sqrt{2}x, \ \frac{x}{2}=2n\pi-\frac{\pi}{2}+\sqrt{2}x\)
\(\Rightarrow \frac{x}{2}+\sqrt{2}x=2n\pi+\frac{\pi}{2}, \ \frac{x}{2}-\sqrt{2}x=2n\pi-\frac{\pi}{2}\)
\(\Rightarrow \left(\frac{1}{2}+\sqrt{2}\right)x=(4n+1)\frac{\pi}{2}, \ \left(\frac{1}{2}-\sqrt{2}\right)x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow \frac{1+2\sqrt{2}}{2}x=(4n+1)\frac{\pi}{2}, \ \frac{1-2\sqrt{2}}{2}x=(4n-1)\frac{\pi}{2}\)
\(\Rightarrow x=(4n+1)\frac{\pi}{2}\times\frac{2}{1+2\sqrt{2}}, \ x=(4n-1)\frac{\pi}{2}\times\frac{2}{1-2\sqrt{2}}\)
\(\Rightarrow x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ x=\frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{(4n+1)\pi}{1+2\sqrt{2}}, \ \frac{(4n-1)\pi}{1-2\sqrt{2}}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.4.(xvi)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\sin^2{x}=\frac{1}{2}\sin{2x}, \ 0\le{x}\lt{2\pi} \)
উত্তরঃ \(0, \ \frac{\pi}{4}, \ \pi, \ \frac{5\pi}{4}\)
উত্তরঃ \(0, \ \frac{\pi}{4}, \ \pi, \ \frac{5\pi}{4}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\sin^2{x}=\frac{1}{2}\sin{2x}, \ 0\le{x}\lt{2\pi}\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে \(y=\sin^2{x}\) এবং \(y=\frac{1}{2}\sin{2x}\) ফাংশন দুইটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\sin^2{x}\) এবং \(y=\frac{1}{2}\sin{2x}\) ফাংশন দুইটির লেখচিত্র উক্ত ব্যাবধির মধ্যে \(x\) অক্ষকে এবং \(x\) অক্ষের সমান্তরাল সরলরেখাকে চারটি বিন্দুতে ছেদ করে। সুতরাং এর চারটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(0, 0\right);\) \(\left(\frac{\pi}{4}, \frac{1}{2}\right);\) \(\left(\pi, 0\right)\) এবং \(\left(\frac{5\pi}{4}, \frac{1}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে সমাধান, \(x=0, \ \frac{\pi}{4}, \ \pi, \ \frac{5\pi}{4}\)
\(\sin^2{x}=\frac{1}{2}\sin{2x}, \ 0\le{x}\lt{2\pi}\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে \(y=\sin^2{x}\) এবং \(y=\frac{1}{2}\sin{2x}\) ফাংশন দুইটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\sin^2{x}\) এবং \(y=\frac{1}{2}\sin{2x}\) ফাংশন দুইটির লেখচিত্র উক্ত ব্যাবধির মধ্যে \(x\) অক্ষকে এবং \(x\) অক্ষের সমান্তরাল সরলরেখাকে চারটি বিন্দুতে ছেদ করে। সুতরাং এর চারটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(0, 0\right);\) \(\left(\frac{\pi}{4}, \frac{1}{2}\right);\) \(\left(\pi, 0\right)\) এবং \(\left(\frac{5\pi}{4}, \frac{1}{2}\right)\) নির্ণয় করি।

\(Q.4.(xvii)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi} \)
উত্তরঃ \(\frac{\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{8}, \ \frac{\pi}{2}, \ \frac{5\pi}{8}, \ \frac{3\pi}{4}\) এবং \(\frac{7\pi}{8}\)
উত্তরঃ \(\frac{\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{8}, \ \frac{\pi}{2}, \ \frac{5\pi}{8}, \ \frac{3\pi}{4}\) এবং \(\frac{7\pi}{8}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{7\theta}+\cos{\theta}+\cos{5\theta}+\cos{3\theta}=0\)
\(\Rightarrow 2\cos{\frac{7\theta+\theta}{2}}\cos{\frac{7\theta-\theta}{2}}+2\cos{\frac{5\theta+3\theta}{2}}\cos{\frac{5\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{8\theta}{2}}\cos{\frac{6\theta}{2}}+2\cos{\frac{8\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{4\theta}\cos{3\theta}+2\cos{4\theta}\cos{\theta}=0\)
\(\Rightarrow 2\cos{4\theta}(\cos{3\theta}+\cos{\theta})=0\)
\(\Rightarrow 2\cos{4\theta}\times2\cos{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 4\cos{4\theta}\cos{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 4\cos{4\theta}\cos{2\theta}\cos{\theta}=0\)
\(\therefore \cos{\theta}=0, \ \cos{2\theta}=0, \ \cos{4\theta}=0\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে \(y=\cos{\theta},\) \(y=\cos{2\theta}\) এবং \(y=\cos{4\theta}\) ফাংশন তিনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\cos{\theta},\) \(y=\cos{2\theta}\) এবং \(y=\cos{4\theta}\) ফাংশন তিনটির লেখচিত্র উক্ত ব্যাবধির মধ্যে \(x\) অক্ষকে সাতটি বিন্দুতে ছেদ করে। সুতরাং এর সাতটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{8}, 0\right);\) \(\left(\frac{\pi}{4}, 0\right);\) \(\left(\frac{3\pi}{8}, 0\right);\) \(\left(\frac{\pi}{2}, 0\right);\) \(\left(\frac{5\pi}{8}, 0\right);\) \(\left(\frac{3\pi}{4}, 0\right)\) এবং \(\left(\frac{7\pi}{8}, 0\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে সমাধান, \(x=\frac{\pi}{8}, \ \frac{\pi}{4}, \ \frac{3\pi}{8}, \ \frac{\pi}{2}, \ \frac{5\pi}{8}, \ \frac{3\pi}{4}\) এবং \(\frac{7\pi}{8}\)
\(\cos{\theta}+\cos{3\theta}+\cos{5\theta}+\cos{7\theta}=0, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \cos{7\theta}+\cos{\theta}+\cos{5\theta}+\cos{3\theta}=0\)
\(\Rightarrow 2\cos{\frac{7\theta+\theta}{2}}\cos{\frac{7\theta-\theta}{2}}+2\cos{\frac{5\theta+3\theta}{2}}\cos{\frac{5\theta-3\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{8\theta}{2}}\cos{\frac{6\theta}{2}}+2\cos{\frac{8\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{4\theta}\cos{3\theta}+2\cos{4\theta}\cos{\theta}=0\)
\(\Rightarrow 2\cos{4\theta}(\cos{3\theta}+\cos{\theta})=0\)
\(\Rightarrow 2\cos{4\theta}\times2\cos{\frac{3\theta+\theta}{2}}\cos{\frac{3\theta-\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 4\cos{4\theta}\cos{\frac{4\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 4\cos{4\theta}\cos{2\theta}\cos{\theta}=0\)
\(\therefore \cos{\theta}=0, \ \cos{2\theta}=0, \ \cos{4\theta}=0\)
\((a) \ 0\le{x}\le{360^{o}}\) ব্যবধিতে \(y=\cos{\theta},\) \(y=\cos{2\theta}\) এবং \(y=\cos{4\theta}\) ফাংশন তিনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\cos{\theta},\) \(y=\cos{2\theta}\) এবং \(y=\cos{4\theta}\) ফাংশন তিনটির লেখচিত্র উক্ত ব্যাবধির মধ্যে \(x\) অক্ষকে সাতটি বিন্দুতে ছেদ করে। সুতরাং এর সাতটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{8}, 0\right);\) \(\left(\frac{\pi}{4}, 0\right);\) \(\left(\frac{3\pi}{8}, 0\right);\) \(\left(\frac{\pi}{2}, 0\right);\) \(\left(\frac{5\pi}{8}, 0\right);\) \(\left(\frac{3\pi}{4}, 0\right)\) এবং \(\left(\frac{7\pi}{8}, 0\right)\) নির্ণয় করি।

\(Q.4.(xviii)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -\pi\lt{\theta}\lt{\pi} \)
উত্তরঃ \(-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
উত্তরঃ \(-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -\pi\lt{\theta}\lt{\pi} \)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\therefore \cos{\left(\theta+\frac{\pi}{4}\right)}=\frac{1}{2}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\((a) \ -\pi\lt{\theta}\lt{\pi}\) ব্যবধিতে \(y=\cos{\left(\theta+\frac{\pi}{4}\right)}\) ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\cos{\left(\theta+\frac{\pi}{4}\right)}\) ফাংশনটির লেখচিত্র \(x\) অক্ষের সমান্তরাল সরলরেখাকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(-\frac{7\pi}{12}, \frac{1}{2}\right);\) \(\left(\frac{\pi}{12}, \frac{1}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে সমাধান, \(x=-\frac{7\pi}{12}, \ \frac{\pi}{12}\)
\(\cos{\theta}-\sin{\theta}=\frac{1}{\sqrt{2}}, \ -\pi\lt{\theta}\lt{\pi} \)
\(\Rightarrow \cos{\theta}\frac{1}{\sqrt{2}}-\sin{\theta}\frac{1}{\sqrt{2}}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-1)^2}=\sqrt{2}\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{4}}-\sin{\theta}\sin{\frac{\pi}{4}}=\frac{1}{2}\) ➜ \(\because \frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}\)
\(\therefore \cos{\left(\theta+\frac{\pi}{4}\right)}=\frac{1}{2}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\((a) \ -\pi\lt{\theta}\lt{\pi}\) ব্যবধিতে \(y=\cos{\left(\theta+\frac{\pi}{4}\right)}\) ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\cos{\left(\theta+\frac{\pi}{4}\right)}\) ফাংশনটির লেখচিত্র \(x\) অক্ষের সমান্তরাল সরলরেখাকে দুইটি বিন্দুতে ছেদ করে। সুতরাং এর দুইটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(-\frac{7\pi}{12}, \frac{1}{2}\right);\) \(\left(\frac{\pi}{12}, \frac{1}{2}\right)\) নির্ণয় করি।

\(Q.4.(xix)\) লেখচিত্রের সাহায্যে সমাধান করঃ \(\sin{4\theta}=\cos{3\theta}+\sin{2\theta}, \ 0\lt{\theta}\lt{\pi}\)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{5\pi}{6}\)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{5\pi}{6}\)
সমাধানঃ
প্রদত্ত ফাংশনঃ
\(\sin{4\theta}=\cos{3\theta}+\sin{2\theta}, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \sin{4\theta}-\sin{2\theta}=\cos{3\theta}\)
\(\Rightarrow 2\cos{\frac{4\theta+2\theta}{2}}\sin{\frac{4\theta-2\theta}{2}}=\cos{3\theta}\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{6\theta}{2}}\sin{\frac{2\theta}{2}}=\cos{3\theta}\)
\(\Rightarrow 2\cos{3\theta}\sin{\theta}-\cos{3\theta}=0\)
\(\Rightarrow \cos{3\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow \cos{3\theta}=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow \cos{3\theta}=0, \ 2\sin{\theta}=1\)
\(\therefore \cos{3\theta}=0, \ \sin{\theta}=\frac{1}{2}\)
\((a) \ -\pi\lt{\theta}\lt{\pi}\) ব্যবধিতে \(y=\cos{3\theta}, \ y=\sin{\theta}\) এবং \(y=\frac{1}{2}\) ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\cos{3\theta}, \ y=\sin{\theta}\) ফাংশন দুইটির লেখচিত্র উক্ত ব্যাবধির মধ্যে \(x\) অক্ষকে এবং \(x\) অক্ষের সমান্তরাল সরলরেখাকে যথাক্রমে তিনটি ও দুইটি বিন্দুতে ছেদ করে। সুতরাং এর পাঁচটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{6}, 0\right);\) \(\left(\frac{\pi}{2}, 0\right);\) \(\left(\frac{5\pi}{6}, 0\right);\) \(\left(\frac{\pi}{6}, \frac{1}{2}\right)\) এবং \(\left(\frac{5\pi}{6}, \frac{1}{2}\right)\) নির্ণয় করি।
\(\therefore\) প্রদত্ত ব্যবধিতে সমাধান, \(x=\frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{5\pi}{6}\)
\(\sin{4\theta}=\cos{3\theta}+\sin{2\theta}, \ 0\lt{\theta}\lt{\pi}\)
\(\Rightarrow \sin{4\theta}-\sin{2\theta}=\cos{3\theta}\)
\(\Rightarrow 2\cos{\frac{4\theta+2\theta}{2}}\sin{\frac{4\theta-2\theta}{2}}=\cos{3\theta}\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{6\theta}{2}}\sin{\frac{2\theta}{2}}=\cos{3\theta}\)
\(\Rightarrow 2\cos{3\theta}\sin{\theta}-\cos{3\theta}=0\)
\(\Rightarrow \cos{3\theta}(2\sin{\theta}-1)=0\)
\(\Rightarrow \cos{3\theta}=0, \ 2\sin{\theta}-1=0\)
\(\Rightarrow \cos{3\theta}=0, \ 2\sin{\theta}=1\)
\(\therefore \cos{3\theta}=0, \ \sin{\theta}=\frac{1}{2}\)
\((a) \ -\pi\lt{\theta}\lt{\pi}\) ব্যবধিতে \(y=\cos{3\theta}, \ y=\sin{\theta}\) এবং \(y=\frac{1}{2}\) ফাংশনটির লেখচিত্র অংকন করি।
স্কেলঃ \(x\) অক্ষ বরাবর \(1\) বর্গ ঘর \(=10^{o}\)
\(y\) অক্ষ বরাবর \(10\) বর্গ ঘর \(=1\) একক।
\((b) \ y=\cos{3\theta}, \ y=\sin{\theta}\) ফাংশন দুইটির লেখচিত্র উক্ত ব্যাবধির মধ্যে \(x\) অক্ষকে এবং \(x\) অক্ষের সমান্তরাল সরলরেখাকে যথাক্রমে তিনটি ও দুইটি বিন্দুতে ছেদ করে। সুতরাং এর পাঁচটি সমাধান বিদ্যমান।
\((c)\) প্রতিসমতা ব্যবহার করে, বিন্দুগুলির স্থানাংক \(\left(\frac{\pi}{6}, 0\right);\) \(\left(\frac{\pi}{2}, 0\right);\) \(\left(\frac{5\pi}{6}, 0\right);\) \(\left(\frac{\pi}{6}, \frac{1}{2}\right)\) এবং \(\left(\frac{5\pi}{6}, \frac{1}{2}\right)\) নির্ণয় করি।

\(Q.4.(xx)\) \(h(x)=\sin{x}\) হলে, \(0\le{\theta}\le{2\pi}\) ব্যবধিতে \(2h(\theta).h(3\theta)=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}\) এবং \(\frac{11\pi}{6}\)
উত্তরঃ \(\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}\) এবং \(\frac{11\pi}{6}\)
ঢাঃ২০১৯।
সমাধানঃ
দেওয়া আছে,
\(h(x)=\sin{x}\)
প্রদত্ত ফাংশনঃ
\(2h(\theta).h(3\theta)=1, \ 0\le{\theta}\le{2\pi}\)
\(\Rightarrow 2\sin{\theta}\sin{(3\theta)}=1\) ➜ \(\because h(x)=\sin{x}\)
\(\therefore 2h(\theta).h(3\theta)=1\)
\(\Rightarrow 2\sin{\theta}\sin{(3\theta)}=1\)
\(\Rightarrow \cos{(3\theta-\theta)}-\cos{(3\theta+\theta)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=1\)
\(\Rightarrow \cos{2\theta}=1+\cos{4\theta}\)
\(\Rightarrow \cos{2\theta}=2\cos^2{2\theta}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2\theta}-2\cos^2{2\theta}=0\)
\(\Rightarrow \cos{2\theta}(1-2\cos{2\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ 1-2\cos{2\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ -2\cos{2\theta}=-1\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{\pi}{4}, \ -\frac{5\pi}{6}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=-2,\) \(\theta=-\frac{3\pi}{4}, \ -2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{3\pi}{4}, \ -\frac{11\pi}{6}, \ -\frac{13\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=3,\) \(\theta=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\)কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}\) এবং \(\frac{11\pi}{6}\)
\(h(x)=\sin{x}\)
প্রদত্ত ফাংশনঃ
\(2h(\theta).h(3\theta)=1, \ 0\le{\theta}\le{2\pi}\)
\(\Rightarrow 2\sin{\theta}\sin{(3\theta)}=1\) ➜ \(\because h(x)=\sin{x}\)
\(\therefore 2h(\theta).h(3\theta)=1\)
\(\Rightarrow 2\sin{\theta}\sin{(3\theta)}=1\)
\(\Rightarrow \cos{(3\theta-\theta)}-\cos{(3\theta+\theta)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=1\)
\(\Rightarrow \cos{2\theta}=1+\cos{4\theta}\)
\(\Rightarrow \cos{2\theta}=2\cos^2{2\theta}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2\theta}-2\cos^2{2\theta}=0\)
\(\Rightarrow \cos{2\theta}(1-2\cos{2\theta})=0\)
\(\Rightarrow \cos{2\theta}=0, \ 1-2\cos{2\theta}=0\)
\(\Rightarrow \cos{2\theta}=0, \ -2\cos{2\theta}=-1\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{4}, \ \frac{\pi}{6}, \ -\frac{\pi}{6}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{\pi}{4}, \ -\frac{5\pi}{6}, \ -\frac{7\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=-2,\) \(\theta=-\frac{3\pi}{4}, \ -2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{3\pi}{4}, \ -\frac{11\pi}{6}, \ -\frac{13\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=3,\) \(\theta=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\)কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}\) এবং \(\frac{11\pi}{6}\)
\(Q.4.(xxi)\) যদি \(\sin{(\pi\cot{\theta})}=\cos{(\pi\tan{\theta})}\) হয়, তবে প্রমাণ কর যে, \(cosec \ {2\theta}\) বা \(\cot{2\theta}\) এর মান \(\left(n+\frac{1}{4}\right)\) এর সমান, যখন \(n\) এর মান একটি পূর্ণ সংখ্যা হয়।
সমাধানঃ
দেওয়া আছে,
\(\sin{(\pi\cot{\theta})}=\cos{(\pi\tan{\theta})}\)
\(\Rightarrow \cos{(\pi\tan{\theta})}=\sin{(\pi\cot{\theta})}\)
\(\Rightarrow \cos{(\pi\tan{\theta})}=\cos{\left(\frac{\pi}{2}-\pi\cot{\theta}\right)}\)
\(\Rightarrow \pi\tan{\theta}=2n\pi\pm\left(\frac{\pi}{2}-\pi\cot{\theta}\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \pi\tan{\theta}=\pi\left\{2n\pm\left(\frac{1}{2}-\cot{\theta}\right)\right\}\)
\(\Rightarrow \tan{\theta}=2n\pm\left(\frac{1}{2}-\cot{\theta}\right)\)
\(\therefore \tan{\theta}=2n+\left(\frac{1}{2}-\cot{\theta}\right) ........(1)\) ধনাত্মক চিহ্ন ব্যবহার করে,
এবং \(\tan{\theta}=2n-\left(\frac{1}{2}-\cot{\theta}\right) ........(2)\) ঋনাত্মক চিহ্ন ব্যবহার করে,
\((1)\) হতে,
\(\tan{\theta}=2n+\frac{1}{2}-\cot{\theta}\)
\(\Rightarrow \tan{\theta}+\cot{\theta}=2n+\frac{1}{2}\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=2n+\frac{1}{2}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}+\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=2n+\frac{1}{2}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=2n+\frac{1}{2}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow \frac{1}{2\sin{\theta}\cos{\theta}}=n+\frac{1}{4}\) ➜ উভয় পার্শে \(2\) ভাগ করে,
\(\Rightarrow \frac{1}{\sin{2\theta}}=n+\frac{1}{4}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\therefore cosec \ {2\theta}=n+\frac{1}{4}\) ➜ \(\because \frac{1}{\sin{A}}=cosec \ {A}\)
\((2)\) হতে,
\(\tan{\theta}=2n-\frac{1}{2}+\cot{\theta}\)
\(\Rightarrow \tan{\theta}-\cot{\theta}=2n-\frac{1}{2}\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}-\frac{\cos{\theta}}{\sin{\theta}}=2n-\frac{1}{2}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}-\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=2n-\frac{1}{2}\)
\(\Rightarrow -\frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=2n-\frac{1}{2}\)
\(\Rightarrow -\frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=2n-\frac{1}{2}\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow \frac{\cos{2\theta}}{2\sin{\theta}\cos{\theta}}=-n+\frac{1}{4}\) ➜ উভয় পার্শে \(-2\) ভাগ করে,
\(\Rightarrow \frac{\cos{2\theta}}{\sin{2\theta}}=-n+\frac{1}{4}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \cot{2\theta}=-n+\frac{1}{4}\) ➜ \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)
\(\therefore \cot{2\theta}=n+\frac{1}{4}, \ \because n\in{\mathbb{Z}}\)
(প্রমাণিত)
\(\sin{(\pi\cot{\theta})}=\cos{(\pi\tan{\theta})}\)
\(\Rightarrow \cos{(\pi\tan{\theta})}=\sin{(\pi\cot{\theta})}\)
\(\Rightarrow \cos{(\pi\tan{\theta})}=\cos{\left(\frac{\pi}{2}-\pi\cot{\theta}\right)}\)
\(\Rightarrow \pi\tan{\theta}=2n\pi\pm\left(\frac{\pi}{2}-\pi\cot{\theta}\right)\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \pi\tan{\theta}=\pi\left\{2n\pm\left(\frac{1}{2}-\cot{\theta}\right)\right\}\)
\(\Rightarrow \tan{\theta}=2n\pm\left(\frac{1}{2}-\cot{\theta}\right)\)
\(\therefore \tan{\theta}=2n+\left(\frac{1}{2}-\cot{\theta}\right) ........(1)\) ধনাত্মক চিহ্ন ব্যবহার করে,
এবং \(\tan{\theta}=2n-\left(\frac{1}{2}-\cot{\theta}\right) ........(2)\) ঋনাত্মক চিহ্ন ব্যবহার করে,
\((1)\) হতে,
\(\tan{\theta}=2n+\frac{1}{2}-\cot{\theta}\)
\(\Rightarrow \tan{\theta}+\cot{\theta}=2n+\frac{1}{2}\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}=2n+\frac{1}{2}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}+\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=2n+\frac{1}{2}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=2n+\frac{1}{2}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
\(\Rightarrow \frac{1}{2\sin{\theta}\cos{\theta}}=n+\frac{1}{4}\) ➜ উভয় পার্শে \(2\) ভাগ করে,
\(\Rightarrow \frac{1}{\sin{2\theta}}=n+\frac{1}{4}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\therefore cosec \ {2\theta}=n+\frac{1}{4}\) ➜ \(\because \frac{1}{\sin{A}}=cosec \ {A}\)
\((2)\) হতে,
\(\tan{\theta}=2n-\frac{1}{2}+\cot{\theta}\)
\(\Rightarrow \tan{\theta}-\cot{\theta}=2n-\frac{1}{2}\)
\(\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}-\frac{\cos{\theta}}{\sin{\theta}}=2n-\frac{1}{2}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\sin^2{\theta}-\cos^2{\theta}}{\sin{\theta}\cos{\theta}}=2n-\frac{1}{2}\)
\(\Rightarrow -\frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=2n-\frac{1}{2}\)
\(\Rightarrow -\frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=2n-\frac{1}{2}\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow \frac{\cos{2\theta}}{2\sin{\theta}\cos{\theta}}=-n+\frac{1}{4}\) ➜ উভয় পার্শে \(-2\) ভাগ করে,
\(\Rightarrow \frac{\cos{2\theta}}{\sin{2\theta}}=-n+\frac{1}{4}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \cot{2\theta}=-n+\frac{1}{4}\) ➜ \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)
\(\therefore \cot{2\theta}=n+\frac{1}{4}, \ \because n\in{\mathbb{Z}}\)
(প্রমাণিত)
\(Q.4.(xxii)\) যদি \(\sin{A}=\sin{B}\) এবং \(\cos{A}=\cos{B}\) হয়, তবে প্রমাণ কর যে, \(A=B\) অথবা এদের পার্থক্য চার সমকোণের যেকোনো গুণিতকের সমান।
সমাধানঃ
দেওয়া আছে,
\(\sin{A}=\sin{B}\)
\(\Rightarrow \sin{A}-\sin{B}=0\)
\(\Rightarrow 2\cos{\frac{A+B}{2}}\sin{\frac{A-B}{2}}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\therefore \cos{\frac{A+B}{2}}\sin{\frac{A-B}{2}}=0 ........(1)\)
আবার,
\(\cos{A}=\cos{B}\)
\(\Rightarrow \cos{B}=\cos{A}\)
\(\Rightarrow \cos{B}-\cos{A}=0\)
\(\Rightarrow 2\sin{\frac{B+A}{2}}\sin{\frac{A-B}{2}}=0\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\therefore \sin{\frac{A+B}{2}}\sin{\frac{A-B}{2}}=0 ........(2)\)
\((1)\) ও \((2)\) সমীকরণদ্বয় একই সময় সিদ্ধ হতে পারে কেবল এবং কেবল যদি \(\sin{\frac{A-B}{2}}=0\) হয়।
কেননা, \(\sin{\frac{A+B}{2}}\) ও \(\cos{\frac{A+B}{2}}\) উৎপাদকদ্বয় একই সঙ্গে শুন্য হতে পারে না।
সুতরাং, \(\sin{\frac{A-B}{2}}=0\)
\(\Rightarrow \frac{A-B}{2}=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore A-B=2n\pi\)
যখন, \(n=0,\)
\(\Rightarrow A-B=0\)
\(\therefore A=B\)
যখন, \(n\) যে কোনো পূর্ণ সংখ্যা
\(\therefore A-B=2n\pi\) অর্থাৎ চার সমকোণের গুণিতকের সমান।
(প্রমাণিত)
\(\sin{A}=\sin{B}\)
\(\Rightarrow \sin{A}-\sin{B}=0\)
\(\Rightarrow 2\cos{\frac{A+B}{2}}\sin{\frac{A-B}{2}}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\therefore \cos{\frac{A+B}{2}}\sin{\frac{A-B}{2}}=0 ........(1)\)
আবার,
\(\cos{A}=\cos{B}\)
\(\Rightarrow \cos{B}=\cos{A}\)
\(\Rightarrow \cos{B}-\cos{A}=0\)
\(\Rightarrow 2\sin{\frac{B+A}{2}}\sin{\frac{A-B}{2}}=0\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\therefore \sin{\frac{A+B}{2}}\sin{\frac{A-B}{2}}=0 ........(2)\)
\((1)\) ও \((2)\) সমীকরণদ্বয় একই সময় সিদ্ধ হতে পারে কেবল এবং কেবল যদি \(\sin{\frac{A-B}{2}}=0\) হয়।
কেননা, \(\sin{\frac{A+B}{2}}\) ও \(\cos{\frac{A+B}{2}}\) উৎপাদকদ্বয় একই সঙ্গে শুন্য হতে পারে না।
সুতরাং, \(\sin{\frac{A-B}{2}}=0\)
\(\Rightarrow \frac{A-B}{2}=n\pi\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore A-B=2n\pi\)
যখন, \(n=0,\)
\(\Rightarrow A-B=0\)
\(\therefore A=B\)
যখন, \(n\) যে কোনো পূর্ণ সংখ্যা
\(\therefore A-B=2n\pi\) অর্থাৎ চার সমকোণের গুণিতকের সমান।
(প্রমাণিত)
\(Q.4.(xxiii)\) যদি \(a\cos{\theta}+b\sin{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\) হয়, তবে প্রমাণ কর যে, \(\sin{(\alpha+\beta)}=\frac{2ab}{a^2+b^2}\)
সমাধানঃ
দেওয়া আছে,
\(a\cos{\theta}+b\sin{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\)
\(\therefore a\cos{\alpha}+b\sin{\alpha}=c .......(1)\)
এবং \(a\cos{\beta}+b\sin{\beta}=c ..........(2)\)
\((1)\) ও \((2)\) নং সমীকরণ হতে,
\(a\cos{\alpha}+b\sin{\alpha}=a\cos{\beta}+b\sin{\beta}\)
\(\Rightarrow a\cos{\alpha}-a\cos{\beta}+b\sin{\alpha}-b\sin{\beta}=0\)
\(\Rightarrow -a(\cos{\beta}-\cos{\alpha})+b(\sin{\alpha}-\sin{\beta})=0\)
\(\Rightarrow -a.2\sin{\frac{\beta+\alpha}{2}}\sin{\frac{\alpha-\beta}{2}}+b.2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
এবং \(\cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{\alpha-\beta}{2}}\left(b\cos{\frac{\alpha+\beta}{2}}-a\sin{\frac{\alpha+\beta}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{\alpha-\beta}{2}}\ne{0}, \ b\cos{\frac{\alpha+\beta}{2}}-a\sin{\frac{\alpha+\beta}{2}}=0, \ \because \alpha\ne{\beta}\)
\(\Rightarrow b\cos{\frac{\alpha+\beta}{2}}-a\sin{\frac{\alpha+\beta}{2}}=0\)
\(\Rightarrow b\cos{\frac{\alpha+\beta}{2}}=a\sin{\frac{\alpha+\beta}{2}}\)
\(\Rightarrow a\sin{\frac{\alpha+\beta}{2}}=b\cos{\frac{\alpha+\beta}{2}}\)
\(\Rightarrow \frac{\sin{\frac{\alpha+\beta}{2}}}{\cos{\frac{\alpha+\beta}{2}}}=\frac{b}{a}\)
\(\therefore \tan{\frac{\alpha+\beta}{2}}=\frac{b}{a}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
এখন,
\(\sin{(\alpha+\beta)}=\frac{2\tan{\frac{\alpha+\beta}{2}}}{1+\tan^2{\frac{\alpha+\beta}{2}}}\) ➜ \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\frac{2\times\frac{b}{a}}{1+\left(\frac{b}{a}\right)^2}\) ➜ \(\because \tan{\frac{\alpha+\beta}{2}}=\frac{b}{a}\)
\(=\frac{\frac{2b}{a}}{1+\frac{b^2}{a^2}}\)
\(=\frac{2ab}{a^2+b^2}\) ➜ লব ও হরকে \(a^2\) দ্বারা গুণ করে,
\(\therefore \sin{(\alpha+\beta)}=\frac{2ab}{a^2+b^2}\)
(প্রমাণিত)
\(a\cos{\theta}+b\sin{\theta}=c\) সমীকরণের দুইটি সমাধান \(\alpha\) ও \(\beta\)
\(\therefore a\cos{\alpha}+b\sin{\alpha}=c .......(1)\)
এবং \(a\cos{\beta}+b\sin{\beta}=c ..........(2)\)
\((1)\) ও \((2)\) নং সমীকরণ হতে,
\(a\cos{\alpha}+b\sin{\alpha}=a\cos{\beta}+b\sin{\beta}\)
\(\Rightarrow a\cos{\alpha}-a\cos{\beta}+b\sin{\alpha}-b\sin{\beta}=0\)
\(\Rightarrow -a(\cos{\beta}-\cos{\alpha})+b(\sin{\alpha}-\sin{\beta})=0\)
\(\Rightarrow -a.2\sin{\frac{\beta+\alpha}{2}}\sin{\frac{\alpha-\beta}{2}}+b.2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}=0\) ➜ \(\because \sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
এবং \(\cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{\alpha-\beta}{2}}\left(b\cos{\frac{\alpha+\beta}{2}}-a\sin{\frac{\alpha+\beta}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{\alpha-\beta}{2}}\ne{0}, \ b\cos{\frac{\alpha+\beta}{2}}-a\sin{\frac{\alpha+\beta}{2}}=0, \ \because \alpha\ne{\beta}\)
\(\Rightarrow b\cos{\frac{\alpha+\beta}{2}}-a\sin{\frac{\alpha+\beta}{2}}=0\)
\(\Rightarrow b\cos{\frac{\alpha+\beta}{2}}=a\sin{\frac{\alpha+\beta}{2}}\)
\(\Rightarrow a\sin{\frac{\alpha+\beta}{2}}=b\cos{\frac{\alpha+\beta}{2}}\)
\(\Rightarrow \frac{\sin{\frac{\alpha+\beta}{2}}}{\cos{\frac{\alpha+\beta}{2}}}=\frac{b}{a}\)
\(\therefore \tan{\frac{\alpha+\beta}{2}}=\frac{b}{a}\) ➜ \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
এখন,
\(\sin{(\alpha+\beta)}=\frac{2\tan{\frac{\alpha+\beta}{2}}}{1+\tan^2{\frac{\alpha+\beta}{2}}}\) ➜ \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\frac{2\times\frac{b}{a}}{1+\left(\frac{b}{a}\right)^2}\) ➜ \(\because \tan{\frac{\alpha+\beta}{2}}=\frac{b}{a}\)
\(=\frac{\frac{2b}{a}}{1+\frac{b^2}{a^2}}\)
\(=\frac{2ab}{a^2+b^2}\) ➜ লব ও হরকে \(a^2\) দ্বারা গুণ করে,
\(\therefore \sin{(\alpha+\beta)}=\frac{2ab}{a^2+b^2}\)
(প্রমাণিত)
\(Q.4.(xxiv)\) \(1+\cos{(y-z)}+\cos{(z-x)}+\cos{(x-y)}=0\) হলে, দেখাও যে, \((y-z)\) অথবা, \((z-x)\) অথবা, \((x-y)\) হবে \(\pi\) এর বিজোড় গুণিতকের সমান।
সমাধানঃ
দেওয়া আছে,
\(1+\cos{(y-z)}+\cos{(z-x)}+\cos{(x-y)}=0\)
\(\Rightarrow 2\cos^2{\frac{y-z}{2}}+2\cos{\frac{z-x+x-y}{2}}\cos{\frac{z-x-x+y}{2}}=0\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
এবং \(\cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos^2{\frac{y-z}{2}}+2\cos{\frac{z-y}{2}}\cos{\frac{z-2x+y}{2}}=0\)
\(\Rightarrow 2\cos^2{\frac{y-z}{2}}+2\cos{\frac{y-z}{2}}\cos{\frac{z-2x+y}{2}}=0\) ➜ \(\because \cos{(-A)}=\cos{A}\)
\(\Rightarrow 2\cos{\frac{y-z}{2}}\left(\cos{\frac{y-z}{2}}+\cos{\frac{z-2x+y}{2}}\right)=0\)
\(\Rightarrow \cos{\frac{y-z}{2}}\left\{\cos{\left(\frac{z-x}{2}+\frac{x-y}{2}\right)}+\cos{\left(\frac{z-x}{2}-\frac{x-y}{2}\right)}\right\}=0\)
\(\Rightarrow \cos{\frac{y-z}{2}}.2\cos{\frac{z-x}{2}}\cos{\frac{x-y}{2}}=0\) ➜ \(\because \cos{(A+B)}+\cos{(A-B)}=2\cos{A}\cos{B}\)
\(\Rightarrow \cos{\frac{y-z}{2}}=0, \ \cos{\frac{z-x}{2}}=0, \ \cos{\frac{x-y}{2}}=0\)
\(\Rightarrow \frac{y-z}{2}=(2n+1)\frac{\pi}{2}, \ \frac{z-x}{2}=(2n+1)\frac{\pi}{2}, \ \frac{x-y}{2}=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore y-z=(2n+1)\pi \ \text{অথবা}, \ z-x=(2n+1)\pi \ \text{অথবা}, \ x-y=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\because n\) এর মান শুন্য অথবা যে কোনো পূর্ণ সংখ্যা, সুতরাং \((2n+1)\) এর মান বিজোড়।
\(\therefore (y-z)\) অথবা, \(( z-x)\) অথবা, \((x-y)\) হলো \(\pi\) এর বিজোড় গুণিতকের সমান।
(দেখানো হলো)
\(1+\cos{(y-z)}+\cos{(z-x)}+\cos{(x-y)}=0\)
\(\Rightarrow 2\cos^2{\frac{y-z}{2}}+2\cos{\frac{z-x+x-y}{2}}\cos{\frac{z-x-x+y}{2}}=0\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
এবং \(\cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos^2{\frac{y-z}{2}}+2\cos{\frac{z-y}{2}}\cos{\frac{z-2x+y}{2}}=0\)
\(\Rightarrow 2\cos^2{\frac{y-z}{2}}+2\cos{\frac{y-z}{2}}\cos{\frac{z-2x+y}{2}}=0\) ➜ \(\because \cos{(-A)}=\cos{A}\)
\(\Rightarrow 2\cos{\frac{y-z}{2}}\left(\cos{\frac{y-z}{2}}+\cos{\frac{z-2x+y}{2}}\right)=0\)
\(\Rightarrow \cos{\frac{y-z}{2}}\left\{\cos{\left(\frac{z-x}{2}+\frac{x-y}{2}\right)}+\cos{\left(\frac{z-x}{2}-\frac{x-y}{2}\right)}\right\}=0\)
\(\Rightarrow \cos{\frac{y-z}{2}}.2\cos{\frac{z-x}{2}}\cos{\frac{x-y}{2}}=0\) ➜ \(\because \cos{(A+B)}+\cos{(A-B)}=2\cos{A}\cos{B}\)
\(\Rightarrow \cos{\frac{y-z}{2}}=0, \ \cos{\frac{z-x}{2}}=0, \ \cos{\frac{x-y}{2}}=0\)
\(\Rightarrow \frac{y-z}{2}=(2n+1)\frac{\pi}{2}, \ \frac{z-x}{2}=(2n+1)\frac{\pi}{2}, \ \frac{x-y}{2}=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore y-z=(2n+1)\pi \ \text{অথবা}, \ z-x=(2n+1)\pi \ \text{অথবা}, \ x-y=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\because n\) এর মান শুন্য অথবা যে কোনো পূর্ণ সংখ্যা, সুতরাং \((2n+1)\) এর মান বিজোড়।
\(\therefore (y-z)\) অথবা, \(( z-x)\) অথবা, \((x-y)\) হলো \(\pi\) এর বিজোড় গুণিতকের সমান।
(দেখানো হলো)
অধ্যায় \(7H\) / \(Q.5\)-এর সৃজনশীল প্রশ্নসমূহ
\(Q.5.(i)\) \(f(x)=\frac{2x}{1+x^2}, \ g(y)=\frac{1-y^2}{1+y^2}\) এবং \(h(x)=\sin{x}\)
\((a)\) \(\cos^{-1}{m}+\cos^{-1}{n}=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(m^2+n^2=1\)
\((b)\) \(cosec^{-1}{\frac{1}{f(a)}}+\sec^{-1}{\frac{1}{g(b)}}=2\tan^{-1}{x}\) হলে, দেখাও যে, \(x=\frac{a-b}{1+ab}\)
\((c)\) \(0\le{\theta}\le{2\pi}\) ব্যবধিতে \(2h(\theta)h(3\theta)=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \((a) \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
\(Q.5.(ii)\) দৃশ্যকল্প-১: \(\sec{A}=\sqrt{5}, \ cosec \ {B}=\frac{5}{3}\) এবং \(\cot{C}=3\)
দৃশ্যকল্প-২: \(f(x)=\sin{x}\)
\((a)\) \(cosec^{-1}{\sqrt{17}}+\sec^{-1}{\frac{\sqrt{26}}{5}}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে \(A+C-\frac{1}{2}B\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ দৃশ্যকল্প-২: থেকে \(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \tan^{-1}{\left(\frac{9}{19}\right)}\)
\((b) \ \tan^{-1}{2}\)
\((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(Q.5.(iii)\) \(f(x)=\sin^{-1}{x}\) এবং \(g(x)=\cos{x}\)
\((a)\) \(\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\) এর মান কত?
\((b)\) \(f\left\{\sqrt{2}g\left(\frac{\pi}{2}-\theta\right)\right\}+f\left\{\sqrt{g(2\theta)}\right\}\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ \(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((b) \ \frac{\pi}{2}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
\(Q.5.(iv)\) দৃশ্যকল্প-১: \(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
দৃশ্যকল্প-২: \(A=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\sqrt{\frac{2}{3}}}\)
\((a)\) \(\sec^2{(\cot^{-1}{3})}+cosec^2{(\tan^{-1}{2})}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে সমাধান করঃ \(f(\sin{\theta})=0\)
\((c)\) দৃশ্যকল্প-২: থেকে প্রমান কর যে, \(A=\tan^{-1}{\frac{5}{\sqrt{2}}}\)
উত্তরঃ \((a) \ 2\frac{13}{36}\)
\((b) \ n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
\(Q.5.(v)\) দৃশ্যকল্প-১: \(f(a)=\sec^{-1}{\frac{1}{a}}+\sec^{-1}{\frac{1}{b}}\)
দৃশ্যকল্প-২: \(g(\alpha)=\sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}\)
\((a)\) \(\cot{\left(\sin^{-1}{\frac{1}{\sqrt{5}}}\right)}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-২: থেকে যদি \(g(\alpha)=0\) হয় তবে দেখাও যে, \(\alpha=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) দৃশ্যকল্প-১: থেকে \(f(a)=\alpha\) হলে, প্রমান কর যে, \(\sin{\alpha}=\sqrt{a^2+b^2-2ab\cos{\alpha}}\)
উত্তরঃ \((a) \ 2\)
\(Q.5.(vi)\) \(f(x)=\tan{x}\)
\((a)\) দেখাও যে, \(\tan^{-1}{\frac{5}{3}}=\frac{\pi}{2}-\cos^{-1}{\frac{5}{\sqrt{34}}}\)
\((b)\) প্রমান কর যে, \(\tan^{-1}{\left\{(2+\sqrt{3})f(x)\right\}}+\tan^{-1}{\left\{(2-\sqrt{3})f(x)\right\}}=\tan^{-1}{\{2f(2x)\}}\)
\((c)\) সমাধান করঃ \(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
উত্তরঃ \((c) \ n\pi-\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(Q.5.(vii)\) দৃশ্যকল্প-১: \(\sin^{-1}{\left(\frac{4}{5}\right)}+\cos^{-1}{\left(\frac{2}{\sqrt{5}}\right)}-\cot^{-1}{\left(\frac{2}{11}\right)}\)
দৃশ্যকল্প-২: \(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\lt{\theta}\lt{2\pi}\)
\((a)\) প্রমান কর যে, \(2\sin^{-1}{x}=\sin^{-1}{\left(2x\sqrt{1-x^2}\right)}\)
\((b)\) দৃশ্যকল্প-১ এর মান নির্ণয় কর।
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((b) 0\)
\((c) \ \pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
\(Q.5.(viii)\) দৃশ্যকল্প-১: \(\sec^{-1}{\left(\frac{5}{3}\right)}+\cot^{-1}{\left(\frac{12}{5}\right)}+\sin^{-1}{\left(\frac{16}{65}\right)}\)
দৃশ্যকল্প-২: \(\sqrt{3}\sin{\theta}=2+\cos{\theta}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\sin^{-1}{\frac{2x}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\)
\((c)\) দৃশ্যকল্প-২ এর সমাধান কর যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(Q.5.(ix)\) \(f(x)=\tan{x}\)
\((a)\) \(\cot^{-1}{\cos{\left(cosec^{-1}{\sqrt{\frac{3}{2}}}\right)}}\) এর মুখ্যমান নির্ণয় কর।
\((b)\) উদ্দীপকে উল্লেখিত \(f(x)\) এর জন্য \(f^{-1}(x)+f^{-1}(y)=\pi\) হলে প্রমাণ কর যে, প্রাপ্ত সঞ্চারপথটি একটি সরলরেখা নির্দেশ করে যার ঢাল \(-1\) হবে।
\((c)\) \(\left\{f(x)\right\}^2+f^{\prime}(x)=3f(x)\) হলে, বিশেষ সমাধান নির্ণয় কর যখন \(0\le{x}\le{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{3}\)
\((c) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(Q.5.(x)\) \(f(x)=\sin{x}, \ g(x)=\cos{x}, \ \sin{\theta}=\frac{4}{5}\)
\((a)\) \(cosec^{-1}{\sqrt{5}}+\sec^{-1}{\frac{\sqrt{10}}{3}}\) এর মান নির্ণয় কর।
\((b)\) উদ্দীপকের আলোকে প্রমাণ কর যে, \(\sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{2}\)
\((c)\) উদ্দীপকের আলোকে সমাধান করঃ \(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((c) \ 2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xi)\) দৃশ্যকল্প-১: \(\cot{\theta}-\tan{\theta}=\frac{6}{5}\)
দৃশ্যকল্প-২: \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{(\cot{3x})}+\tan^{-1}{(-\cot{5x})}=2x\)
\((b)\) দৃশ্যকল্প-১ হতে প্রমাণ কর যে, \(\theta=\frac{1}{2}\sin^{-1}{\frac{5}{\sqrt{34}}}\)
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ 2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xii)\) \(g(x)=p\sin^{-1}{x}, \ h(x)=\cos{x}\)
\((a)\) প্রমাণ কর যে, \(\sec^{-1}{\frac{\sqrt{5}}{2}}+\tan^{-1}{\frac{1}{2}}=\cot^{-1}{\frac{3}{4}}\)
\((b)\) \(g(x)\) এর লেখচিত্র অংকন কর, যখন \(p=\frac{1}{2}, \ -1\le{x}\le{1}\)
\((c)\) \(2\left\{h(x)\right\}^2+\left\{h(2x)\right\}^2=2\) সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ \frac{1}{2}(2n\pi\pm\alpha)\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(Q.5.(xiii)\) \(f(x)=\cot^{-1}{y}-\tan^{-1}{x} ........(1)\)
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta} ........(2)\)
\((a)\) \(\sin{\frac{x}{3}}\) এর পর্যায়কাল কত?
\((b)\) \(f(x)=\frac{\pi}{6}\) হলে, প্রমাণ কর যে, \(x+y+\sqrt{3}xy=\sqrt{3}\)
\((c)\) উদ্দীপক-২ এর সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \((a) \ 6\pi\)
\((c) \ \frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((a)\) \(\cos^{-1}{m}+\cos^{-1}{n}=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(m^2+n^2=1\)
\((b)\) \(cosec^{-1}{\frac{1}{f(a)}}+\sec^{-1}{\frac{1}{g(b)}}=2\tan^{-1}{x}\) হলে, দেখাও যে, \(x=\frac{a-b}{1+ab}\)
\((c)\) \(0\le{\theta}\le{2\pi}\) ব্যবধিতে \(2h(\theta)h(3\theta)=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \((a) \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
ঢাঃ২০১৯।
\(Q.5.(ii)\) দৃশ্যকল্প-১: \(\sec{A}=\sqrt{5}, \ cosec \ {B}=\frac{5}{3}\) এবং \(\cot{C}=3\)
দৃশ্যকল্প-২: \(f(x)=\sin{x}\)
\((a)\) \(cosec^{-1}{\sqrt{17}}+\sec^{-1}{\frac{\sqrt{26}}{5}}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে \(A+C-\frac{1}{2}B\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ দৃশ্যকল্প-২: থেকে \(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \tan^{-1}{\left(\frac{9}{19}\right)}\)
\((b) \ \tan^{-1}{2}\)
\((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
রাঃ২০১৯।
\(Q.5.(iii)\) \(f(x)=\sin^{-1}{x}\) এবং \(g(x)=\cos{x}\)
\((a)\) \(\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\) এর মান কত?
\((b)\) \(f\left\{\sqrt{2}g\left(\frac{\pi}{2}-\theta\right)\right\}+f\left\{\sqrt{g(2\theta)}\right\}\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ \(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((b) \ \frac{\pi}{2}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
যঃ২০১৯।
\(Q.5.(iv)\) দৃশ্যকল্প-১: \(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
দৃশ্যকল্প-২: \(A=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\sqrt{\frac{2}{3}}}\)
\((a)\) \(\sec^2{(\cot^{-1}{3})}+cosec^2{(\tan^{-1}{2})}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে সমাধান করঃ \(f(\sin{\theta})=0\)
\((c)\) দৃশ্যকল্প-২: থেকে প্রমান কর যে, \(A=\tan^{-1}{\frac{5}{\sqrt{2}}}\)
উত্তরঃ \((a) \ 2\frac{13}{36}\)
\((b) \ n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
বঃ২০১৯।
\(Q.5.(v)\) দৃশ্যকল্প-১: \(f(a)=\sec^{-1}{\frac{1}{a}}+\sec^{-1}{\frac{1}{b}}\)
দৃশ্যকল্প-২: \(g(\alpha)=\sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}\)
\((a)\) \(\cot{\left(\sin^{-1}{\frac{1}{\sqrt{5}}}\right)}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-২: থেকে যদি \(g(\alpha)=0\) হয় তবে দেখাও যে, \(\alpha=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) দৃশ্যকল্প-১: থেকে \(f(a)=\alpha\) হলে, প্রমান কর যে, \(\sin{\alpha}=\sqrt{a^2+b^2-2ab\cos{\alpha}}\)
উত্তরঃ \((a) \ 2\)
চঃ২০১৯।
\(Q.5.(vi)\) \(f(x)=\tan{x}\)
\((a)\) দেখাও যে, \(\tan^{-1}{\frac{5}{3}}=\frac{\pi}{2}-\cos^{-1}{\frac{5}{\sqrt{34}}}\)
\((b)\) প্রমান কর যে, \(\tan^{-1}{\left\{(2+\sqrt{3})f(x)\right\}}+\tan^{-1}{\left\{(2-\sqrt{3})f(x)\right\}}=\tan^{-1}{\{2f(2x)\}}\)
\((c)\) সমাধান করঃ \(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
উত্তরঃ \((c) \ n\pi-\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সিঃ২০১৯।
\(Q.5.(vii)\) দৃশ্যকল্প-১: \(\sin^{-1}{\left(\frac{4}{5}\right)}+\cos^{-1}{\left(\frac{2}{\sqrt{5}}\right)}-\cot^{-1}{\left(\frac{2}{11}\right)}\)
দৃশ্যকল্প-২: \(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\lt{\theta}\lt{2\pi}\)
\((a)\) প্রমান কর যে, \(2\sin^{-1}{x}=\sin^{-1}{\left(2x\sqrt{1-x^2}\right)}\)
\((b)\) দৃশ্যকল্প-১ এর মান নির্ণয় কর।
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((b) 0\)
\((c) \ \pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
ঢাঃ,যঃ,সিঃ,দিঃ২০১৮।
\(Q.5.(viii)\) দৃশ্যকল্প-১: \(\sec^{-1}{\left(\frac{5}{3}\right)}+\cot^{-1}{\left(\frac{12}{5}\right)}+\sin^{-1}{\left(\frac{16}{65}\right)}\)
দৃশ্যকল্প-২: \(\sqrt{3}\sin{\theta}=2+\cos{\theta}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\sin^{-1}{\frac{2x}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\)
\((c)\) দৃশ্যকল্প-২ এর সমাধান কর যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
ঢাঃ২০১৭।
\(Q.5.(ix)\) \(f(x)=\tan{x}\)
\((a)\) \(\cot^{-1}{\cos{\left(cosec^{-1}{\sqrt{\frac{3}{2}}}\right)}}\) এর মুখ্যমান নির্ণয় কর।
\((b)\) উদ্দীপকে উল্লেখিত \(f(x)\) এর জন্য \(f^{-1}(x)+f^{-1}(y)=\pi\) হলে প্রমাণ কর যে, প্রাপ্ত সঞ্চারপথটি একটি সরলরেখা নির্দেশ করে যার ঢাল \(-1\) হবে।
\((c)\) \(\left\{f(x)\right\}^2+f^{\prime}(x)=3f(x)\) হলে, বিশেষ সমাধান নির্ণয় কর যখন \(0\le{x}\le{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{3}\)
\((c) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
রাঃ২০১৭।
\(Q.5.(x)\) \(f(x)=\sin{x}, \ g(x)=\cos{x}, \ \sin{\theta}=\frac{4}{5}\)
\((a)\) \(cosec^{-1}{\sqrt{5}}+\sec^{-1}{\frac{\sqrt{10}}{3}}\) এর মান নির্ণয় কর।
\((b)\) উদ্দীপকের আলোকে প্রমাণ কর যে, \(\sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{2}\)
\((c)\) উদ্দীপকের আলোকে সমাধান করঃ \(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((c) \ 2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৭।
\(Q.5.(xi)\) দৃশ্যকল্প-১: \(\cot{\theta}-\tan{\theta}=\frac{6}{5}\)
দৃশ্যকল্প-২: \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{(\cot{3x})}+\tan^{-1}{(-\cot{5x})}=2x\)
\((b)\) দৃশ্যকল্প-১ হতে প্রমাণ কর যে, \(\theta=\frac{1}{2}\sin^{-1}{\frac{5}{\sqrt{34}}}\)
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ 2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৭।
\(Q.5.(xii)\) \(g(x)=p\sin^{-1}{x}, \ h(x)=\cos{x}\)
\((a)\) প্রমাণ কর যে, \(\sec^{-1}{\frac{\sqrt{5}}{2}}+\tan^{-1}{\frac{1}{2}}=\cot^{-1}{\frac{3}{4}}\)
\((b)\) \(g(x)\) এর লেখচিত্র অংকন কর, যখন \(p=\frac{1}{2}, \ -1\le{x}\le{1}\)
\((c)\) \(2\left\{h(x)\right\}^2+\left\{h(2x)\right\}^2=2\) সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ \frac{1}{2}(2n\pi\pm\alpha)\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
বঃ২০১৭।
\(Q.5.(xiii)\) \(f(x)=\cot^{-1}{y}-\tan^{-1}{x} ........(1)\)
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta} ........(2)\)
\((a)\) \(\sin{\frac{x}{3}}\) এর পর্যায়কাল কত?
\((b)\) \(f(x)=\frac{\pi}{6}\) হলে, প্রমাণ কর যে, \(x+y+\sqrt{3}xy=\sqrt{3}\)
\((c)\) উদ্দীপক-২ এর সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \((a) \ 6\pi\)
\((c) \ \frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০১৭।
\(Q.5.(xiv)\) \(A=\cos{\theta}, \ B=\sin{\theta}, \ C=\cos{2\theta}, \ D=\sin{2\theta}\)
\((a)\) মান নির্ণয় করঃ \(\tan^{-1}{\sin{\cos^{-1}{\sqrt{\frac{2}{3}}}}}\)
\((b)\) \(A+\sqrt{3}B=\sqrt{2}\) হলে, সমীকরণটি সমাধান কর।
\((c)\) \(A+B=C+D\) হলে, সমীকরণটি \(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে সমাধান আছে কিনা যাচাই কর।
উত্তরঃ \((a) \ \frac{\pi}{6}\)
\((b) \ 2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 0, \ \frac{\pi}{6}\)
\(Q.5.(xv)\) \(f(x)=\tan{\theta}\) এবং \(\sin{\theta}=\frac{12}{13}\)
\((a)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{1}{\sqrt{x}}}=\frac{1}{2}\sec^{-1}{\frac{1+x}{1-x}}\)
\((b)\) সমাধান করঃ \(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\((c)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{3}{4}}+\frac{\theta}{2}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
উত্তরঃ \((b) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) এবং \(\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(Q.5.(xvi)\) \(f(x)=\cos{x}\) এবং \(g(x)=\sin{x}\)
\((a)\) \(f\left(\cot^{-1}{\frac{3}{4}}\right)\) এর মান কত?
\((b)\) \(f^{-1}\left(\frac{x}{a}\right)+f^{-1}\left(\frac{y}{b}\right)=\theta\) হলে, দেখাও যে, \(\frac{x^2}{a^2}-\frac{2xy}{ab}f(\theta)+\frac{y^2}{b^2}=1-\left\{f(x)\right\}^2\)
\((c)\) সমাধান করঃ \(f(\theta)-f(7\theta)=g(4\theta)\)
উত্তরঃ \((a) \ \frac{3}{5}\)
\((c) \ \frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xvii)\) দৃশ্যকল্প-১: \(\cos^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{5}{13}}+cosec^{-1}{\frac{65}{16}}\)
দৃশ্যকল্প-২: \(f(x)=\tan{x}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\cos^{-1}{\frac{1-x^2}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\) হবে।
\((c)\) দৃশ্যকল্প-২: থেকে সমাধান করঃ \(f(x)+f(2x)+f(3x)=f(x)f(2x)f(3x)\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{3}, \ \frac{2\pi}{3}, \ \pi, \ \frac{4\pi}{3}, \ \frac{5\pi}{3}\)
\(Q.5.(xviii)\) \(f(x)=\cos{x}\)
\((a)\) \(\cos{3x}=-\frac{\sqrt{3}}{2}\) হলে, \(x\) এর মান নির্ণয় কর।
\((b)\) সমাধান করঃ \(f(x)+f(2x)+f(3x)=0\)
\((c)\) সমাধান করঃ \(4f(x)f(2x)f(3x)=1\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{2n\pi}{3}\pm\frac{7\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ (2n+1)\frac{\pi}{4}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
\(Q.5.(xix)\) \(f(x)=\tan^{-1}{x}, \ h(x)=\sin^{-1}{x}, \ g(x)=\sec{x}\)
\((a)\) দেখাও যে, \(f\left(\frac{5}{6}\right)-f\left(\frac{49}{71}\right)=f\left(\frac{1}{11}\right)\)
\((b)\) দেখাও যে, \(h\left(\sqrt{2}\sin{\theta}\right)+h\left(\sqrt{\cos{2\theta}}\right)=\frac{\pi}{2}\)
\((c)\) সমাধান করঃ \(g(4\theta)-g(2\theta)=2\) যখন, \(0\lt{x}\lt{180^{o}}\)
উত্তরঃ \((c) \ 18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}\) এবং \(162^{o}\)
\(Q.5.(xx)\) \(\cot{\theta}+\tan{\theta}=2\sec{\theta}\) একটি ত্রিকোণমিতিক সমীকরণ এবং \(f(x)=\sin^{-1}{\left\{\sqrt{2(1-x^2)}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\((a)\) প্রমাণ কর যে, \(\sin{\cot^{-1}{\tan{\cos^{-1}{\frac{3}{4}}}}}=\frac{3}{4}\)
\((b)\) দেখাও যে, \(f(\cos{\theta})=\frac{\pi}{2}\)
\((c)\) \(-2\pi\lt{\theta}\lt{2\pi}\) সীমার মধ্যে ত্রিকোণমিতিক সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
\(Q.5.(xxi)\) \(A=\sin^{-1}{\frac{x}{y}}, \ B=\cos^{-1}{\frac{a}{x}}\) এবং \(C=\cos^{-1}{\frac{b}{y}}\)
\((a)\) দেখাও যে, \(\{\cos{(\sin^{-1}{x})}\}^2=\{\sin{(\cos^{-1}{x})}\}^2\)
\((b)\) দেখাও যে, \(B+C=\cos^{-1}{\left\{\frac{ab}{xy}-\frac{1}{xy}\sqrt{(x^2-a^2)(y^2-b^2)}\right\}}\)
\((c)\) \(A+B+C=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(x^4-x^2y^2-2abx^2+a^2y^2+b^2x^2=0\)
\(Q.5.(xxii)\) \(P=\sin^{-1}{\frac{1}{\sqrt{5}}}-\frac{1}{2}\cos^{-1}{\frac{4}{5}}+\tan^{-1}{\frac{1}{3}}\) এবং \(\tan^{-1}{x}=A\)
\((a)\) \(\frac{1}{2}\sin^{-1}{x}\) কে \(\tan^{-1}{y}\) এ রুপান্তর কর, যেখানে \(y=\frac{x}{1+\sqrt{1-x^2}}\)
\((b)\) দেখাও যে, \(5A=\tan^{-1}{\left\{\frac{x^5-10x^3+5x}{5x^4-10x^2+1}\right\}}\)
\((c)\) প্রমাণ কর যে, \(P=\tan^{-1}{\frac{1}{2}}\)
\(Q.5.(xxiii)\) \(f(t)=\sqrt{3}\sin{t}+\cos{t}\) এবং \(g(t)=\tan^{-1}{t}\)
\((a)\) প্রমাণ কর যে, \(\cot^{-1}{x}+\cot^{-1}{y}=\cot{\frac{xy-1}{x+y}}\)
\((b)\) \(g\left(\frac{p}{q}\right)+g\left(\frac{r}{s}\right)=\theta\) হলে, প্রমাণ কর যে, \((sq-rp)\sin{\theta}=(rq+ps)\cos{\theta}\)
\((c)\) \(f(t)=\sqrt{3}\) সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ 2n\pi+\frac{\pi}{2}, \ 2n\pi+\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xxiv)\) দেওয়া আছে, \(f(x)=\sin^{-1}{x}, \ g(x)=\cos^{-1}{\sqrt{1-x^2}}\) এবং \(h(x)=\tan^{-1}{\frac{x}{\sqrt{1-x^2}}}\)
\((a)\) দেখাও যে, \(f(x)=g(x)\)
\((b)\) প্রমাণ কর যে, \(f(x)g(x)h(x)=\left\{f(x)\right\}^3\)
\((c)\) \(\left\{f(x)\right\}^2+\left\{g(x)\right\}^2+\left\{h(x)\right\}^2=\frac{3}{16}\pi^2\) হলে, প্রমাণ কর যে, \(x=\pm\frac{1}{\sqrt{2}}\)
\(Q.5.(xxv)\) দেওয়া আছে, \(f(x)=cosec \ {x}, \ g(x)=\tan{x}\)
\((a)\) যদি \(x=\frac{1}{2}\cos^{-1}{\frac{3}{5}}\) হয় তাহলে, প্রমাণ কর যে, \(\tan{x}=\frac{1}{2}\)
\((b)\) \(2\tan^{-1}{\left\{f(x)\right\}}=\cot^{-1}{\left(\frac{\cos{x}}{2}\right)}\) এর সাধারণ সমাধান নির্ণয় কর।
\((c)\) প্রমাণ কর যে, \(\tan^{-1}{\left\{f\left(\cos^{-1}{x}\right)\right\}}-\tan^{-1}{\left\{g\left(\sin^{-1}{x}\right)\right\}}=\tan^{-1}{\left\{\frac{(1-x)\sqrt{1-x^2}}{1+x-x^2}\right\}}\)
উত্তরঃ \((b) \ (2n+1)\frac{\pi}{2}, \ n\pi+\frac{3\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xxvi)\) দেওয়া আছে, \(f(x)=\cos^{-1}{x}\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}f\left(\frac{1-x}{1+x}\right)\)
\((b)\) \(f(x)+f(y)+f(z)=\pi\) হলে দেখাও যে, \(x^2+y^2+z^2+2xyz=1\)
\((c)\) \(f(x)=\theta\) হলে, \(4(1-x^2+x)=5\) এর সমাধান নির্ণয় কর যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((b) \ \pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
\(Q.5.(xxvii)\) দৃশ্যকল্প-১: \(\sin^{-1}{\frac{2a}{1+a^2}}-\cos^{-1}{\frac{1-b^2}{1+b^2}}=2\tan^{-1}{x}\)
দৃশ্যকল্প-২: \(2\sin{\theta}\sin{3\theta}=1, \ 0\le{\theta}\le{\pi}\)
\((a)\) দেখাও যে, \(\cot{\cos^{-1}{\sin{\tan^{-1}{\frac{3}{4}}}}}=\frac{3}{4}\)
\((b)\) দৃশ্যকল্প-১ থেকে প্রমাণ কর যে, \(x=\frac{a-b}{1+ab}\)
\((c)\) দৃশ্যকল্প-২ থেকে সমীকরণটির সাধারণ সমাধান নির্ণয় কর এবং প্রদত্ত সীমার মধ্যে \(\theta\) এর মান নির্ণয় কর।
উত্তরঃ \((c) \ n\pi\pm\frac{\pi}{6}, \ (2n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\)
\(Q.5.(xxviii)\) \(f(x)=\sin{x}; \ \triangle{ABC}\)-এ \(\cot^{-1}{\frac{1}{2}}=B, \ \cot^{-1}{\frac{1}{3}}=C\)
\((a)\) প্রমাণ কর যে, \(cosec \ {\sin^{-1}{\tan{\sec^{-1}{\frac{x}{y}}}}}=\frac{y}{\sqrt{x^2-y^2}}\)
\((b)\) \(2+f\left(\frac{\pi}{2}-x\right)=\sqrt{3}f(x); \ -2\pi\lt{x}\lt{2\pi}\) সমীকরণটি সমাধান কর
\((c)\) \(\angle{A}\) নির্ণয় কর।
উত্তরঃ \((b) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\((c) \ \frac{\pi}{4}\)
\((a)\) মান নির্ণয় করঃ \(\tan^{-1}{\sin{\cos^{-1}{\sqrt{\frac{2}{3}}}}}\)
\((b)\) \(A+\sqrt{3}B=\sqrt{2}\) হলে, সমীকরণটি সমাধান কর।
\((c)\) \(A+B=C+D\) হলে, সমীকরণটি \(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে সমাধান আছে কিনা যাচাই কর।
উত্তরঃ \((a) \ \frac{\pi}{6}\)
\((b) \ 2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 0, \ \frac{\pi}{6}\)
দিঃ২০১৭।
\(Q.5.(xv)\) \(f(x)=\tan{\theta}\) এবং \(\sin{\theta}=\frac{12}{13}\)
\((a)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{1}{\sqrt{x}}}=\frac{1}{2}\sec^{-1}{\frac{1+x}{1-x}}\)
\((b)\) সমাধান করঃ \(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\((c)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{3}{4}}+\frac{\theta}{2}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
উত্তরঃ \((b) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) এবং \(\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(Q.5.(xvi)\) \(f(x)=\cos{x}\) এবং \(g(x)=\sin{x}\)
\((a)\) \(f\left(\cot^{-1}{\frac{3}{4}}\right)\) এর মান কত?
\((b)\) \(f^{-1}\left(\frac{x}{a}\right)+f^{-1}\left(\frac{y}{b}\right)=\theta\) হলে, দেখাও যে, \(\frac{x^2}{a^2}-\frac{2xy}{ab}f(\theta)+\frac{y^2}{b^2}=1-\left\{f(x)\right\}^2\)
\((c)\) সমাধান করঃ \(f(\theta)-f(7\theta)=g(4\theta)\)
উত্তরঃ \((a) \ \frac{3}{5}\)
\((c) \ \frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xvii)\) দৃশ্যকল্প-১: \(\cos^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{5}{13}}+cosec^{-1}{\frac{65}{16}}\)
দৃশ্যকল্প-২: \(f(x)=\tan{x}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\cos^{-1}{\frac{1-x^2}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\) হবে।
\((c)\) দৃশ্যকল্প-২: থেকে সমাধান করঃ \(f(x)+f(2x)+f(3x)=f(x)f(2x)f(3x)\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{3}, \ \frac{2\pi}{3}, \ \pi, \ \frac{4\pi}{3}, \ \frac{5\pi}{3}\)
\(Q.5.(xviii)\) \(f(x)=\cos{x}\)
\((a)\) \(\cos{3x}=-\frac{\sqrt{3}}{2}\) হলে, \(x\) এর মান নির্ণয় কর।
\((b)\) সমাধান করঃ \(f(x)+f(2x)+f(3x)=0\)
\((c)\) সমাধান করঃ \(4f(x)f(2x)f(3x)=1\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{2n\pi}{3}\pm\frac{7\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ (2n+1)\frac{\pi}{4}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
\(Q.5.(xix)\) \(f(x)=\tan^{-1}{x}, \ h(x)=\sin^{-1}{x}, \ g(x)=\sec{x}\)
\((a)\) দেখাও যে, \(f\left(\frac{5}{6}\right)-f\left(\frac{49}{71}\right)=f\left(\frac{1}{11}\right)\)
\((b)\) দেখাও যে, \(h\left(\sqrt{2}\sin{\theta}\right)+h\left(\sqrt{\cos{2\theta}}\right)=\frac{\pi}{2}\)
\((c)\) সমাধান করঃ \(g(4\theta)-g(2\theta)=2\) যখন, \(0\lt{x}\lt{180^{o}}\)
উত্তরঃ \((c) \ 18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}\) এবং \(162^{o}\)
\(Q.5.(xx)\) \(\cot{\theta}+\tan{\theta}=2\sec{\theta}\) একটি ত্রিকোণমিতিক সমীকরণ এবং \(f(x)=\sin^{-1}{\left\{\sqrt{2(1-x^2)}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\((a)\) প্রমাণ কর যে, \(\sin{\cot^{-1}{\tan{\cos^{-1}{\frac{3}{4}}}}}=\frac{3}{4}\)
\((b)\) দেখাও যে, \(f(\cos{\theta})=\frac{\pi}{2}\)
\((c)\) \(-2\pi\lt{\theta}\lt{2\pi}\) সীমার মধ্যে ত্রিকোণমিতিক সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
\(Q.5.(xxi)\) \(A=\sin^{-1}{\frac{x}{y}}, \ B=\cos^{-1}{\frac{a}{x}}\) এবং \(C=\cos^{-1}{\frac{b}{y}}\)
\((a)\) দেখাও যে, \(\{\cos{(\sin^{-1}{x})}\}^2=\{\sin{(\cos^{-1}{x})}\}^2\)
\((b)\) দেখাও যে, \(B+C=\cos^{-1}{\left\{\frac{ab}{xy}-\frac{1}{xy}\sqrt{(x^2-a^2)(y^2-b^2)}\right\}}\)
\((c)\) \(A+B+C=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(x^4-x^2y^2-2abx^2+a^2y^2+b^2x^2=0\)
\(Q.5.(xxii)\) \(P=\sin^{-1}{\frac{1}{\sqrt{5}}}-\frac{1}{2}\cos^{-1}{\frac{4}{5}}+\tan^{-1}{\frac{1}{3}}\) এবং \(\tan^{-1}{x}=A\)
\((a)\) \(\frac{1}{2}\sin^{-1}{x}\) কে \(\tan^{-1}{y}\) এ রুপান্তর কর, যেখানে \(y=\frac{x}{1+\sqrt{1-x^2}}\)
\((b)\) দেখাও যে, \(5A=\tan^{-1}{\left\{\frac{x^5-10x^3+5x}{5x^4-10x^2+1}\right\}}\)
\((c)\) প্রমাণ কর যে, \(P=\tan^{-1}{\frac{1}{2}}\)
\(Q.5.(xxiii)\) \(f(t)=\sqrt{3}\sin{t}+\cos{t}\) এবং \(g(t)=\tan^{-1}{t}\)
\((a)\) প্রমাণ কর যে, \(\cot^{-1}{x}+\cot^{-1}{y}=\cot{\frac{xy-1}{x+y}}\)
\((b)\) \(g\left(\frac{p}{q}\right)+g\left(\frac{r}{s}\right)=\theta\) হলে, প্রমাণ কর যে, \((sq-rp)\sin{\theta}=(rq+ps)\cos{\theta}\)
\((c)\) \(f(t)=\sqrt{3}\) সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ 2n\pi+\frac{\pi}{2}, \ 2n\pi+\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xxiv)\) দেওয়া আছে, \(f(x)=\sin^{-1}{x}, \ g(x)=\cos^{-1}{\sqrt{1-x^2}}\) এবং \(h(x)=\tan^{-1}{\frac{x}{\sqrt{1-x^2}}}\)
\((a)\) দেখাও যে, \(f(x)=g(x)\)
\((b)\) প্রমাণ কর যে, \(f(x)g(x)h(x)=\left\{f(x)\right\}^3\)
\((c)\) \(\left\{f(x)\right\}^2+\left\{g(x)\right\}^2+\left\{h(x)\right\}^2=\frac{3}{16}\pi^2\) হলে, প্রমাণ কর যে, \(x=\pm\frac{1}{\sqrt{2}}\)
\(Q.5.(xxv)\) দেওয়া আছে, \(f(x)=cosec \ {x}, \ g(x)=\tan{x}\)
\((a)\) যদি \(x=\frac{1}{2}\cos^{-1}{\frac{3}{5}}\) হয় তাহলে, প্রমাণ কর যে, \(\tan{x}=\frac{1}{2}\)
\((b)\) \(2\tan^{-1}{\left\{f(x)\right\}}=\cot^{-1}{\left(\frac{\cos{x}}{2}\right)}\) এর সাধারণ সমাধান নির্ণয় কর।
\((c)\) প্রমাণ কর যে, \(\tan^{-1}{\left\{f\left(\cos^{-1}{x}\right)\right\}}-\tan^{-1}{\left\{g\left(\sin^{-1}{x}\right)\right\}}=\tan^{-1}{\left\{\frac{(1-x)\sqrt{1-x^2}}{1+x-x^2}\right\}}\)
উত্তরঃ \((b) \ (2n+1)\frac{\pi}{2}, \ n\pi+\frac{3\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xxvi)\) দেওয়া আছে, \(f(x)=\cos^{-1}{x}\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{\sqrt{x}}=\frac{1}{2}f\left(\frac{1-x}{1+x}\right)\)
\((b)\) \(f(x)+f(y)+f(z)=\pi\) হলে দেখাও যে, \(x^2+y^2+z^2+2xyz=1\)
\((c)\) \(f(x)=\theta\) হলে, \(4(1-x^2+x)=5\) এর সমাধান নির্ণয় কর যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((b) \ \pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
\(Q.5.(xxvii)\) দৃশ্যকল্প-১: \(\sin^{-1}{\frac{2a}{1+a^2}}-\cos^{-1}{\frac{1-b^2}{1+b^2}}=2\tan^{-1}{x}\)
দৃশ্যকল্প-২: \(2\sin{\theta}\sin{3\theta}=1, \ 0\le{\theta}\le{\pi}\)
\((a)\) দেখাও যে, \(\cot{\cos^{-1}{\sin{\tan^{-1}{\frac{3}{4}}}}}=\frac{3}{4}\)
\((b)\) দৃশ্যকল্প-১ থেকে প্রমাণ কর যে, \(x=\frac{a-b}{1+ab}\)
\((c)\) দৃশ্যকল্প-২ থেকে সমীকরণটির সাধারণ সমাধান নির্ণয় কর এবং প্রদত্ত সীমার মধ্যে \(\theta\) এর মান নির্ণয় কর।
উত্তরঃ \((c) \ n\pi\pm\frac{\pi}{6}, \ (2n+1)\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}\)
\(Q.5.(xxviii)\) \(f(x)=\sin{x}; \ \triangle{ABC}\)-এ \(\cot^{-1}{\frac{1}{2}}=B, \ \cot^{-1}{\frac{1}{3}}=C\)
\((a)\) প্রমাণ কর যে, \(cosec \ {\sin^{-1}{\tan{\sec^{-1}{\frac{x}{y}}}}}=\frac{y}{\sqrt{x^2-y^2}}\)
\((b)\) \(2+f\left(\frac{\pi}{2}-x\right)=\sqrt{3}f(x); \ -2\pi\lt{x}\lt{2\pi}\) সমীকরণটি সমাধান কর
\((c)\) \(\angle{A}\) নির্ণয় কর।
উত্তরঃ \((b) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\((c) \ \frac{\pi}{4}\)
\(Q.5.(i)\) \(f(x)=\frac{2x}{1+x^2}, \ g(y)=\frac{1-y^2}{1+y^2}\) এবং \(h(x)=\sin{x}\)
\((a)\) \(\cos^{-1}{m}+\cos^{-1}{n}=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(m^2+n^2=1\)
\((b)\) \(cosec^{-1}{\frac{1}{f(a)}}+\sec^{-1}{\frac{1}{g(b)}}=2\tan^{-1}{x}\) হলে, দেখাও যে, \(x=\frac{a-b}{1+ab}\)
\((c)\) \(0\le{\theta}\le{2\pi}\) ব্যবধিতে \(2h(\theta)h(3\theta)=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \((a) \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
\((a)\) \(\cos^{-1}{m}+\cos^{-1}{n}=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(m^2+n^2=1\)
\((b)\) \(cosec^{-1}{\frac{1}{f(a)}}+\sec^{-1}{\frac{1}{g(b)}}=2\tan^{-1}{x}\) হলে, দেখাও যে, \(x=\frac{a-b}{1+ab}\)
\((c)\) \(0\le{\theta}\le{2\pi}\) ব্যবধিতে \(2h(\theta)h(3\theta)=1\) সমীকরণটির সমাধান কর।
উত্তরঃ \((a) \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}\) এবং \(\frac{11\pi}{6}\)
ঢাঃ২০১৯।
সমাধানঃ
\((a)\)
দেওয়া আছে,
\(\cos^{-1}{m}+\cos^{-1}{n}=\frac{\pi}{2}\)
\(\Rightarrow \cos^{-1}{\{mn-\sqrt{(1-m^2)(1-n^2)}\}}=\frac{\pi}{2}\) ➜ \(\because \cos^{-1}{A}+\cos^{-1}{B}=\cos^{-1}{\{AB-\sqrt{(1-A^2)(1-B^2)}\}}\)
\(\Rightarrow mn-\sqrt{(1-m^2)(1-n^2)}=\cos{\frac{\pi}{2}}\)
\(\Rightarrow mn-\sqrt{(1-m^2)(1-n^2)}=0\) ➜ \(\because \cos{\frac{\pi}{2}}=0\)
\(\Rightarrow mn=\sqrt{(1-m^2)(1-n^2)}\)
\(\Rightarrow m^2n^2=(1-m^2)(1-n^2)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow m^2n^2=1-m^2-n^2+m^2n^2\)
\(\Rightarrow m^2n^2+m^2+n^2-m^2n^2=1\)
\(\therefore m^2+n^2=1\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(f(x)=\frac{2x}{1+x^2}, \ g(y)=\frac{1-y^2}{1+y^2}\)
\(\Rightarrow f(a)=\frac{2a}{1+a^2}, \ g(b)=\frac{1-b^2}{1+b^2}\)
আবার,
\(cosec^{-1}{\frac{1}{f(a)}}+\sec^{-1}{\frac{1}{g(b)}}=2\tan^{-1}{x}\)
\(\Rightarrow \sin^{-1}{f(a)}+\cos^{-1}{g(b)}=2\tan^{-1}{x}\) ➜ \(\because cosec^{-1}{\frac{1}{x}}=\sin^{-1}{x}\)
এবং \(\sec^{-1}{\frac{1}{x}}=\cos^{-1}{x}\)
\(\Rightarrow \sin^{-1}{\frac{2a}{1+a^2}}-\cos^{-1}{\frac{1-b^2}{1+b^2}}=2\tan^{-1}{x}\)➜ \(\because f(a)=\frac{2a}{1+a^2}\)
এবং \(g(b)=\frac{1-b^2}{1+b^2}\)
\(\Rightarrow 2\tan^{-1}{a}-2\tan^{-1}{b}=2\tan^{-1}{x}\) ➜ \(\because \sin^{-1}{\frac{2x}{1+x^2}}=2\tan^{-1}{x}\)
এবং \(\cos^{-1}{\frac{1-x^2}{1+x^2}}=2\tan^{-1}{x}\)
\(\Rightarrow 2\left(\tan^{-1}{a}-\tan^{-1}{b}\right)=2\tan^{-1}{x}\)
\(\Rightarrow \tan^{-1}{a}-\tan^{-1}{b}=\tan^{-1}{x}\)
\(\Rightarrow \tan^{-1}{x}=\tan^{-1}{a}-\tan^{-1}{b}\)
\(\Rightarrow \tan^{-1}{x}=\tan^{-1}{\left(\frac{a-b}{1+ab}\right)}\) ➜ \(\because \tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left(\frac{x-y}{1+xy}\right)}\)
\(\therefore x=\frac{a-b}{1+ab}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(h(x)=\sin{x}\)
\(\Rightarrow h(\theta)=\sin{\theta}, \ h(3\theta)=\sin{3\theta}\)
প্রদত্ত সমীকরণ,
\(2h(\theta)h(3\theta)=1\)
\(\Rightarrow 2\sin{\theta}\sin{3\theta}=1\) ➜ \(\because h(\theta)=\sin{\theta}, \ h(3\theta)=\sin{3\theta}\)
\(\Rightarrow \cos{(3\theta-\theta)}-\cos{(3\theta+\theta)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=1\)
\(\Rightarrow \cos{2\theta}=1+\cos{4\theta}\)
\(\Rightarrow \cos{2\theta}=2\cos^2{2\theta}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2\theta}-2\cos^2{2\theta}=0\)
\(\Rightarrow -\cos{2\theta}(2\cos{2\theta}-1)=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}-1=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সকল মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
দেওয়া আছে,
\(\cos^{-1}{m}+\cos^{-1}{n}=\frac{\pi}{2}\)
\(\Rightarrow \cos^{-1}{\{mn-\sqrt{(1-m^2)(1-n^2)}\}}=\frac{\pi}{2}\) ➜ \(\because \cos^{-1}{A}+\cos^{-1}{B}=\cos^{-1}{\{AB-\sqrt{(1-A^2)(1-B^2)}\}}\)
\(\Rightarrow mn-\sqrt{(1-m^2)(1-n^2)}=\cos{\frac{\pi}{2}}\)
\(\Rightarrow mn-\sqrt{(1-m^2)(1-n^2)}=0\) ➜ \(\because \cos{\frac{\pi}{2}}=0\)
\(\Rightarrow mn=\sqrt{(1-m^2)(1-n^2)}\)
\(\Rightarrow m^2n^2=(1-m^2)(1-n^2)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow m^2n^2=1-m^2-n^2+m^2n^2\)
\(\Rightarrow m^2n^2+m^2+n^2-m^2n^2=1\)
\(\therefore m^2+n^2=1\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(f(x)=\frac{2x}{1+x^2}, \ g(y)=\frac{1-y^2}{1+y^2}\)
\(\Rightarrow f(a)=\frac{2a}{1+a^2}, \ g(b)=\frac{1-b^2}{1+b^2}\)
আবার,
\(cosec^{-1}{\frac{1}{f(a)}}+\sec^{-1}{\frac{1}{g(b)}}=2\tan^{-1}{x}\)
\(\Rightarrow \sin^{-1}{f(a)}+\cos^{-1}{g(b)}=2\tan^{-1}{x}\) ➜ \(\because cosec^{-1}{\frac{1}{x}}=\sin^{-1}{x}\)
এবং \(\sec^{-1}{\frac{1}{x}}=\cos^{-1}{x}\)
\(\Rightarrow \sin^{-1}{\frac{2a}{1+a^2}}-\cos^{-1}{\frac{1-b^2}{1+b^2}}=2\tan^{-1}{x}\)➜ \(\because f(a)=\frac{2a}{1+a^2}\)
এবং \(g(b)=\frac{1-b^2}{1+b^2}\)
\(\Rightarrow 2\tan^{-1}{a}-2\tan^{-1}{b}=2\tan^{-1}{x}\) ➜ \(\because \sin^{-1}{\frac{2x}{1+x^2}}=2\tan^{-1}{x}\)
এবং \(\cos^{-1}{\frac{1-x^2}{1+x^2}}=2\tan^{-1}{x}\)
\(\Rightarrow 2\left(\tan^{-1}{a}-\tan^{-1}{b}\right)=2\tan^{-1}{x}\)
\(\Rightarrow \tan^{-1}{a}-\tan^{-1}{b}=\tan^{-1}{x}\)
\(\Rightarrow \tan^{-1}{x}=\tan^{-1}{a}-\tan^{-1}{b}\)
\(\Rightarrow \tan^{-1}{x}=\tan^{-1}{\left(\frac{a-b}{1+ab}\right)}\) ➜ \(\because \tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left(\frac{x-y}{1+xy}\right)}\)
\(\therefore x=\frac{a-b}{1+ab}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(h(x)=\sin{x}\)
\(\Rightarrow h(\theta)=\sin{\theta}, \ h(3\theta)=\sin{3\theta}\)
প্রদত্ত সমীকরণ,
\(2h(\theta)h(3\theta)=1\)
\(\Rightarrow 2\sin{\theta}\sin{3\theta}=1\) ➜ \(\because h(\theta)=\sin{\theta}, \ h(3\theta)=\sin{3\theta}\)
\(\Rightarrow \cos{(3\theta-\theta)}-\cos{(3\theta+\theta)}=1\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=1\)
\(\Rightarrow \cos{2\theta}=1+\cos{4\theta}\)
\(\Rightarrow \cos{2\theta}=2\cos^2{2\theta}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(\Rightarrow \cos{2\theta}-2\cos^2{2\theta}=0\)
\(\Rightarrow -\cos{2\theta}(2\cos{2\theta}-1)=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}-1=0\)
\(\Rightarrow \cos{2\theta}=0, \ 2\cos{2\theta}=1\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{2\theta}=0, \ \cos{2\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow 2\theta=(2n+1)\frac{\pi}{2}, \ 2\theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\frac{\pi}{4}, \ \theta=n\pi\pm\frac{\pi}{6}\)
\(\therefore \theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{4}, \ n\pi\pm\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{4}, \ \frac{\pi}{6}, \ -\frac{\pi}{6}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{4}, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{4}, \ \pi+\frac{\pi}{6}, \ \pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\) সকল মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{3\pi}{4}, \ \frac{7\pi}{6}, \ \frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{4}, \ -\pi+\frac{\pi}{6}, \ -\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{4}, \ 2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{4}, \ \frac{13\pi}{6}, \ \frac{11\pi}{6}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{4}, \ \frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{4}, \ 3\pi+\frac{\pi}{6}, \ 3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{7\pi}{4}, \ \frac{19\pi}{6}, \ \frac{17\pi}{6}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{7\pi}{4}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{4}, \ 4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=\frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)
\(Q.5.(ii)\) দৃশ্যকল্প-১: \(\sec{A}=\sqrt{5}, \ cosec \ {B}=\frac{5}{3}\) এবং \(\cot{C}=3\)
দৃশ্যকল্প-২: \(f(x)=\sin{x}\)
\((a)\) \(cosec^{-1}{\sqrt{17}}+\sec^{-1}{\frac{\sqrt{26}}{5}}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে \(A+C-\frac{1}{2}B\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ দৃশ্যকল্প-২: থেকে \(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \tan^{-1}{\left(\frac{9}{19}\right)}\)
\((b) \ \tan^{-1}{2}\)
\((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
দৃশ্যকল্প-২: \(f(x)=\sin{x}\)
\((a)\) \(cosec^{-1}{\sqrt{17}}+\sec^{-1}{\frac{\sqrt{26}}{5}}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে \(A+C-\frac{1}{2}B\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ দৃশ্যকল্প-২: থেকে \(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \tan^{-1}{\left(\frac{9}{19}\right)}\)
\((b) \ \tan^{-1}{2}\)
\((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
রাঃ২০১৯।
সমাধানঃ
\((a)\)
প্রদত্ত রাশি,
\(cosec^{-1}{\sqrt{17}}+\sec^{-1}{\frac{\sqrt{26}}{5}}\)
\(=\tan^{-1}{\frac{1}{4}}+\tan^{-1}{\frac{1}{5}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{17}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{17})^2-1^2} \)
\(=\sqrt{17-1} \)
\(=\sqrt{16}\)
\(=4\)
\(\therefore cosec^{-1}{\sqrt{17}}=\tan^{-1}{\frac{1}{4}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=5 , \ \text{অতিভুজ}=\sqrt{26}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{26})^2-5^2} \)
\(=\sqrt{26-25} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \sec^{-1}{\frac{\sqrt{26}}{5}}=\tan^{-1}{\frac{1}{5}}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4}\times\frac{1}{5}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{20}}\right)}\)
\(=\tan^{-1}{\left(\frac{5+4}{20-1}\right)}\) ➜ লব ও হরকে \(20\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{9}{19}\right)}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(\sec{A}=\sqrt{5}, \ cosec \ {B}=\frac{5}{3}\) এবং \(\cot{C}=3\)
\(\Rightarrow A=\sec^{-1}{\sqrt{5}}, \ B=cosec^{-1}{\frac{5}{3}}\) এবং \(C=\cot^{-1}{3}\)
প্রদত্ত রাশি,
\(A+C-\frac{1}{2}B\)
\(=\sec^{-1}{\sqrt{5}}+\cot^{-1}{3}-\frac{1}{2}cosec^{-1}{\frac{5}{3}}\) ➜ \(\because A=\sec^{-1}{\sqrt{5}}, \ B=cosec^{-1}{\frac{5}{3}}\)
এবং \(C=\cot^{-1}{3}\)
\(=\tan^{-1}{2}+\tan^{-1}{\frac{1}{3}}-\frac{1}{2}\tan^{-1}{\frac{3}{4}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{ভূমি}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2}\)
\(=\sqrt{5-1}\)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sec^{-1}{\sqrt{5}}=\tan^{-1}{\frac{2}{1}}\)
\(=\tan^{-1}{2}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\cot^{-1}{x}=\tan^{-1}{\frac{1}{x}}\)
তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=3, \ \text{অতিভুজ}=5\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{5^2-3^2}\)
\(=\sqrt{25-9}\)
\(=\sqrt{16}\)
\(=4\)
\(\therefore cosec^{-1}{\frac{5}{3}}=\tan^{-1}{\frac{3}{4}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{2\tan^{-1}{\frac{1}{3}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{2\times\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}}-\tan^{-1}{\frac{3}{4}}\right\}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\frac{2x}{1-x^2}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{\frac{2}{3}}{1-\frac{1}{9}}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{6}{9-1}}-\tan^{-1}{\frac{3}{4}}\right\}\) ➜ দ্বিতীয় পদের
লব ও হরকে \(9\) দ্বারা গুন করে।
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{6}{8}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{3}{4}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\{0\}\)
\(=\tan^{-1}{2}+0\)
\(=\tan^{-1}{2}\)
ইহাই নির্ণেয় মান।
\((c)\)
দেওয়া আছে,
\(f(x)=\sin{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}+x\right)}=2\) ➜ \(\because f(x)=\sin{x}\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}\times1+x\right)}=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\cos{x}=2\) ➜
দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow \cos{x}-\sqrt{3}\sin{x}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{x}\frac{1}{2}-\sin{x}\frac{\sqrt{3}}{2}=-\frac{2}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{3}}-\sin{x}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow x=(6n+3-1)\frac{\pi}{3}\)
\(\Rightarrow x=(6n+2)\frac{\pi}{3}\)
\(\Rightarrow x=2(3n+1)\frac{\pi}{3}\)
\(\therefore x=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(x=4\frac{2\pi}{3}\)
\(\Rightarrow x=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\frac{2\pi}{3}\)
\(\Rightarrow x=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{4\pi}{3}\)
যখন, \(n=2,\) \(x=\frac{14\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(x=-\frac{10\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
প্রদত্ত রাশি,
\(cosec^{-1}{\sqrt{17}}+\sec^{-1}{\frac{\sqrt{26}}{5}}\)
\(=\tan^{-1}{\frac{1}{4}}+\tan^{-1}{\frac{1}{5}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{17}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{17})^2-1^2} \)
\(=\sqrt{17-1} \)
\(=\sqrt{16}\)
\(=4\)
\(\therefore cosec^{-1}{\sqrt{17}}=\tan^{-1}{\frac{1}{4}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=5 , \ \text{অতিভুজ}=\sqrt{26}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{26})^2-5^2} \)
\(=\sqrt{26-25} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \sec^{-1}{\frac{\sqrt{26}}{5}}=\tan^{-1}{\frac{1}{5}}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4}\times\frac{1}{5}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{20}}\right)}\)
\(=\tan^{-1}{\left(\frac{5+4}{20-1}\right)}\) ➜ লব ও হরকে \(20\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{9}{19}\right)}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(\sec{A}=\sqrt{5}, \ cosec \ {B}=\frac{5}{3}\) এবং \(\cot{C}=3\)
\(\Rightarrow A=\sec^{-1}{\sqrt{5}}, \ B=cosec^{-1}{\frac{5}{3}}\) এবং \(C=\cot^{-1}{3}\)
প্রদত্ত রাশি,
\(A+C-\frac{1}{2}B\)
\(=\sec^{-1}{\sqrt{5}}+\cot^{-1}{3}-\frac{1}{2}cosec^{-1}{\frac{5}{3}}\) ➜ \(\because A=\sec^{-1}{\sqrt{5}}, \ B=cosec^{-1}{\frac{5}{3}}\)
এবং \(C=\cot^{-1}{3}\)
\(=\tan^{-1}{2}+\tan^{-1}{\frac{1}{3}}-\frac{1}{2}\tan^{-1}{\frac{3}{4}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{ভূমি}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2}\)
\(=\sqrt{5-1}\)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sec^{-1}{\sqrt{5}}=\tan^{-1}{\frac{2}{1}}\)
\(=\tan^{-1}{2}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\cot^{-1}{x}=\tan^{-1}{\frac{1}{x}}\)
তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=3, \ \text{অতিভুজ}=5\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{5^2-3^2}\)
\(=\sqrt{25-9}\)
\(=\sqrt{16}\)
\(=4\)
\(\therefore cosec^{-1}{\frac{5}{3}}=\tan^{-1}{\frac{3}{4}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{2\tan^{-1}{\frac{1}{3}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{2\times\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}}-\tan^{-1}{\frac{3}{4}}\right\}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\frac{2x}{1-x^2}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{\frac{2}{3}}{1-\frac{1}{9}}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{6}{9-1}}-\tan^{-1}{\frac{3}{4}}\right\}\) ➜ দ্বিতীয় পদের
লব ও হরকে \(9\) দ্বারা গুন করে।
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{6}{8}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{3}{4}}-\tan^{-1}{\frac{3}{4}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\{0\}\)
\(=\tan^{-1}{2}+0\)
\(=\tan^{-1}{2}\)
ইহাই নির্ণেয় মান।
\((c)\)
দেওয়া আছে,
\(f(x)=\sin{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}f(x)-f\left(\frac{\pi}{2}+x\right)=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}+x\right)}=2\) ➜ \(\because f(x)=\sin{x}\)
\(\Rightarrow \sqrt{3}\sin{x}-\sin{\left(\frac{\pi}{2}\times1+x\right)}=2\)
\(\Rightarrow \sqrt{3}\sin{x}-\cos{x}=2\) ➜

দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং সাইন অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং সাইন অনুপাতের পরিবর্তন হয়ে কোসাইন হয়েছে।
\(\Rightarrow \cos{x}-\sqrt{3}\sin{x}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{x}\frac{1}{2}-\sin{x}\frac{\sqrt{3}}{2}=-\frac{2}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{3}}-\sin{x}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow x=(6n+3-1)\frac{\pi}{3}\)
\(\Rightarrow x=(6n+2)\frac{\pi}{3}\)
\(\Rightarrow x=2(3n+1)\frac{\pi}{3}\)
\(\therefore x=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{2\pi}{3}\) মানটি গ্রহনযোগ্য।\(\therefore x=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(x=4\frac{2\pi}{3}\)
\(\Rightarrow x=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(x=-2\frac{2\pi}{3}\)
\(\Rightarrow x=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{4\pi}{3}\)
যখন, \(n=2,\) \(x=\frac{14\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(x=-\frac{10\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(Q.5.(iii)\) \(f(x)=\sin^{-1}{x}\) এবং \(g(x)=\cos{x}\)
\((a)\) \(\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\) এর মান কত?
\((b)\) \(f\left\{\sqrt{2}g\left(\frac{\pi}{2}-\theta\right)\right\}+f\left\{\sqrt{g(2\theta)}\right\}\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ \(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((b) \ \frac{\pi}{2}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
\((a)\) \(\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\) এর মান কত?
\((b)\) \(f\left\{\sqrt{2}g\left(\frac{\pi}{2}-\theta\right)\right\}+f\left\{\sqrt{g(2\theta)}\right\}\) এর মান নির্ণয় কর।
\((c)\) সমাধান করঃ \(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\) যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((b) \ \frac{\pi}{2}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
যঃ২০১৯।
সমাধানঃ
\((a)\)
প্রদত্ত রাশি,
\(\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)}\)
\(=\tan^{-1}{\left(\frac{3+2}{6-1}\right)}\) ➜ লব ও হরকে \(6\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{5}{5}\right)}\)
\(=\tan^{-1}{\left(1\right)}\)
\(=\tan^{-1}{\left(\tan{\frac{\pi}{4}}\right)}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(=\frac{\pi}{4}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(f(x)=\sin^{-1}{x}\) এবং \(g(x)=\cos{x}\)
\(\Rightarrow g\left(\frac{\pi}{2}-\theta\right)=\cos{\left(\frac{\pi}{2}-\theta\right)}, \ g(2\theta)=\cos{(2\theta)}\)
\(\Rightarrow f\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}, \ f\left\{\sqrt{\cos{(2\theta)}}\right\}=\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
প্রদত্ত রাশি,
\(f\left\{\sqrt{2}g\left(\frac{\pi}{2}-\theta\right)\right\}+f\left\{\sqrt{g(2\theta)}\right\}\)
\(=f\left\{\sqrt{2}\cos{\left(\frac{\pi}{2}-\theta\right)}\right\}+f\left\{\sqrt{\cos{(2\theta)}}\right\}\) ➜ \(\because g\left(\frac{\pi}{2}-\theta\right)=\cos{\left(\frac{\pi}{2}-\theta\right)}\)
এবং \(g(2\theta)=\cos{(2\theta)}\)
\(=f\left\{\sqrt{2}\sin{\theta}\right\}+f\left\{\sqrt{\cos{(2\theta)}}\right\}\)
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)➜ \(\because f\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}\)
এবং \(f\left\{\sqrt{\cos{(2\theta)}}\right\}=\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-(\sqrt{\cos{2\theta}})^2}+\sqrt{\cos{2\theta}}\sqrt{1-(\sqrt{2}\sin{\theta})^2}\right\}}\) ➜ \(\because \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{\left\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-\cos{2\theta}}+\sqrt{\cos{2\theta}}\sqrt{1-2\sin^2{\theta}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2\sin^2{\theta}}+\sqrt{\cos{2\theta}}\sqrt{\cos{2\theta}}\right\}}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1-2\sin^2{A}=\cos{2A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2}\sin{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{2\sin^2{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{1-\cos{2\theta}+\cos{2\theta}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\sin^{-1}{\left\{1\right\}}\)
\(=\sin^{-1}{\left\{\sin{\frac{\pi}{2}}\right\}}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
ইহাই নির্ণেয় মান।
\((c)\)
দেওয়া আছে,
\(g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}+x\right)}=1\) ➜ \(\because g(x)=\cos{x}\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}\times1+x\right)}=1\)
\(\Rightarrow \sqrt{3}\cos{x}-\sin{x}=1\) ➜
দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।
\(\Rightarrow \cos{x}\times\frac{\sqrt{3}}{2}-\sin{x}\times\frac{1}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+(-1)^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}-\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{6}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{6}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}-\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi-\pi}{6}, \ x=2n\pi-\frac{2\pi+\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{3\pi}{6}\)
\(\therefore x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{6}, \ -\frac{\pi}{2}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=\frac{13\pi}{6}, \ \frac{3\pi}{2}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{2}\)
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=-\frac{11\pi}{6}, \ -\frac{5\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{11\pi}{6}\)
যখন, \(n=2,\) \(x=4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{2}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
প্রদত্ত রাশি,
\(\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)}\)
\(=\tan^{-1}{\left(\frac{3+2}{6-1}\right)}\) ➜ লব ও হরকে \(6\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{5}{5}\right)}\)
\(=\tan^{-1}{\left(1\right)}\)
\(=\tan^{-1}{\left(\tan{\frac{\pi}{4}}\right)}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(=\frac{\pi}{4}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(f(x)=\sin^{-1}{x}\) এবং \(g(x)=\cos{x}\)
\(\Rightarrow g\left(\frac{\pi}{2}-\theta\right)=\cos{\left(\frac{\pi}{2}-\theta\right)}, \ g(2\theta)=\cos{(2\theta)}\)
\(\Rightarrow f\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}, \ f\left\{\sqrt{\cos{(2\theta)}}\right\}=\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
প্রদত্ত রাশি,
\(f\left\{\sqrt{2}g\left(\frac{\pi}{2}-\theta\right)\right\}+f\left\{\sqrt{g(2\theta)}\right\}\)
\(=f\left\{\sqrt{2}\cos{\left(\frac{\pi}{2}-\theta\right)}\right\}+f\left\{\sqrt{\cos{(2\theta)}}\right\}\) ➜ \(\because g\left(\frac{\pi}{2}-\theta\right)=\cos{\left(\frac{\pi}{2}-\theta\right)}\)
এবং \(g(2\theta)=\cos{(2\theta)}\)
\(=f\left\{\sqrt{2}\sin{\theta}\right\}+f\left\{\sqrt{\cos{(2\theta)}}\right\}\)
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)➜ \(\because f\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}\)
এবং \(f\left\{\sqrt{\cos{(2\theta)}}\right\}=\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-(\sqrt{\cos{2\theta}})^2}+\sqrt{\cos{2\theta}}\sqrt{1-(\sqrt{2}\sin{\theta})^2}\right\}}\) ➜ \(\because \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{\left\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-\cos{2\theta}}+\sqrt{\cos{2\theta}}\sqrt{1-2\sin^2{\theta}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2\sin^2{\theta}}+\sqrt{\cos{2\theta}}\sqrt{\cos{2\theta}}\right\}}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1-2\sin^2{A}=\cos{2A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2}\sin{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{2\sin^2{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{1-\cos{2\theta}+\cos{2\theta}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\sin^{-1}{\left\{1\right\}}\)
\(=\sin^{-1}{\left\{\sin{\frac{\pi}{2}}\right\}}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
ইহাই নির্ণেয় মান।
\((c)\)
দেওয়া আছে,
\(g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+g\left(\frac{\pi}{2}+x\right)=1\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}+x\right)}=1\) ➜ \(\because g(x)=\cos{x}\)
\(\Rightarrow \sqrt{3}\cos{x}+\cos{\left(\frac{\pi}{2}\times1+x\right)}=1\)
\(\Rightarrow \sqrt{3}\cos{x}-\sin{x}=1\) ➜

দ্বিতীয় পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি দ্বিতীয় চতুর্ভাগে অবস্থিত
সুতরাং কোসাইন অনুপাত ঋনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং কোসাইন অনুপাতের পরিবর্তন হয়ে সাইন হয়েছে।
\(\Rightarrow \cos{x}\times\frac{\sqrt{3}}{2}-\sin{x}\times\frac{1}{2}=\frac{1}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+(-1)^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}-\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(x+\frac{\pi}{6}\right)}=\cos{\frac{\pi}{3}}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow x+\frac{\pi}{6}=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}-\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi-\pi}{6}, \ x=2n\pi-\frac{2\pi+\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{3\pi}{6}\)
\(\therefore x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{6}, \ -\frac{\pi}{2}\) সবগুলি মান গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{6}, \ -\frac{\pi}{2}\)
যখন, \(n=1,\) \(x=2\pi+\frac{\pi}{6}, \ 2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=\frac{13\pi}{6}, \ \frac{3\pi}{2}\) দ্বিতীয় মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{2}\)
যখন, \(n=-1,\) \(x=-2\pi+\frac{\pi}{6}, \ -2\pi-\frac{\pi}{2}\)
\(\Rightarrow x=-\frac{11\pi}{6}, \ -\frac{5\pi}{2}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=-\frac{11\pi}{6}\)
যখন, \(n=2,\) \(x=4\pi+\frac{\pi}{6}, \ 4\pi-\frac{\pi}{2}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ -2\pi\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=-\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
\(Q.5.(iv)\) দৃশ্যকল্প-১: \(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
দৃশ্যকল্প-২: \(A=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\sqrt{\frac{2}{3}}}\)
\((a)\) \(\sec^2{(\cot^{-1}{3})}+cosec^2{(\tan^{-1}{2})}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে সমাধান করঃ \(f(\sin{\theta})=0\)
\((c)\) দৃশ্যকল্প-২: থেকে প্রমান কর যে, \(A=\tan^{-1}{\frac{5}{\sqrt{2}}}\)
উত্তরঃ \((a) \ 2\frac{13}{36}\)
\((b) \ n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
দৃশ্যকল্প-২: \(A=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\sqrt{\frac{2}{3}}}\)
\((a)\) \(\sec^2{(\cot^{-1}{3})}+cosec^2{(\tan^{-1}{2})}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-১: থেকে সমাধান করঃ \(f(\sin{\theta})=0\)
\((c)\) দৃশ্যকল্প-২: থেকে প্রমান কর যে, \(A=\tan^{-1}{\frac{5}{\sqrt{2}}}\)
উত্তরঃ \((a) \ 2\frac{13}{36}\)
\((b) \ n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ -\frac{\pi}{2}, \ -\frac{11\pi}{6}, \ \frac{\pi}{6}, \ \frac{3\pi}{2}\)
বঃ২০১৯।
সমাধানঃ
\((a)\)
প্রদত্ত রাশি,
\(\sec^2{\left(\cot^{-1}{3}\right)}+cosec^2{\left(\tan^{-1}{2}\right)}\)
\(=\sec^2{\left(\tan^{-1}{\frac{1}{3}}\right)}+cosec^2{\left(\cot^{-1}{\frac{1}{2}}\right)}\) ➜ \(\because \cot^{-1}{x}=\tan^{-1}{\frac{1}{x}}\)
এবং \(\tan^{-1}{x}=\cot^{-1}{\frac{1}{x}}\)
\(=1+\tan^2{\left(\tan^{-1}{\frac{1}{3}}\right)}+1+\cot^2{\left(\cot^{-1}{\frac{1}{2}}\right)}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
এবং \(cosec^2{A}=1+\cot^2{A}\)
\(=2+\tan^2{\left(\tan^{-1}{\frac{1}{3}}\right)}+\cot^2{\left(\cot^{-1}{\frac{1}{2}}\right)}\)
\(=2+\left\{\tan{\left(\tan^{-1}{\frac{1}{3}}\right)}\right\}^2+\left\{\cot{\left(\cot^{-1}{\frac{1}{2}}\right)}\right\}^2\)
\(=2+\left\{\frac{1}{3}\right\}^2+\left\{\frac{1}{2}\right\}^2\)
\(=2+\frac{1}{9}+\frac{1}{4}\)
\(=\frac{72+4+9}{36}\)
\(=\frac{85}{36}\)
\(=2\frac{13}{36}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
প্রদত্ত সমীকরণ,
\(f(\sin{\theta})=0\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\) ➜ \(\because f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
\(\Rightarrow f(\sin{\theta})=\sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}(\sin{\theta}-\sqrt{2})-1(\sin{\theta}-\sqrt{2})=0\)
\(\Rightarrow (\sin{\theta}-\sqrt{2})(\sqrt{2}\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}-\sqrt{2}\ne{0}, \ \sqrt{2}\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sqrt{2}\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
দেওয়া আছে,
\(A=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\sqrt{\frac{2}{3}}}\)
\(=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\frac{\sqrt{2}}{\sqrt{3}}}\)
\(=2\tan^{-1}{\frac{1}{2\sqrt{2}}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{লম্ব}=1 , \ \text{অতিভুজ}=3\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{3^2-1^2} \)
\(=\sqrt{9-1} \)
\(=\sqrt{8}\)
\(=2\sqrt{2}\)
\(\therefore \sin^{-1}{\frac{1}{3}}=\tan^{-1}{\frac{1}{2\sqrt{2}}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=\sqrt{2}, \ \text{অতিভুজ}=\sqrt{3}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{3})^2-(\sqrt{2})^2} \)
\(=\sqrt{3-2} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \cos^{-1}{x}=\tan^{-1}{\frac{1}{\sqrt{2}}}\)
\(=\tan^{-1}{\left\{\frac{2\times\frac{1}{2\sqrt{2}}}{1-\left(\frac{1}{2\sqrt{2}}\right)^2}\right\}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{1}{\sqrt{2}}}{1-\frac{1}{8}}\right\}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\)
\(=\tan^{-1}{\left\{\frac{4\sqrt{2}}{8-1}\right\}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\) ➜ প্রথম পদের
লব ও হরকে \(8\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{4\sqrt{2}}{7}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\)
\(=\tan^{-1}{\left(\frac{\frac{4\sqrt{2}}{7}+\frac{1}{\sqrt{2}}}{1-\frac{4\sqrt{2}}{7}\times\frac{1}{\sqrt{2}}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{4\sqrt{2}}{7}+\frac{1}{\sqrt{2}}}{1-\frac{4\sqrt{2}}{7\sqrt{2}}}\right)}\)
\(=\tan^{-1}{\left(\frac{8+7}{7\sqrt{2}-4\sqrt{2}}\right)}\) ➜ লব ও হরকে \(7\sqrt{2}\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{15}{3\sqrt{2}}\right)}\)
\(\therefore A=\tan^{-1}{\frac{5}{\sqrt{2}}}\)
(প্রমাণিত)
প্রদত্ত রাশি,
\(\sec^2{\left(\cot^{-1}{3}\right)}+cosec^2{\left(\tan^{-1}{2}\right)}\)
\(=\sec^2{\left(\tan^{-1}{\frac{1}{3}}\right)}+cosec^2{\left(\cot^{-1}{\frac{1}{2}}\right)}\) ➜ \(\because \cot^{-1}{x}=\tan^{-1}{\frac{1}{x}}\)
এবং \(\tan^{-1}{x}=\cot^{-1}{\frac{1}{x}}\)
\(=1+\tan^2{\left(\tan^{-1}{\frac{1}{3}}\right)}+1+\cot^2{\left(\cot^{-1}{\frac{1}{2}}\right)}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
এবং \(cosec^2{A}=1+\cot^2{A}\)
\(=2+\tan^2{\left(\tan^{-1}{\frac{1}{3}}\right)}+\cot^2{\left(\cot^{-1}{\frac{1}{2}}\right)}\)
\(=2+\left\{\tan{\left(\tan^{-1}{\frac{1}{3}}\right)}\right\}^2+\left\{\cot{\left(\cot^{-1}{\frac{1}{2}}\right)}\right\}^2\)
\(=2+\left\{\frac{1}{3}\right\}^2+\left\{\frac{1}{2}\right\}^2\)
\(=2+\frac{1}{9}+\frac{1}{4}\)
\(=\frac{72+4+9}{36}\)
\(=\frac{85}{36}\)
\(=2\frac{13}{36}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
প্রদত্ত সমীকরণ,
\(f(\sin{\theta})=0\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\) ➜ \(\because f(x)=\sqrt{2}x^2-3x+\sqrt{2}\)
\(\Rightarrow f(\sin{\theta})=\sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}\)
\(\Rightarrow \sqrt{2}\sin^2{\theta}-3\sin{\theta}+\sqrt{2}=0\)
\(\Rightarrow \sqrt{2}\sin{\theta}(\sin{\theta}-\sqrt{2})-1(\sin{\theta}-\sqrt{2})=0\)
\(\Rightarrow (\sin{\theta}-\sqrt{2})(\sqrt{2}\sin{\theta}-1)=0\)
\(\Rightarrow \sin{\theta}-\sqrt{2}\ne{0}, \ \sqrt{2}\sin{\theta}-1=0, \ \because -1\le{\sin{\theta}}\le{1}\)
\(\Rightarrow \sqrt{2}\sin{\theta}=1\)
\(\Rightarrow \sin{\theta}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{4}}\) ➜ \(\because \frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\)
\(\therefore \theta=n\pi+(-1)^n\frac{\pi}{4}\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{4}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
দেওয়া আছে,
\(A=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\sqrt{\frac{2}{3}}}\)
\(=2\sin^{-1}{\frac{1}{3}}+\cos^{-1}{\frac{\sqrt{2}}{\sqrt{3}}}\)
\(=2\tan^{-1}{\frac{1}{2\sqrt{2}}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{লম্ব}=1 , \ \text{অতিভুজ}=3\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{3^2-1^2} \)
\(=\sqrt{9-1} \)
\(=\sqrt{8}\)
\(=2\sqrt{2}\)
\(\therefore \sin^{-1}{\frac{1}{3}}=\tan^{-1}{\frac{1}{2\sqrt{2}}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=\sqrt{2}, \ \text{অতিভুজ}=\sqrt{3}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{3})^2-(\sqrt{2})^2} \)
\(=\sqrt{3-2} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \cos^{-1}{x}=\tan^{-1}{\frac{1}{\sqrt{2}}}\)
\(=\tan^{-1}{\left\{\frac{2\times\frac{1}{2\sqrt{2}}}{1-\left(\frac{1}{2\sqrt{2}}\right)^2}\right\}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{1}{\sqrt{2}}}{1-\frac{1}{8}}\right\}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\)
\(=\tan^{-1}{\left\{\frac{4\sqrt{2}}{8-1}\right\}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\) ➜ প্রথম পদের
লব ও হরকে \(8\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{4\sqrt{2}}{7}}+\tan^{-1}{\frac{1}{\sqrt{2}}}\)
\(=\tan^{-1}{\left(\frac{\frac{4\sqrt{2}}{7}+\frac{1}{\sqrt{2}}}{1-\frac{4\sqrt{2}}{7}\times\frac{1}{\sqrt{2}}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{4\sqrt{2}}{7}+\frac{1}{\sqrt{2}}}{1-\frac{4\sqrt{2}}{7\sqrt{2}}}\right)}\)
\(=\tan^{-1}{\left(\frac{8+7}{7\sqrt{2}-4\sqrt{2}}\right)}\) ➜ লব ও হরকে \(7\sqrt{2}\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{15}{3\sqrt{2}}\right)}\)
\(\therefore A=\tan^{-1}{\frac{5}{\sqrt{2}}}\)
(প্রমাণিত)
\(Q.5.(v)\) দৃশ্যকল্প-১: \(f(a)=\sec^{-1}{\frac{1}{a}}+\sec^{-1}{\frac{1}{b}}\)
দৃশ্যকল্প-২: \(g(\alpha)=\sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}\)
\((a)\) \(\cot{\left(\sin^{-1}{\frac{1}{\sqrt{5}}}\right)}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-২: থেকে যদি \(g(\alpha)=0\) হয় তবে দেখাও যে, \(\alpha=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) দৃশ্যকল্প-১: থেকে \(f(a)=\alpha\) হলে, প্রমান কর যে, \(\sin{\alpha}=\sqrt{a^2+b^2-2ab\cos{\alpha}}\)
উত্তরঃ \((a) \ 2\)
দৃশ্যকল্প-২: \(g(\alpha)=\sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}\)
\((a)\) \(\cot{\left(\sin^{-1}{\frac{1}{\sqrt{5}}}\right)}\) এর মান নির্ণয় কর।
\((b)\) দৃশ্যকল্প-২: থেকে যদি \(g(\alpha)=0\) হয় তবে দেখাও যে, \(\alpha=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
\((c)\) দৃশ্যকল্প-১: থেকে \(f(a)=\alpha\) হলে, প্রমান কর যে, \(\sin{\alpha}=\sqrt{a^2+b^2-2ab\cos{\alpha}}\)
উত্তরঃ \((a) \ 2\)
চঃ২০১৯।
সমাধানঃ
\((a)\)
প্রদত্ত রাশি,
\(\cot{\left(\sin^{-1}{\frac{1}{\sqrt{5}}}\right)}\)
\(=\cot{\left(\cot^{-1}{2}\right)}\) ➜ এখানে,
\(\text{লম্ব}=1 , \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{2}{1}}\)
\(=\cot^{-1}{2}\)
\(=2\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(g(\alpha)=\sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}\)
এবং \(g(\alpha)=0\) \(\Rightarrow \sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}=0\)
\(\Rightarrow \sin{(\pi\cos{\alpha})}=\sin{\left(\frac{\pi}{2}\pm\pi\sin{\alpha}\right)}\) ➜ \(\because \cos{(\pi\sin{\alpha})}=\sin{\left(\frac{\pi}{2}\pm\pi\sin{\alpha}\right)}\)
\(\Rightarrow \pi\cos{\alpha}=\frac{\pi}{2}\pm\pi\sin{\alpha}\)
\(\Rightarrow \pi\cos{\alpha}=\pi\left(\frac{1}{2}\pm\sin{\alpha}\right)\)
\(\Rightarrow \cos{\alpha}=\frac{1}{2}\pm\sin{\alpha}\)
\(\Rightarrow \cos{\alpha}\pm\sin{\alpha}=\frac{1}{2}\)
\(\Rightarrow (\cos{\alpha}\pm\sin{\alpha})^2=\frac{1}{4}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \cos^2{\alpha}\pm2\sin{\alpha}\cos{\alpha}+\sin^2{\alpha}=\frac{1}{4}\)
\(\Rightarrow \sin^2{\alpha}+\cos^2{\alpha}\pm2\sin{\alpha}\cos{\alpha}=\frac{1}{4}\)
\(\Rightarrow 1\pm\sin{2\alpha}=\frac{1}{4}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \pm\sin{2\alpha}=\frac{1}{4}-1\)
\(\Rightarrow \pm\sin{2\alpha}=\frac{1-4}{4}\)
\(\Rightarrow \pm\sin{2\alpha}=\frac{-3}{4}\)
\(\Rightarrow \pm\sin{2\alpha}=-\frac{3}{4}\)
\(\Rightarrow \sin{2\alpha}=\pm\frac{3}{4}\)
\(\Rightarrow 2\alpha=\pm\sin^{-1}{\frac{3}{4}}\)
\(\therefore \alpha=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(f(a)=\sec^{-1}{\frac{1}{a}}+\sec^{-1}{\frac{1}{b}}\)
এবং \(f(a)=\alpha\)
\(\Rightarrow \sec^{-1}{\frac{1}{a}}+\sec^{-1}{\frac{1}{b}}=\alpha\)
\(\Rightarrow \cos^{-1}{a}+\cos^{-1}{b}=\alpha\) ➜ \(\because \sec^{-1}{\frac{1}{x}}=\cos^{-1}{x}\)
\(\Rightarrow \cos^{-1}{\{ab-\sqrt{(1-a^2)(1-b^2)}\}}=\alpha\) ➜ \(\because \cos^{-1}{x}+\cos^{-1}{y}=\cos^{-1}{\{xy-\sqrt{(1-x^2)(1-y^2)}\}}\)
\(\Rightarrow ab-\sqrt{(1-a^2)(1-b^2)}=\cos{\alpha}\)
\(\Rightarrow ab-\cos{\alpha}=\sqrt{(1-a^2)(1-b^2)}\)
\(\Rightarrow (ab-\cos{\alpha})^2=(1-a^2)(1-b^2)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow a^2b^2-2ab\cos{\alpha}+\cos^2{\alpha}=1-a^2-b^2+a^2b^2\) ➜ \(\because (x-y)^2=x^2-2xy+y^2\)
\(\Rightarrow a^2b^2-2ab\cos{\alpha}+a^2+b^2-a^2b^2=1-\cos^2{\alpha}\)
\(\Rightarrow a^2+b^2-2ab\cos{\alpha}=\sin^2{\alpha}\)
\(\Rightarrow \sin^2{\alpha}=a^2+b^2-2ab\cos{\alpha}\)
\(\therefore \sin{\alpha}=\sqrt{a^2+b^2-2ab\cos{\alpha}}\)
(প্রমাণিত)
প্রদত্ত রাশি,
\(\cot{\left(\sin^{-1}{\frac{1}{\sqrt{5}}}\right)}\)
\(=\cot{\left(\cot^{-1}{2}\right)}\) ➜ এখানে,
\(\text{লম্ব}=1 , \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{2}{1}}\)
\(=\cot^{-1}{2}\)
\(=2\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(g(\alpha)=\sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}\)
এবং \(g(\alpha)=0\) \(\Rightarrow \sin{(\pi\cos{\alpha})}-\cos{(\pi\sin{\alpha})}=0\)
\(\Rightarrow \sin{(\pi\cos{\alpha})}=\sin{\left(\frac{\pi}{2}\pm\pi\sin{\alpha}\right)}\) ➜ \(\because \cos{(\pi\sin{\alpha})}=\sin{\left(\frac{\pi}{2}\pm\pi\sin{\alpha}\right)}\)
\(\Rightarrow \pi\cos{\alpha}=\frac{\pi}{2}\pm\pi\sin{\alpha}\)
\(\Rightarrow \pi\cos{\alpha}=\pi\left(\frac{1}{2}\pm\sin{\alpha}\right)\)
\(\Rightarrow \cos{\alpha}=\frac{1}{2}\pm\sin{\alpha}\)
\(\Rightarrow \cos{\alpha}\pm\sin{\alpha}=\frac{1}{2}\)
\(\Rightarrow (\cos{\alpha}\pm\sin{\alpha})^2=\frac{1}{4}\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \cos^2{\alpha}\pm2\sin{\alpha}\cos{\alpha}+\sin^2{\alpha}=\frac{1}{4}\)
\(\Rightarrow \sin^2{\alpha}+\cos^2{\alpha}\pm2\sin{\alpha}\cos{\alpha}=\frac{1}{4}\)
\(\Rightarrow 1\pm\sin{2\alpha}=\frac{1}{4}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \pm\sin{2\alpha}=\frac{1}{4}-1\)
\(\Rightarrow \pm\sin{2\alpha}=\frac{1-4}{4}\)
\(\Rightarrow \pm\sin{2\alpha}=\frac{-3}{4}\)
\(\Rightarrow \pm\sin{2\alpha}=-\frac{3}{4}\)
\(\Rightarrow \sin{2\alpha}=\pm\frac{3}{4}\)
\(\Rightarrow 2\alpha=\pm\sin^{-1}{\frac{3}{4}}\)
\(\therefore \alpha=\pm\frac{1}{2}\sin^{-1}{\frac{3}{4}}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(f(a)=\sec^{-1}{\frac{1}{a}}+\sec^{-1}{\frac{1}{b}}\)
এবং \(f(a)=\alpha\)
\(\Rightarrow \sec^{-1}{\frac{1}{a}}+\sec^{-1}{\frac{1}{b}}=\alpha\)
\(\Rightarrow \cos^{-1}{a}+\cos^{-1}{b}=\alpha\) ➜ \(\because \sec^{-1}{\frac{1}{x}}=\cos^{-1}{x}\)
\(\Rightarrow \cos^{-1}{\{ab-\sqrt{(1-a^2)(1-b^2)}\}}=\alpha\) ➜ \(\because \cos^{-1}{x}+\cos^{-1}{y}=\cos^{-1}{\{xy-\sqrt{(1-x^2)(1-y^2)}\}}\)
\(\Rightarrow ab-\sqrt{(1-a^2)(1-b^2)}=\cos{\alpha}\)
\(\Rightarrow ab-\cos{\alpha}=\sqrt{(1-a^2)(1-b^2)}\)
\(\Rightarrow (ab-\cos{\alpha})^2=(1-a^2)(1-b^2)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow a^2b^2-2ab\cos{\alpha}+\cos^2{\alpha}=1-a^2-b^2+a^2b^2\) ➜ \(\because (x-y)^2=x^2-2xy+y^2\)
\(\Rightarrow a^2b^2-2ab\cos{\alpha}+a^2+b^2-a^2b^2=1-\cos^2{\alpha}\)
\(\Rightarrow a^2+b^2-2ab\cos{\alpha}=\sin^2{\alpha}\)
\(\Rightarrow \sin^2{\alpha}=a^2+b^2-2ab\cos{\alpha}\)
\(\therefore \sin{\alpha}=\sqrt{a^2+b^2-2ab\cos{\alpha}}\)
(প্রমাণিত)
\(Q.5.(vi)\) \(f(x)=\tan{x}\)
\((a)\) দেখাও যে, \(\tan^{-1}{\frac{5}{3}}=\frac{\pi}{2}-\cos^{-1}{\frac{5}{\sqrt{34}}}\)
\((b)\) প্রমান কর যে, \(\tan^{-1}{\left\{(2+\sqrt{3})f(x)\right\}}+\tan^{-1}{\left\{(2-\sqrt{3})f(x)\right\}}=\tan^{-1}{\{2f(2x)\}}\)
\((c)\) সমাধান করঃ \(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
উত্তরঃ \((c) \ n\pi-\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\((a)\) দেখাও যে, \(\tan^{-1}{\frac{5}{3}}=\frac{\pi}{2}-\cos^{-1}{\frac{5}{\sqrt{34}}}\)
\((b)\) প্রমান কর যে, \(\tan^{-1}{\left\{(2+\sqrt{3})f(x)\right\}}+\tan^{-1}{\left\{(2-\sqrt{3})f(x)\right\}}=\tan^{-1}{\{2f(2x)\}}\)
\((c)\) সমাধান করঃ \(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
উত্তরঃ \((c) \ n\pi-\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
সিঃ২০১৯।
সমাধানঃ
\((a)\)
\(R.S=\frac{\pi}{2}-\cos^{-1}{\frac{5}{\sqrt{34}}}\)
\(=\frac{\pi}{2}-\cot^{-1}{\frac{5}{3}}\) ➜ দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=5, \ \text{অতিভুজ}=\sqrt{34}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{34})^2-5^2} \)
\(=\sqrt{34-25} \)
\(=\sqrt{9}\)
\(=3\)
\(\therefore \cos^{-1}{\frac{5}{\sqrt{34}}}=\cot^{-1}{\frac{5}{3}}\)
\(=\tan^{-1}{\frac{5}{3}}\) ➜ \(\because \frac{\pi}{2}=\tan^{-1}{x}+\cot^{-1}{x}\)
\(\therefore \frac{\pi}{2}-\cot^{-1}{x}=\tan^{-1}{x}\)
\(=L.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
\(L.S=\tan^{-1}{\left\{(2+\sqrt{3})f(x)\right\}}+\tan^{-1}{\left\{(2-\sqrt{3})f(x)\right\}}\)
\(=\tan^{-1}{\left\{\frac{(2+\sqrt{3})\tan{x}+(2-\sqrt{3})\tan{x}}{1-(2+\sqrt{3})\tan{x}(2-\sqrt{3})\tan{x}}\right\}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\frac{x+y}{1-xy}}\)
\(=\tan^{-1}{\left[\frac{2\tan{x}+\sqrt{3}\tan{x}+2\tan{x}-\sqrt{3}\tan{x}}{1-\{2^2-(\sqrt{3})^2\}\tan^2{x}}\right]}\)
\(=\tan^{-1}{\left[\frac{4\tan{x}}{1-\{4-3\}\tan^2{x}}\right]}\)
\(=\tan^{-1}{\left(2\frac{2\tan{x}}{1-\tan^2{x}}\right)}\)
\(=\tan^{-1}{\left(2\tan{2x}\right)}\) ➜ \(\because \frac{2\tan{A}}{1-\tan^2{A}}=\tan{2A}\)
\(=\tan^{-1}{\{2f(2x)\}}\) ➜ \(\because f(x)=\tan{x}\)
\(\therefore f(2x)=\tan{2x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
প্রদত্ত সমীকরণ,
\(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}-2x\right)}=\cos{x}+\sin{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}\times1-2x\right)}=\cos{x}+\sin{x}\)
\(\Rightarrow \cot{2x}=\cos{x}+\sin{x}\) ➜
প্রথম পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং ট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং ট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে কোট্যানজেন্ট হয়েছে।
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\cos{x}+\sin{x}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos^2{2x}}{\sin^2{2x}}=(\cos{x}+\sin{x})^2\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=\cos^2{x}+\sin^2{x}+2\sin{x}\cos{x}\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
এবং \((a+b)^2=a^2+b^2+2ab\)
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=1+\sin{2x}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})=1-\sin^2{2x}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1-\sin^2{2x})=0\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1+\sin{2x})(1-\sin{2x})=0\) ➜ \(\because a^2-b^2=(a+b)(a-b)\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}-1+\sin{2x})=0\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}+\sin{2x}-1)=0\)
\(\Rightarrow 1+\sin{2x}=0, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=-1, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow 2x=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n-1)\frac{\pi}{4}\)
আবার,
\(\sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1+\sqrt{5}}{2}, \ \sin{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\sin{2x}}\le{1}\)
\(\Rightarrow \sin{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi-\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(R.S=\frac{\pi}{2}-\cos^{-1}{\frac{5}{\sqrt{34}}}\)
\(=\frac{\pi}{2}-\cot^{-1}{\frac{5}{3}}\) ➜ দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=5, \ \text{অতিভুজ}=\sqrt{34}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{34})^2-5^2} \)
\(=\sqrt{34-25} \)
\(=\sqrt{9}\)
\(=3\)
\(\therefore \cos^{-1}{\frac{5}{\sqrt{34}}}=\cot^{-1}{\frac{5}{3}}\)
\(=\tan^{-1}{\frac{5}{3}}\) ➜ \(\because \frac{\pi}{2}=\tan^{-1}{x}+\cot^{-1}{x}\)
\(\therefore \frac{\pi}{2}-\cot^{-1}{x}=\tan^{-1}{x}\)
\(=L.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
\(L.S=\tan^{-1}{\left\{(2+\sqrt{3})f(x)\right\}}+\tan^{-1}{\left\{(2-\sqrt{3})f(x)\right\}}\)
\(=\tan^{-1}{\left\{\frac{(2+\sqrt{3})\tan{x}+(2-\sqrt{3})\tan{x}}{1-(2+\sqrt{3})\tan{x}(2-\sqrt{3})\tan{x}}\right\}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\frac{x+y}{1-xy}}\)
\(=\tan^{-1}{\left[\frac{2\tan{x}+\sqrt{3}\tan{x}+2\tan{x}-\sqrt{3}\tan{x}}{1-\{2^2-(\sqrt{3})^2\}\tan^2{x}}\right]}\)
\(=\tan^{-1}{\left[\frac{4\tan{x}}{1-\{4-3\}\tan^2{x}}\right]}\)
\(=\tan^{-1}{\left(2\frac{2\tan{x}}{1-\tan^2{x}}\right)}\)
\(=\tan^{-1}{\left(2\tan{2x}\right)}\) ➜ \(\because \frac{2\tan{A}}{1-\tan^2{A}}=\tan{2A}\)
\(=\tan^{-1}{\{2f(2x)\}}\) ➜ \(\because f(x)=\tan{x}\)
\(\therefore f(2x)=\tan{2x}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
প্রদত্ত সমীকরণ,
\(f\left(\frac{\pi}{2}-2x\right)=\cos{x}+\sin{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}-2x\right)}=\cos{x}+\sin{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan{\left(\frac{\pi}{2}\times1-2x\right)}=\cos{x}+\sin{x}\)
\(\Rightarrow \cot{2x}=\cos{x}+\sin{x}\) ➜

প্রথম পদে,
\(\because\) কোণ উৎপন্নকারী রেখাটি প্রথম চতুর্ভাগে অবস্থিত
সুতরাং ট্যানজেন্ট অনুপাত ধনাত্মক।
আবার, \(\frac{\pi}{2}\) এর সহগুণক \(1\) একটি বিজোড় সংখ্যা,
সুতরাং ট্যানজেন্ট অনুপাতের পরিবর্তন হয়ে কোট্যানজেন্ট হয়েছে।
\(\Rightarrow \frac{\cos{2x}}{\sin{2x}}=\cos{x}+\sin{x}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\Rightarrow \frac{\cos^2{2x}}{\sin^2{2x}}=(\cos{x}+\sin{x})^2\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=\cos^2{x}+\sin^2{x}+2\sin{x}\cos{x}\) ➜ \(\because \cos^2{A}=1-\sin^2{A}\)
এবং \((a+b)^2=a^2+b^2+2ab\)
\(\Rightarrow \frac{1-\sin^2{2x}}{\sin^2{2x}}=1+\sin{2x}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
এবং \(2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})=1-\sin^2{2x}\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1-\sin^2{2x})=0\)
\(\Rightarrow \sin^2{2x}(1+\sin{2x})-(1+\sin{2x})(1-\sin{2x})=0\) ➜ \(\because a^2-b^2=(a+b)(a-b)\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}-1+\sin{2x})=0\)
\(\Rightarrow (1+\sin{2x})(\sin^2{2x}+\sin{2x}-1)=0\)
\(\Rightarrow 1+\sin{2x}=0, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=-1, \ \sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow 2x=(4n-1)\frac{\pi}{2}\) ➜ \(\because \sin{A}=-1\)
\(\Rightarrow A=(4n-1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(4n-1)\frac{\pi}{4}\)
আবার,
\(\sin^2{2x}+\sin{2x}-1=0\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \sin{2x}=\frac{-1+\sqrt{5}}{2}, \ \sin{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\sin{2x}}\le{1}\)
\(\Rightarrow \sin{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \sin{2x}=\sin{\alpha}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=n\pi+(-1)^n\alpha\) ➜ \(\because \sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi-\frac{\pi}{4}, \ \frac{n\pi}{2}+(-1)^n\frac{\alpha}{2}\) যেখানে, \(\sin{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(Q.5.(vii)\) দৃশ্যকল্প-১: \(\sin^{-1}{\left(\frac{4}{5}\right)}+\cos^{-1}{\left(\frac{2}{\sqrt{5}}\right)}-\cot^{-1}{\left(\frac{2}{11}\right)}\)
দৃশ্যকল্প-২: \(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\lt{\theta}\lt{2\pi}\)
\((a)\) প্রমান কর যে, \(2\sin^{-1}{x}=\sin^{-1}{\left(2x\sqrt{1-x^2}\right)}\)
\((b)\) দৃশ্যকল্প-১ এর মান নির্ণয় কর।
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((b) 0\)
\((c) \ \pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
দৃশ্যকল্প-২: \(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\lt{\theta}\lt{2\pi}\)
\((a)\) প্রমান কর যে, \(2\sin^{-1}{x}=\sin^{-1}{\left(2x\sqrt{1-x^2}\right)}\)
\((b)\) দৃশ্যকল্প-১ এর মান নির্ণয় কর।
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((b) 0\)
\((c) \ \pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
ঢাঃ,যঃ,সিঃ,দিঃ২০১৮।
সমাধানঃ
\((a)\)
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\sin{2\theta}=2\sin{\theta}\cos{\theta}\)
\(\Rightarrow \sin{2\theta}=2\sin{\theta}\sqrt{1-\sin^2{\theta}}\)
\(\Rightarrow 2\theta=\sin^{-1}{(2\sin{\theta}\sqrt{1-\sin^2{\theta}})}\)
\(\therefore 2\sin^{-1}{x}=\sin^{-1}{(2x\sqrt{1-x^2})}\) ➜ \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)
(প্রমাণিত)
\((b)\)
প্রদত্ত রাশি,
\(\sin^{-1}{\left(\frac{4}{5}\right)}+\cos^{-1}{\left(\frac{2}{\sqrt{5}}\right)}-\cot^{-1}{\left(\frac{2}{11}\right)}\)
\(=\tan^{-1}{\left(\frac{4}{3}\right)}+\tan^{-1}{\left(\frac{1}{2}\right)}-\cot^{-1}{\left(\frac{2}{11}\right)}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{লম্ব}=4, \ \text{অতিভুজ}=5\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{5^2-4^2} \)
\(=\sqrt{25-16} \)
\(=\sqrt{9}\)
\(=3\)
\(\therefore \sin^{-1}{\frac{4}{5}}=\tan^{-1}{\frac{4}{3}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=2 , \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-2^2} \)
\(=\sqrt{5-4} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \cos^{-1}{\frac{2}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3}\times\frac{1}{2}}\right)}-\cot^{-1}{\frac{2}{11}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{6}}\right)}-\cot^{-1}{\frac{2}{11}}\)
\(=\tan^{-1}{\left(\frac{8+3}{6-4}\right)}\) ➜ লব ও হরকে \(6\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{11}{2}\right)}-\cot^{-1}{\frac{2}{11}}\)
\(=\tan^{-1}{\frac{11}{2}}-\tan^{-1}{\frac{11}{2}}\) ➜ \(\because \cot^{-1}{x}=\tan{\frac{1}{x}}\)
\(=0\)
ইহাই নির্ণেয় মান।
\((c)\)
প্রদত্ত সমীকরণ,
\(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow 4(1-\cos^2{\theta}+\cos{\theta})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-4\cos^2{\theta}+4\cos{\theta}-5=0\)
\(\Rightarrow -4\cos^2{\theta}+4\cos{\theta}-1=0\)
\(\Rightarrow 4\cos^2{\theta}-4\cos{\theta}+1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow (2\cos{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\cos{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{3}, \ -\frac{\pi}{3}\) উভয়ে গ্রহনযোগ্য।
যখন, \(n=1,\) \(\theta=2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=-\frac{5\pi}{3}\)
যখন, \(n=2,\) \(\theta=4\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{13\pi}{3}, \ \frac{11\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\sin{2\theta}=2\sin{\theta}\cos{\theta}\)
\(\Rightarrow \sin{2\theta}=2\sin{\theta}\sqrt{1-\sin^2{\theta}}\)
\(\Rightarrow 2\theta=\sin^{-1}{(2\sin{\theta}\sqrt{1-\sin^2{\theta}})}\)
\(\therefore 2\sin^{-1}{x}=\sin^{-1}{(2x\sqrt{1-x^2})}\) ➜ \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)
(প্রমাণিত)
\((b)\)
প্রদত্ত রাশি,
\(\sin^{-1}{\left(\frac{4}{5}\right)}+\cos^{-1}{\left(\frac{2}{\sqrt{5}}\right)}-\cot^{-1}{\left(\frac{2}{11}\right)}\)
\(=\tan^{-1}{\left(\frac{4}{3}\right)}+\tan^{-1}{\left(\frac{1}{2}\right)}-\cot^{-1}{\left(\frac{2}{11}\right)}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{লম্ব}=4, \ \text{অতিভুজ}=5\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{5^2-4^2} \)
\(=\sqrt{25-16} \)
\(=\sqrt{9}\)
\(=3\)
\(\therefore \sin^{-1}{\frac{4}{5}}=\tan^{-1}{\frac{4}{3}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=2 , \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-2^2} \)
\(=\sqrt{5-4} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \cos^{-1}{\frac{2}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3}\times\frac{1}{2}}\right)}-\cot^{-1}{\frac{2}{11}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{6}}\right)}-\cot^{-1}{\frac{2}{11}}\)
\(=\tan^{-1}{\left(\frac{8+3}{6-4}\right)}\) ➜ লব ও হরকে \(6\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{11}{2}\right)}-\cot^{-1}{\frac{2}{11}}\)
\(=\tan^{-1}{\frac{11}{2}}-\tan^{-1}{\frac{11}{2}}\) ➜ \(\because \cot^{-1}{x}=\tan{\frac{1}{x}}\)
\(=0\)
ইহাই নির্ণেয় মান।
\((c)\)
প্রদত্ত সমীকরণ,
\(4(\sin^2{\theta}+\cos{\theta})=5, \ -2\pi\le{\theta}\le{2\pi}\)
\(\Rightarrow 4(1-\cos^2{\theta}+\cos{\theta})=5\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow 4-4\cos^2{\theta}+4\cos{\theta}-5=0\)
\(\Rightarrow -4\cos^2{\theta}+4\cos{\theta}-1=0\)
\(\Rightarrow 4\cos^2{\theta}-4\cos{\theta}+1=0\) ➜ উভয় পার্শে \(-1\) গুন করে,
\(\Rightarrow (2\cos{\theta}-1)^2=0\) ➜ \(\because a^2-2ab+b^2=(a-b)^2\)
\(\Rightarrow 2\cos{\theta}-1=0\)
\(\Rightarrow 2\cos{\theta}=1\)
\(\Rightarrow \cos{\theta}=\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{\pi}{3}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{3}\) ➜ \(\because \cos{x}=\cos{\alpha}\)
\(\Rightarrow x=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\pm\frac{\pi}{3}\)\(\therefore \theta=\frac{\pi}{3}, \ -\frac{\pi}{3}\) উভয়ে গ্রহনযোগ্য।
যখন, \(n=1,\) \(\theta=2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{7\pi}{3}, \ \frac{5\pi}{3}\) দ্বিতীয় মান গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{3}\)
যখন, \(n=-1,\) \(\theta=-2\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=-2\pi+\frac{\pi}{3}, \ -2\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=-\frac{5\pi}{3}, \ -\frac{7\pi}{3}\) প্রথম মান গ্রহনযোগ্য।
\(\therefore \theta=-\frac{5\pi}{3}\)
যখন, \(n=2,\) \(\theta=4\pi\pm\frac{\pi}{3}\)
\(\Rightarrow \theta=4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{13\pi}{3}, \ \frac{11\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\pm\frac{\pi}{3}, \ \pm\frac{5\pi}{3}\)
\(Q.5.(viii)\) দৃশ্যকল্প-১: \(\sec^{-1}{\left(\frac{5}{3}\right)}+\cot^{-1}{\left(\frac{12}{5}\right)}+\sin^{-1}{\left(\frac{16}{65}\right)}\)
দৃশ্যকল্প-২: \(\sqrt{3}\sin{\theta}=2+\cos{\theta}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\sin^{-1}{\frac{2x}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\)
\((c)\) দৃশ্যকল্প-২ এর সমাধান কর যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
দৃশ্যকল্প-২: \(\sqrt{3}\sin{\theta}=2+\cos{\theta}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\sin^{-1}{\frac{2x}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\)
\((c)\) দৃশ্যকল্প-২ এর সমাধান কর যখন, \(-2\pi\lt{\theta}\lt{2\pi}\)
উত্তরঃ \((c) \ -\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
ঢাঃ২০১৭।
সমাধানঃ
\((a)\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
এখন,
\(\sin{2\theta}=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\sin^{-1}{\left(\frac{2\tan{\theta}}{1+\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\sin^{-1}{\left(\frac{2x}{1+x^2}\right)}\) ➜ \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)
(দেখানো হলো)
\((b)\)
প্রদত্ত রাশি,
\(\sec^{-1}{\frac{5}{3}}+\cot^{-1}{\frac{12}{5}}+\sin^{-1}{\frac{16}{65}}\)
\(=\tan^{-1}{\frac{4}{3}}+\tan^{-1}{\frac{5}{12}}+\tan^{-1}{\frac{16}{63}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{ভূমি}=3 , \ \text{অতিভুজ}=5 \)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{5^2-3^2} \)
\(=\sqrt{25-9} \)
\(=\sqrt{16} \)
\(=4 \)
\(\therefore \sec^{-1}{\frac{5}{3}}=\tan^{-1}{\frac{4}{3}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\cot^{-1}{x}=\tan^{-1}{\frac{1}{x}}\)
এবং তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=16 , \ \text{অতিভুজ}=65\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(65)^2-(16)^2} \)
\(=\sqrt{4225-256} \)
\(=\sqrt{3969}\)
\(=63\)
\(\therefore cosec^{-1}{\frac{\sqrt{5}}{2}}=\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}}\right\}}+\tan^{-1}{\frac{16}{63}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{20}{36}}\right\}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left(\frac{48+15}{36-20}\right)}+\tan^{-1}{\frac{16}{63}}\) ➜ লব ও হরকে \(36\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{63}{16}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-\frac{63}{16}\times\frac{16}{63}}\right\}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-1}\right\}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{0}\right\}}\)
\(=\tan^{-1}{(\infty)}\)
\(=\tan^{-1}{(\tan{\frac{\pi}{2}})}\)➜ \(\because \infty=\tan{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
(দেখানো হলো)
\((c)\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}\sin{\theta}=2+\cos{\theta}, \ -\pi\lt{\theta}\lt{\pi}\)
\(\Rightarrow \sqrt{3}\sin{\theta}-\cos{\theta}=2\)
\(\Rightarrow \cos{\theta}-\sqrt{3}\sin{\theta}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{\theta}\frac{1}{2}-\sin{\theta}\frac{\sqrt{3}}{2}=-1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}-\sin{\theta}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \theta+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\{3(2n+1)-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=\{6n+3-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=(6n+2)\frac{\pi}{3}\)
\(\therefore \theta=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(\theta=4\frac{2\pi}{3}\)
\(\Rightarrow \theta=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\frac{2\pi}{3}\)
\(\Rightarrow \theta=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{4\pi}{3}\)
\(\therefore -\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
এখন,
\(\sin{2\theta}=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\sin^{-1}{\left(\frac{2\tan{\theta}}{1+\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\sin^{-1}{\left(\frac{2x}{1+x^2}\right)}\) ➜ \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)
(দেখানো হলো)
\((b)\)
প্রদত্ত রাশি,
\(\sec^{-1}{\frac{5}{3}}+\cot^{-1}{\frac{12}{5}}+\sin^{-1}{\frac{16}{65}}\)
\(=\tan^{-1}{\frac{4}{3}}+\tan^{-1}{\frac{5}{12}}+\tan^{-1}{\frac{16}{63}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{ভূমি}=3 , \ \text{অতিভুজ}=5 \)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{5^2-3^2} \)
\(=\sqrt{25-9} \)
\(=\sqrt{16} \)
\(=4 \)
\(\therefore \sec^{-1}{\frac{5}{3}}=\tan^{-1}{\frac{4}{3}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\cot^{-1}{x}=\tan^{-1}{\frac{1}{x}}\)
এবং তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=16 , \ \text{অতিভুজ}=65\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(65)^2-(16)^2} \)
\(=\sqrt{4225-256} \)
\(=\sqrt{3969}\)
\(=63\)
\(\therefore cosec^{-1}{\frac{\sqrt{5}}{2}}=\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}}\right\}}+\tan^{-1}{\frac{16}{63}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{20}{36}}\right\}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left(\frac{48+15}{36-20}\right)}+\tan^{-1}{\frac{16}{63}}\) ➜ লব ও হরকে \(36\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{63}{16}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-\frac{63}{16}\times\frac{16}{63}}\right\}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-1}\right\}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{0}\right\}}\)
\(=\tan^{-1}{(\infty)}\)
\(=\tan^{-1}{(\tan{\frac{\pi}{2}})}\)➜ \(\because \infty=\tan{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
(দেখানো হলো)
\((c)\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}\sin{\theta}=2+\cos{\theta}, \ -\pi\lt{\theta}\lt{\pi}\)
\(\Rightarrow \sqrt{3}\sin{\theta}-\cos{\theta}=2\)
\(\Rightarrow \cos{\theta}-\sqrt{3}\sin{\theta}=-2\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow \cos{\theta}\frac{1}{2}-\sin{\theta}\frac{\sqrt{3}}{2}=-1\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(-\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}-\sin{\theta}\sin{\frac{\pi}{3}}=-1\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}\)
এবং \(\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(\theta+\frac{\pi}{3}\right)}=-1\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\Rightarrow \theta+\frac{\pi}{3}=(2n+1)\pi\) ➜ \(\because \cos{A}=-1\)
\(\Rightarrow A=(2n+1)\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=(2n+1)\pi-\frac{\pi}{3}\)
\(\Rightarrow \theta=\{3(2n+1)-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=\{6n+3-1\}\frac{\pi}{3}\)
\(\Rightarrow \theta=(6n+2)\frac{\pi}{3}\)
\(\therefore \theta=(3n+1)\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(3n+1)\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{2\pi}{3}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{2\pi}{3}\)
যখন, \(n=1,\) \(\theta=4\frac{2\pi}{3}\)
\(\Rightarrow \theta=\frac{8\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-1,\) \(\theta=-2\frac{2\pi}{3}\)
\(\Rightarrow \theta=-\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{4\pi}{3}\)
\(\therefore -\pi\lt{\theta}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=-\frac{4\pi}{3}, \ \frac{2\pi}{3}\)
\(Q.5.(ix)\) \(f(x)=\tan{x}\)
\((a)\) \(\cot^{-1}{\cos{\left(cosec^{-1}{\sqrt{\frac{3}{2}}}\right)}}\) এর মুখ্যমান নির্ণয় কর।
\((b)\) উদ্দীপকে উল্লেখিত \(f(x)\) এর জন্য \(f^{-1}(x)+f^{-1}(y)=\pi\) হলে প্রমাণ কর যে, প্রাপ্ত সঞ্চারপথটি একটি সরলরেখা নির্দেশ করে যার ঢাল \(-1\) হবে।
\((c)\) \(\left\{f(x)\right\}^2+f^{\prime}(x)=3f(x)\) হলে, বিশেষ সমাধান নির্ণয় কর যখন \(0\le{x}\le{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{3}\)
\((c) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\((a)\) \(\cot^{-1}{\cos{\left(cosec^{-1}{\sqrt{\frac{3}{2}}}\right)}}\) এর মুখ্যমান নির্ণয় কর।
\((b)\) উদ্দীপকে উল্লেখিত \(f(x)\) এর জন্য \(f^{-1}(x)+f^{-1}(y)=\pi\) হলে প্রমাণ কর যে, প্রাপ্ত সঞ্চারপথটি একটি সরলরেখা নির্দেশ করে যার ঢাল \(-1\) হবে।
\((c)\) \(\left\{f(x)\right\}^2+f^{\prime}(x)=3f(x)\) হলে, বিশেষ সমাধান নির্ণয় কর যখন \(0\le{x}\le{2\pi}\)
উত্তরঃ \((a) \ \frac{\pi}{3}\)
\((c) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
রাঃ২০১৭।
সমাধানঃ
\((a)\)
প্রদত্ত রাশি,
\(\cot^{-1}{\cos{\left(cosec^{-1}{\sqrt{\frac{3}{2}}}\right)}}\)
\(=\cot^{-1}{\cos{\left(cosec^{-1}{\frac{\sqrt{3}}{\sqrt{2}}}\right)}}\)
\(=\cot^{-1}{\cos{\left(\cos^{-1}{\frac{1}{\sqrt{3}}}\right)}}\) ➜ এখানে,
\(\text{লম্ব}=\sqrt{2}, \ \text{অতিভুজ}=\sqrt{3}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{3})^2-(\sqrt{2})^2}\)
\(=\sqrt{3-2}\)
\(=\sqrt{1}\)
\(=1\)
\(\therefore cosec^{-1}{\frac{\sqrt{3}}{\sqrt{2}}}=\cos^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\cot^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\cot^{-1}{\left(\cot{\frac{\pi}{3}}\right)}\) ➜ \(\because \frac{1}{\sqrt{3}}=\cot{\frac{\pi}{3}}\)
\(=\frac{\pi}{3}\)
ইহাই নির্ণেয় মুখ্যমান।
\((b)\)
দেওয়া আছে,
\(f(x)\) এর জন্য \(f^{-1}(x)+f^{-1}(y)=\pi\)
\(\Rightarrow f^{-1}{x}=\tan^{-1}{x}, \ f^{-1}{y}=\tan^{-1}{y}\)
এখন,
\(f^{-1}{x}+f^{-1}{y}=\pi\)
\(\Rightarrow \tan^{-1}{x}+\tan^{-1}{y}=\pi\)
\(\Rightarrow \tan^{-1}{\frac{x+y}{1-xy}}=\pi\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\frac{x+y}{1-xy}}\)
\(\Rightarrow \frac{x+y}{1-xy}=\tan{\pi}\)
\(\Rightarrow \frac{x+y}{1-xy}=0\) ➜ \(\because \tan{\pi}=0\)
\(\Rightarrow x+y=0\)
\(\because x+y=0, \ x\) এবং \(y\) এর একঘাত সমীকরণ।
অতএব ইহা একটি সরলরেখার সমীকরণ নির্দেশ করে।
যার ঢাল \(=-\frac{1}{1}\) ➜ \(\because ax+by+c=0\) সমীকরণের
ঢাল \(=-\frac{a}{b}\)
\(=-1\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
প্রদত্ত সমীকরণ,
\(\left\{f(x)\right\}^2+f^{\prime}(x)=3f(x)\) যখন \(0\le{x}\le{2\pi}\)
\(\Rightarrow \{f(x)\}^2+\frac{d}{dx}\{f(x)\}=3f(x)\)
\(\Rightarrow \{\tan{x}\}^2+\frac{d}{dx}(\tan{x})=3\tan{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan^2{x}+\sec^2{x}=3\tan{x}\) ➜ \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
প্রদত্ত রাশি,
\(\cot^{-1}{\cos{\left(cosec^{-1}{\sqrt{\frac{3}{2}}}\right)}}\)
\(=\cot^{-1}{\cos{\left(cosec^{-1}{\frac{\sqrt{3}}{\sqrt{2}}}\right)}}\)
\(=\cot^{-1}{\cos{\left(\cos^{-1}{\frac{1}{\sqrt{3}}}\right)}}\) ➜ এখানে,
\(\text{লম্ব}=\sqrt{2}, \ \text{অতিভুজ}=\sqrt{3}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{3})^2-(\sqrt{2})^2}\)
\(=\sqrt{3-2}\)
\(=\sqrt{1}\)
\(=1\)
\(\therefore cosec^{-1}{\frac{\sqrt{3}}{\sqrt{2}}}=\cos^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\cot^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\cot^{-1}{\left(\cot{\frac{\pi}{3}}\right)}\) ➜ \(\because \frac{1}{\sqrt{3}}=\cot{\frac{\pi}{3}}\)
\(=\frac{\pi}{3}\)
ইহাই নির্ণেয় মুখ্যমান।
\((b)\)
দেওয়া আছে,
\(f(x)\) এর জন্য \(f^{-1}(x)+f^{-1}(y)=\pi\)
\(\Rightarrow f^{-1}{x}=\tan^{-1}{x}, \ f^{-1}{y}=\tan^{-1}{y}\)
এখন,
\(f^{-1}{x}+f^{-1}{y}=\pi\)
\(\Rightarrow \tan^{-1}{x}+\tan^{-1}{y}=\pi\)
\(\Rightarrow \tan^{-1}{\frac{x+y}{1-xy}}=\pi\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\frac{x+y}{1-xy}}\)
\(\Rightarrow \frac{x+y}{1-xy}=\tan{\pi}\)
\(\Rightarrow \frac{x+y}{1-xy}=0\) ➜ \(\because \tan{\pi}=0\)
\(\Rightarrow x+y=0\)
\(\because x+y=0, \ x\) এবং \(y\) এর একঘাত সমীকরণ।
অতএব ইহা একটি সরলরেখার সমীকরণ নির্দেশ করে।
যার ঢাল \(=-\frac{1}{1}\) ➜ \(\because ax+by+c=0\) সমীকরণের
ঢাল \(=-\frac{a}{b}\)
\(=-1\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
প্রদত্ত সমীকরণ,
\(\left\{f(x)\right\}^2+f^{\prime}(x)=3f(x)\) যখন \(0\le{x}\le{2\pi}\)
\(\Rightarrow \{f(x)\}^2+\frac{d}{dx}\{f(x)\}=3f(x)\)
\(\Rightarrow \{\tan{x}\}^2+\frac{d}{dx}(\tan{x})=3\tan{x}\) ➜ \(\because f(x)=\tan{x}\)
\(\Rightarrow \tan^2{x}+\sec^2{x}=3\tan{x}\) ➜ \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{x}\le{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(Q.5.(x)\) \(f(x)=\sin{x}, \ g(x)=\cos{x}, \ \sin{\theta}=\frac{4}{5}\)
\((a)\) \(cosec^{-1}{\sqrt{5}}+\sec^{-1}{\frac{\sqrt{10}}{3}}\) এর মান নির্ণয় কর।
\((b)\) উদ্দীপকের আলোকে প্রমাণ কর যে, \(\sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{2}\)
\((c)\) উদ্দীপকের আলোকে সমাধান করঃ \(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((c) \ 2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((a)\) \(cosec^{-1}{\sqrt{5}}+\sec^{-1}{\frac{\sqrt{10}}{3}}\) এর মান নির্ণয় কর।
\((b)\) উদ্দীপকের আলোকে প্রমাণ কর যে, \(\sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{2}\)
\((c)\) উদ্দীপকের আলোকে সমাধান করঃ \(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
উত্তরঃ \((a) \ \frac{\pi}{4}\)
\((c) \ 2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
কুঃ২০১৭।
সমাধানঃ
\((a)\)
প্রদত্ত রাশি,
\(cosec^{-1}{\sqrt{5}}+\sec^{-1}{\frac{\sqrt{10}}{3}}\)
\(=\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore cosec^{-1}{\sqrt{5}}=\tan^{-1}{\frac{1}{2}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=3 , \ \text{অতিভুজ}=\sqrt{10}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{10})^2-3^2} \)
\(=\sqrt{10-9} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \sec^{-1}{\frac{\sqrt{10}}{3}}=\tan^{-1}{\frac{1}{3}}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)}\)
\(=\tan^{-1}{\left(\frac{3+2}{6-1}\right)}\) ➜ লব ও হরকে \(6\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{5}{5}\right)}\)
\(=\tan^{-1}{\left(1\right)}\)
\(=\tan^{-1}{\left(\tan{\frac{\pi}{4}}\right)}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(=\frac{\pi}{4}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(\sin{\theta}=\frac{4}{5}\)
\(\Rightarrow \theta=\sin^{-1}{\frac{4}{5}}\)
\(L.S=\sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}\)
\(=\sec^{-1}{\sqrt{5}}+\frac{1}{2}\sin^{-1}{\frac{4}{5}}-\sin^{-1}{\frac{1}{\sqrt{5}}}\) ➜ \(\because \theta=\sin^{-1}{\frac{4}{5}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{1}{2}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{ভূমি}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2}\)
\(=\sqrt{5-1}\)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sec^{-1}{\sqrt{5}}=\tan^{-1}{\frac{2}{1}}\)
\(=\tan^{-1}{2}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=4, \ \text{অতিভুজ}=5\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{5^2-4^2}\)
\(=\sqrt{25-16}\)
\(=\sqrt{9}\)
\(=3\)
\(\therefore \sin^{-1}{\frac{4}{5}}=\tan^{-1}{\frac{4}{3}}\)
তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2}\)
\(=\sqrt{5-1}\)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-2\tan^{-1}{\frac{1}{2}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{2\times\frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}}\right\}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\frac{2x}{1-x^2}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{1}{1-\frac{1}{4}}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{4}{4-1}}\right\}\) ➜ তৃতীয় পদের
লব ও হরকে \(4\) দ্বারা গুন করে।
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{4}{3}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\{0\}\)
\(=\tan^{-1}{2}+0\)
\(=\tan^{-1}{2}\)
\(=R.S\)
\(\therefore \sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{2}\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(f(x)=\sin{x}, \ g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{3}\) ➜ \(\because g(x)=\cos{x}\)
এবং \(f(x)=\sin{x}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{3}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+1^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{6}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi}{6}, \ x=2n\pi\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}, \ x=2n\pi\)
\(\therefore x=2n\pi+\frac{\pi}{3}, \ 2n\pi\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
প্রদত্ত রাশি,
\(cosec^{-1}{\sqrt{5}}+\sec^{-1}{\frac{\sqrt{10}}{3}}\)
\(=\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{3}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore cosec^{-1}{\sqrt{5}}=\tan^{-1}{\frac{1}{2}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=3 , \ \text{অতিভুজ}=\sqrt{10}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{10})^2-3^2} \)
\(=\sqrt{10-9} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \sec^{-1}{\frac{\sqrt{10}}{3}}=\tan^{-1}{\frac{1}{3}}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}\right)}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)}\)
\(=\tan^{-1}{\left(\frac{3+2}{6-1}\right)}\) ➜ লব ও হরকে \(6\) দ্বারা গুন করে,
\(=\tan^{-1}{\left(\frac{5}{5}\right)}\)
\(=\tan^{-1}{\left(1\right)}\)
\(=\tan^{-1}{\left(\tan{\frac{\pi}{4}}\right)}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(=\frac{\pi}{4}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(\sin{\theta}=\frac{4}{5}\)
\(\Rightarrow \theta=\sin^{-1}{\frac{4}{5}}\)
\(L.S=\sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}\)
\(=\sec^{-1}{\sqrt{5}}+\frac{1}{2}\sin^{-1}{\frac{4}{5}}-\sin^{-1}{\frac{1}{\sqrt{5}}}\) ➜ \(\because \theta=\sin^{-1}{\frac{4}{5}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{1}{2}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{ভূমি}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2}\)
\(=\sqrt{5-1}\)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sec^{-1}{\sqrt{5}}=\tan^{-1}{\frac{2}{1}}\)
\(=\tan^{-1}{2}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=4, \ \text{অতিভুজ}=5\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{5^2-4^2}\)
\(=\sqrt{25-16}\)
\(=\sqrt{9}\)
\(=3\)
\(\therefore \sin^{-1}{\frac{4}{5}}=\tan^{-1}{\frac{4}{3}}\)
তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2}\)
\(=\sqrt{5-1}\)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-2\tan^{-1}{\frac{1}{2}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{2\times\frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}}\right\}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\frac{2x}{1-x^2}}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{1}{1-\frac{1}{4}}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{4}{4-1}}\right\}\) ➜ তৃতীয় পদের
লব ও হরকে \(4\) দ্বারা গুন করে।
\(=\tan^{-1}{2}+\frac{1}{2}\left\{\tan^{-1}{\frac{4}{3}}-\tan^{-1}{\frac{4}{3}}\right\}\)
\(=\tan^{-1}{2}+\frac{1}{2}\{0\}\)
\(=\tan^{-1}{2}+0\)
\(=\tan^{-1}{2}\)
\(=R.S\)
\(\therefore \sec^{-1}{\sqrt{5}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{2}\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(f(x)=\sin{x}, \ g(x)=\cos{x}\)
প্রদত্ত সমীকরণ,
\(\sqrt{3}g(x)+f(x)=\sqrt{3}\)
\(\Rightarrow \sqrt{3}\cos{x}+\sin{x}=\sqrt{3}\) ➜ \(\because g(x)=\cos{x}\)
এবং \(f(x)=\sin{x}\)
\(\Rightarrow \cos{x}\frac{\sqrt{3}}{2}+\sin{x}\frac{1}{2}=\frac{\sqrt{3}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{(\sqrt{3})^2+1^2}=2\) ভাগ করে,
\(\Rightarrow \cos{x}\cos{\frac{\pi}{6}}+\sin{x}\sin{\frac{\pi}{6}}=\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\left(x-\frac{\pi}{6}\right)}=\cos{\frac{\pi}{6}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
\(\Rightarrow x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=2n\pi\pm\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}+\frac{\pi}{6}, \ x=2n\pi-\frac{\pi}{6}+\frac{\pi}{6}\)
\(\Rightarrow x=2n\pi+\frac{2\pi}{6}, \ x=2n\pi\)
\(\Rightarrow x=2n\pi+\frac{\pi}{3}, \ x=2n\pi\)
\(\therefore x=2n\pi+\frac{\pi}{3}, \ 2n\pi\)
\(\therefore\) সাধারণ সমাধান, \(x=2n\pi, \ 2n\pi+\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xi)\) দৃশ্যকল্প-১: \(\cot{\theta}-\tan{\theta}=\frac{6}{5}\)
দৃশ্যকল্প-২: \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{(\cot{3x})}+\tan^{-1}{(-\cot{5x})}=2x\)
\((b)\) দৃশ্যকল্প-১ হতে প্রমাণ কর যে, \(\theta=\frac{1}{2}\sin^{-1}{\frac{5}{\sqrt{34}}}\)
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ 2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
দৃশ্যকল্প-২: \(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\((a)\) প্রমাণ কর যে, \(\tan^{-1}{(\cot{3x})}+\tan^{-1}{(-\cot{5x})}=2x\)
\((b)\) দৃশ্যকল্প-১ হতে প্রমাণ কর যে, \(\theta=\frac{1}{2}\sin^{-1}{\frac{5}{\sqrt{34}}}\)
\((c)\) দৃশ্যকল্প-২ এ বর্ণিত সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ 2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
যঃ২০১৭।
সমাধানঃ
\((a)\)
\(L.S=\tan^{-1}{(\cot{3x})}+\tan^{-1}{(-\cot{5x})}\)
\(=\tan^{-1}{(\cot{3x})}-\tan^{-1}{(\cot{5x})}\)
\(=\tan^{-1}{\left\{\tan{\left(\frac{\pi}{2}-3x\right)}\right\}}-\tan^{-1}{\left\{\tan{\left(\frac{\pi}{2}-5x\right)}\right\}}\) ➜ \(\because \cot{A}=\tan{\left(\frac{\pi}{2}-A\right)}\)
\(=\frac{\pi}{2}-3x-\left(\frac{\pi}{2}-5x\right)\)
\(=\frac{\pi}{2}-3x-\frac{\pi}{2}+5x\)
\(=2x\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(\cot{\theta}-\tan{\theta}=\frac{6}{5}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=\frac{6}{5}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{6}{5}\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=\frac{6}{5}\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow \frac{2\cos{2\theta}}{2\sin{\theta}\cos{\theta}}=\frac{6}{5}\) ➜ লব ও হরকে \(2\) দ্বারা গুন করে,
\(\Rightarrow \frac{2\cos{2\theta}}{\sin{2\theta}}=\frac{6}{5}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{2\theta}}=\frac{3}{5}\)
\(\Rightarrow \cot{2\theta}=\frac{3}{5}\) ➜ \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)
\(\Rightarrow 2\theta=\cot^{-1}{\frac{3}{5}}\)
\(\Rightarrow 2\theta=\sin^{-1}{\frac{5}{\sqrt{34}}}\) ➜ এখানে,
\(\text{ভূমি}=3, \ \text{লম্ব}=5\)
\(\therefore \text{অতিভুজ}=\sqrt{(\text{লম্ব})^2+(\text{ভূমি})^2} \)
\(=\sqrt{5^2+3^2}\)
\(=\sqrt{25+9}\)
\(=\sqrt{34}\)
\(\therefore \cot^{-1}{\frac{3}{5}}=\sin^{-1}{\frac{5}{\sqrt{34}}}\)
\(\therefore \theta=\frac{1}{2}\sin^{-1}{\frac{5}{\sqrt{34}}}\)
(প্রমাণিত)
\((c)\)
প্রদত্ত সমীকরণ,
\(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\(\Rightarrow 2\times2\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}+1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}+1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=-1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=-\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\sin{\theta}=\sin{(\pi+\theta)}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=\sin{\frac{7\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=n\pi+(-1)^n\frac{7\pi}{6}\) ➜ \(\because \cos{A}=cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(L.S=\tan^{-1}{(\cot{3x})}+\tan^{-1}{(-\cot{5x})}\)
\(=\tan^{-1}{(\cot{3x})}-\tan^{-1}{(\cot{5x})}\)
\(=\tan^{-1}{\left\{\tan{\left(\frac{\pi}{2}-3x\right)}\right\}}-\tan^{-1}{\left\{\tan{\left(\frac{\pi}{2}-5x\right)}\right\}}\) ➜ \(\because \cot{A}=\tan{\left(\frac{\pi}{2}-A\right)}\)
\(=\frac{\pi}{2}-3x-\left(\frac{\pi}{2}-5x\right)\)
\(=\frac{\pi}{2}-3x-\frac{\pi}{2}+5x\)
\(=2x\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(\cot{\theta}-\tan{\theta}=\frac{6}{5}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=\frac{6}{5}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
এবং \(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}-\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{6}{5}\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{\theta}\cos{\theta}}=\frac{6}{5}\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow \frac{2\cos{2\theta}}{2\sin{\theta}\cos{\theta}}=\frac{6}{5}\) ➜ লব ও হরকে \(2\) দ্বারা গুন করে,
\(\Rightarrow \frac{2\cos{2\theta}}{\sin{2\theta}}=\frac{6}{5}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow \frac{\cos{2\theta}}{\sin{2\theta}}=\frac{3}{5}\)
\(\Rightarrow \cot{2\theta}=\frac{3}{5}\) ➜ \(\because \frac{\cos{A}}{\sin{A}}=\cot{A}\)
\(\Rightarrow 2\theta=\cot^{-1}{\frac{3}{5}}\)
\(\Rightarrow 2\theta=\sin^{-1}{\frac{5}{\sqrt{34}}}\) ➜ এখানে,
\(\text{ভূমি}=3, \ \text{লম্ব}=5\)
\(\therefore \text{অতিভুজ}=\sqrt{(\text{লম্ব})^2+(\text{ভূমি})^2} \)
\(=\sqrt{5^2+3^2}\)
\(=\sqrt{25+9}\)
\(=\sqrt{34}\)
\(\therefore \cot^{-1}{\frac{3}{5}}=\sin^{-1}{\frac{5}{\sqrt{34}}}\)
\(\therefore \theta=\frac{1}{2}\sin^{-1}{\frac{5}{\sqrt{34}}}\)
(প্রমাণিত)
\((c)\)
প্রদত্ত সমীকরণ,
\(2\sin{2\theta}+2(\sin{\theta}+\cos{\theta})+1=0\)
\(\Rightarrow 2\times2\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(\Rightarrow 4\sin{\theta}\cos{\theta}+2\sin{\theta}+2\cos{\theta}+1=0\)
\(\Rightarrow (2\cos{\theta}+1)(2\sin{\theta}+1)=0\)
\(\Rightarrow 2\cos{\theta}+1=0, \ 2\sin{\theta}+1=0\)
\(\Rightarrow 2\cos{\theta}=-1, \ 2\sin{\theta}=-1\)
\(\Rightarrow \cos{\theta}=-\frac{1}{2}, \ \sin{\theta}=-\frac{1}{2}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=-\sin{\frac{\pi}{6}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
এবং \(\frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=\sin{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\sin{\theta}=\sin{(\pi+\theta)}\)
\(\Rightarrow \cos{\theta}=\cos{\frac{2\pi}{3}}, \ \sin{\theta}=\sin{\frac{7\pi}{6}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ \theta=n\pi+(-1)^n\frac{7\pi}{6}\) ➜ \(\because \cos{A}=cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi\pm\frac{2\pi}{3}, \ n\pi+(-1)^n\frac{7\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xii)\) \(g(x)=p\sin^{-1}{x}, \ h(x)=\cos{x}\)
\((a)\) প্রমাণ কর যে, \(\sec^{-1}{\frac{\sqrt{5}}{2}}+\tan^{-1}{\frac{1}{2}}=\cot^{-1}{\frac{3}{4}}\)
\((b)\) \(g(x)\) এর লেখচিত্র অংকন কর, যখন \(p=\frac{1}{2}, \ -1\le{x}\le{1}\)
\((c)\) \(2\left\{h(x)\right\}^2+\left\{h(2x)\right\}^2=2\) সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ \frac{1}{2}(2n\pi\pm\alpha)\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\((a)\) প্রমাণ কর যে, \(\sec^{-1}{\frac{\sqrt{5}}{2}}+\tan^{-1}{\frac{1}{2}}=\cot^{-1}{\frac{3}{4}}\)
\((b)\) \(g(x)\) এর লেখচিত্র অংকন কর, যখন \(p=\frac{1}{2}, \ -1\le{x}\le{1}\)
\((c)\) \(2\left\{h(x)\right\}^2+\left\{h(2x)\right\}^2=2\) সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ \frac{1}{2}(2n\pi\pm\alpha)\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
বঃ২০১৭।
সমাধানঃ
\((a)\)
\(L.S=\sec^{-1}{\frac{\sqrt{5}}{2}}+\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{2}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{ভূমি}=2 , \ \text{অতিভুজ}=\sqrt{5} \)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-2^2} \)
\(=\sqrt{5-4} \)
\(=\sqrt{1} \)
\(=1 \)
\(\therefore \sec^{-1}{\frac{\sqrt{5}}{2}}=\tan^{-1}{\frac{1}{2}}\)
\(=2\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left\{\frac{2\times\frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right\}}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\(=\tan^{-1}{\left\{\frac{1}{1-\frac{1}{4}}\right\}}\)
\(=\tan^{-1}{\left(\frac{4}{4-1}\right)}\) ➜ লব ও হরকে \(4\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{4}{3}}\)
\(=\cot^{-1}{\frac{3}{4}}\) ➜ \(\because \tan^{-1}{x}=\cot^{-1}{\frac{1}{x}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
ধরি,
\(y=g(x)=P\sin^{-1}{x}\)
\(\Rightarrow y=\frac{1}{2}\sin^{-1}{x}\)
\(-1\le{x}\le{1}\) ব্যবধির মধ্যে \(x\) এর ভিন্ন ভিন্ন মানের জন্য \(y\) এর আনুসঙ্গিক মান নির্ণয় করে নিচে একটি টেবিল তৈরী করি।
ছক কাগজে কার্তেসীয় অক্ষদ্বয় \(XOX^{\prime}\) এবং \(YOY^{\prime}\) অঙ্কন করি।
স্কেলঃ ধরি, \(x\) অক্ষ বরাবর ক্ষুদ্রতম \(10\) বর্গ ঘর \(=1\) একক।
\(y\) অক্ষ বরাবর ক্ষুদ্রতম \(6\) বর্গ ঘর \(=15^{o}\) ।
ছক কাগজে স্কেল অনুযায়ী উপরের টেবিল হতে প্রাপ্ত বিন্দুগুলি স্থাপন করি। বিন্দুগুলি সাবলীলভাবে সংযুক্ত করে প্রদত্ত ফাংশনের লেখচিত্র অংকন করি।

\((c)\)
দেওয়া আছে,
\(h(x)=\cos{x}\)
\(\Rightarrow h(2x)=\cos{2x}\)
প্রদত্ত সমীকরণ,
\(2\left\{h(x)\right\}^2+\left\{h(2x)\right\}^2=2\)
\(\Rightarrow 2\left\{\cos{x}\right\}^2+\left\{\cos{2x}\right\}^2=2\) ➜ \(\because h(x)=\cos{x}\)
\(\Rightarrow h(2x)=\cos{2x}\)
\(\Rightarrow 2\cos^2{x}+\cos^2{2x}=2\)
\(\Rightarrow \cos{2x}+1+\cos^2{2x}=2\) ➜ \(\because 2\cos^2{A}=\cos{2A}+1\)
\(\Rightarrow \cos^2{2x}+\cos{2x}+1-2=0\)
\(\Rightarrow \cos^2{2x}+\cos{2x}-1=0\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1+\sqrt{5}}{2}, \ \cos{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \cos{2x}=\cos{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\theta}\)
\(\Rightarrow A=2n\pi\pm\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{1}{2}(2n\pi\pm\alpha)\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(L.S=\sec^{-1}{\frac{\sqrt{5}}{2}}+\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\frac{1}{2}}+\tan^{-1}{\frac{1}{2}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{ভূমি}=2 , \ \text{অতিভুজ}=\sqrt{5} \)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{5})^2-2^2} \)
\(=\sqrt{5-4} \)
\(=\sqrt{1} \)
\(=1 \)
\(\therefore \sec^{-1}{\frac{\sqrt{5}}{2}}=\tan^{-1}{\frac{1}{2}}\)
\(=2\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left\{\frac{2\times\frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right\}}\) ➜ \(\because 2\tan^{-1}{x}=\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}\)
\(=\tan^{-1}{\left\{\frac{1}{1-\frac{1}{4}}\right\}}\)
\(=\tan^{-1}{\left(\frac{4}{4-1}\right)}\) ➜ লব ও হরকে \(4\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{4}{3}}\)
\(=\cot^{-1}{\frac{3}{4}}\) ➜ \(\because \tan^{-1}{x}=\cot^{-1}{\frac{1}{x}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
ধরি,
\(y=g(x)=P\sin^{-1}{x}\)
\(\Rightarrow y=\frac{1}{2}\sin^{-1}{x}\)
\(-1\le{x}\le{1}\) ব্যবধির মধ্যে \(x\) এর ভিন্ন ভিন্ন মানের জন্য \(y\) এর আনুসঙ্গিক মান নির্ণয় করে নিচে একটি টেবিল তৈরী করি।
\(x\) | \(-1\) | \(-0.87\) | \(-0.5\) | \(0\) | \(0.5\) | \(0.87\) | \(1\) |
\(y=\frac{1}{2}\sin^{-1}{x}\) | \(-45^{o}\) | \(-30^{o}\) | \(-15^{o}\) | \(0^{o}\) | \(15^{o}\) | \(30^{o}\) | \(45^{o}\) |
\((x, \ y)\) | \((-1, -45^{o})\) | \((-0.87, -30^{o})\) | \((-0.5, -15^{o})\) | \((0, 0^{o})\) | \((0.5, 15^{o})\) | \((0.87, 30^{o})\) | \((1, 45^{o})\) |
স্কেলঃ ধরি, \(x\) অক্ষ বরাবর ক্ষুদ্রতম \(10\) বর্গ ঘর \(=1\) একক।
\(y\) অক্ষ বরাবর ক্ষুদ্রতম \(6\) বর্গ ঘর \(=15^{o}\) ।
ছক কাগজে স্কেল অনুযায়ী উপরের টেবিল হতে প্রাপ্ত বিন্দুগুলি স্থাপন করি। বিন্দুগুলি সাবলীলভাবে সংযুক্ত করে প্রদত্ত ফাংশনের লেখচিত্র অংকন করি।

\((c)\)
দেওয়া আছে,
\(h(x)=\cos{x}\)
\(\Rightarrow h(2x)=\cos{2x}\)
প্রদত্ত সমীকরণ,
\(2\left\{h(x)\right\}^2+\left\{h(2x)\right\}^2=2\)
\(\Rightarrow 2\left\{\cos{x}\right\}^2+\left\{\cos{2x}\right\}^2=2\) ➜ \(\because h(x)=\cos{x}\)
\(\Rightarrow h(2x)=\cos{2x}\)
\(\Rightarrow 2\cos^2{x}+\cos^2{2x}=2\)
\(\Rightarrow \cos{2x}+1+\cos^2{2x}=2\) ➜ \(\because 2\cos^2{A}=\cos{2A}+1\)
\(\Rightarrow \cos^2{2x}+\cos{2x}+1-2=0\)
\(\Rightarrow \cos^2{2x}+\cos{2x}-1=0\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1^2-4\times1\times(-1)}}{2\times1}\) ➜ \(\because ax^2+bx+c=0\)
\(\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{1+4}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1\pm\sqrt{5}}{2}\)
\(\Rightarrow \cos{2x}=\frac{-1+\sqrt{5}}{2}, \ \cos{2x}\ne{\frac{-1-\sqrt{5}}{2}}, \ \because -1\le{\cos{\theta}}\le{1}\)
\(\Rightarrow \cos{2x}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow \cos{2x}=\cos{\alpha}\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}\)
\(\Rightarrow 2x=2n\pi\pm\alpha\) ➜ \(\because \cos{A}=\cos{\theta}\)
\(\Rightarrow A=2n\pi\pm\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi\pm\frac{\alpha}{2}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{1}{2}(2n\pi\pm\alpha)\) যেখানে, \(\cos{\alpha}=\frac{\sqrt{5}-1}{2}, \ n\in{\mathbb{Z}}\)
\(Q.5.(xiii)\) \(f(x)=\cot^{-1}{y}-\tan^{-1}{x} ........(1)\)
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta} ........(2)\)
\((a)\) \(\sin{\frac{x}{3}}\) এর পর্যায়কাল কত?
\((b)\) \(f(x)=\frac{\pi}{6}\) হলে, প্রমাণ কর যে, \(x+y+\sqrt{3}xy=\sqrt{3}\)
\((c)\) উদ্দীপক-২ এর সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \((a) \ 6\pi\)
\((c) \ \frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta} ........(2)\)
\((a)\) \(\sin{\frac{x}{3}}\) এর পর্যায়কাল কত?
\((b)\) \(f(x)=\frac{\pi}{6}\) হলে, প্রমাণ কর যে, \(x+y+\sqrt{3}xy=\sqrt{3}\)
\((c)\) উদ্দীপক-২ এর সাধারণ সমাধান নির্ণয় কর।
উত্তরঃ \((a) \ 6\pi\)
\((c) \ \frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
চঃ২০১৭।
সমাধানঃ
\((a)\)
দেওয়া আছে,
\(\sin{\frac{x}{3}}\)
\(=\sin{\left(2\pi+\frac{x}{3}\right)}\)
\(=\sin{\left(\frac{6\pi+x}{3}\right)}\)
\(=f(6\pi+x)\)
\(\therefore \) পর্যায়কাল \(=6\pi\)
\((b)\)
দেওয়া আছে,
\(f(x)=\cot^{-1}{y}-\tan^{-1}{x}\)
এবং \(f(x)=\frac{\pi}{6}\)
\(\therefore \cot^{-1}{y}-\tan^{-1}{x}=\frac{\pi}{6}\)
\(\Rightarrow \tan^{-1}{\frac{1}{y}}-\tan^{-1}{x}=\frac{\pi}{6}\) ➜ \(\because \cot^{-1}{A}=\tan^{-1}{\frac{1}{A}}\)
\(\Rightarrow \tan^{-1}{\left(\frac{\frac{1}{y}-x}{1+\frac{1}{y}\times{x}}\right)}=\frac{\pi}{6}\) ➜ \(\because \tan^{-1}{A}-\tan^{-1}{B}=\tan^{-1}{\frac{A-B}{1+AB}}\)
\(\Rightarrow \frac{\frac{1}{y}-x}{1+\frac{x}{y}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \frac{1-xy}{y+x}=\frac{1}{\sqrt{3}}\) ➜ বাম পাশের,
লব ও হরকে \(y\) দ্বারা গুন করে,
এবং \(\tan{\frac{\pi}{6}}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow y+x=\sqrt{3}-\sqrt{3}xy\)
\(\therefore x+y+\sqrt{3}xy=\sqrt{3}\)
(প্রমাণিত)
\((c)\)
প্রদত্ত সমীকরণ,
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+9\theta}{2}}\sin{\frac{9\theta-\theta}{2}}=\sin{5\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{10\theta}{2}}\sin{\frac{8\theta}{2}}=\sin{5\theta}\)
\(\Rightarrow 2\sin{5\theta}\sin{4\theta}-\sin{5\theta}=0\)
\(\Rightarrow \sin{5\theta}(2\sin{4\theta}-1)=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}-1=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}=1\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 5\theta=n\pi, \ 4\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{5}, \ \theta=\frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore \theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
দেওয়া আছে,
\(\sin{\frac{x}{3}}\)
\(=\sin{\left(2\pi+\frac{x}{3}\right)}\)
\(=\sin{\left(\frac{6\pi+x}{3}\right)}\)
\(=f(6\pi+x)\)
\(\therefore \) পর্যায়কাল \(=6\pi\)
\((b)\)
দেওয়া আছে,
\(f(x)=\cot^{-1}{y}-\tan^{-1}{x}\)
এবং \(f(x)=\frac{\pi}{6}\)
\(\therefore \cot^{-1}{y}-\tan^{-1}{x}=\frac{\pi}{6}\)
\(\Rightarrow \tan^{-1}{\frac{1}{y}}-\tan^{-1}{x}=\frac{\pi}{6}\) ➜ \(\because \cot^{-1}{A}=\tan^{-1}{\frac{1}{A}}\)
\(\Rightarrow \tan^{-1}{\left(\frac{\frac{1}{y}-x}{1+\frac{1}{y}\times{x}}\right)}=\frac{\pi}{6}\) ➜ \(\because \tan^{-1}{A}-\tan^{-1}{B}=\tan^{-1}{\frac{A-B}{1+AB}}\)
\(\Rightarrow \frac{\frac{1}{y}-x}{1+\frac{x}{y}}=\tan{\frac{\pi}{6}}\)
\(\Rightarrow \frac{1-xy}{y+x}=\frac{1}{\sqrt{3}}\) ➜ বাম পাশের,
লব ও হরকে \(y\) দ্বারা গুন করে,
এবং \(\tan{\frac{\pi}{6}}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow y+x=\sqrt{3}-\sqrt{3}xy\)
\(\therefore x+y+\sqrt{3}xy=\sqrt{3}\)
(প্রমাণিত)
\((c)\)
প্রদত্ত সমীকরণ,
\(\cos{\theta}-\cos{9\theta}=\sin{5\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+9\theta}{2}}\sin{\frac{9\theta-\theta}{2}}=\sin{5\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{10\theta}{2}}\sin{\frac{8\theta}{2}}=\sin{5\theta}\)
\(\Rightarrow 2\sin{5\theta}\sin{4\theta}-\sin{5\theta}=0\)
\(\Rightarrow \sin{5\theta}(2\sin{4\theta}-1)=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}-1=0\)
\(\Rightarrow \sin{5\theta}=0, \ 2\sin{4\theta}=1\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{5\theta}=0, \ \sin{4\theta}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 5\theta=n\pi, \ 4\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=\frac{n\pi}{5}, \ \theta=\frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore \theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{5}, \ \frac{1}{4}\left\{n\pi+(-1)^n\frac{\pi}{6}\right\}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xiv)\) \(A=\cos{\theta}, \ B=\sin{\theta}, \ C=\cos{2\theta}, \ D=\sin{2\theta}\)
\((a)\) মান নির্ণয় করঃ \(\tan^{-1}{\sin{\cos^{-1}{\sqrt{\frac{2}{3}}}}}\)
\((b)\) \(A+\sqrt{3}B=\sqrt{2}\) হলে, সমীকরণটি সমাধান কর।
\((c)\) \(A+B=C+D\) হলে, সমীকরণটি \(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে সমাধান আছে কিনা যাচাই কর।
উত্তরঃ \((a) \ \frac{\pi}{6}\)
\((b) \ 2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 0, \ \frac{\pi}{6}\)
\((a)\) মান নির্ণয় করঃ \(\tan^{-1}{\sin{\cos^{-1}{\sqrt{\frac{2}{3}}}}}\)
\((b)\) \(A+\sqrt{3}B=\sqrt{2}\) হলে, সমীকরণটি সমাধান কর।
\((c)\) \(A+B=C+D\) হলে, সমীকরণটি \(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে সমাধান আছে কিনা যাচাই কর।
উত্তরঃ \((a) \ \frac{\pi}{6}\)
\((b) \ 2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ 0, \ \frac{\pi}{6}\)
দিঃ২০১৭।
সমাধানঃ
\((a)\)
প্রদত্ত রাশি,
\(\tan^{-1}{\sin{\left(\cos^{-1}{\sqrt{\frac{2}{3}}}\right)}}\)
\(=\tan^{-1}{\sin{\left(\cos^{-1}{\frac{\sqrt{2}}{\sqrt{3}}}\right)}}\)
\(=\tan^{-1}{\sin{\left(\sin^{-1}{\frac{1}{\sqrt{3}}}\right)}}\) ➜ এখানে,
\(\text{ভূমি}=\sqrt{2} , \ \text{অতিভুজ}=\sqrt{3}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{3})^2-(\sqrt{2})^2} \)
\(=\sqrt{3-2} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \cos^{-1}{\frac{\sqrt{2}}{\sqrt{3}}}=\sin^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\tan^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\tan^{-1}{\left(\tan{\frac{\pi}{6}}\right)}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(=\frac{\pi}{6}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(A=\cos{\theta}, \ B=\sin{\theta}\)
এবং \(A+\sqrt{3}B=\sqrt{2}\)
\(\therefore \cos{\theta}+\sqrt{3}\sin{\theta}=\sqrt{2}\)
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\)
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}+\sin{\theta}\sin{\frac{\pi}{3}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}, \ \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(\theta-\frac{\pi}{3}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \theta-\frac{\pi}{3}=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}+\frac{\pi}{3}, \ 2n\pi-\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow \theta=2n\pi+\frac{3\pi+4\pi}{12}, \ 2n\pi+\frac{4\pi-3\pi}{12}\)
\(\therefore \theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
দেওয়া আছে,
\(A=\cos{\theta}, \ B=\sin{\theta}, \ C=\cos{2\theta}, \ D=\sin{2\theta}\)
এবং \(A+B=C+D\)
\(\therefore \cos{\theta}+\sin{\theta}=\cos{2\theta}+\sin{2\theta}\)
\(\Rightarrow \cos{\theta}-\cos{2\theta}=\sin{2\theta}-\sin{\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+2\theta}{2}}\sin{\frac{2\theta-\theta}{2}}=2\cos{\frac{2\theta+\theta}{2}}\sin{\frac{2\theta-\theta}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}-\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}\left(\sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}=\cos{\frac{3\theta}{2}}\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \frac{\sin{\frac{3\theta}{2}}}{\cos{\frac{3\theta}{2}}}=1\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\cos{\frac{3\theta}{2}}\) ভাগ করে,
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=1\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi, \ \frac{3\theta}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi, \ \theta=\frac{2n\pi}{3}+\frac{\pi}{6}\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\frac{2}{3}\) গুণ করে,
\(\Rightarrow \theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=0, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=2\pi, \ \frac{2\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow \theta=2\pi, \ \frac{5\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=0, \ \frac{\pi}{6}\)
প্রদত্ত রাশি,
\(\tan^{-1}{\sin{\left(\cos^{-1}{\sqrt{\frac{2}{3}}}\right)}}\)
\(=\tan^{-1}{\sin{\left(\cos^{-1}{\frac{\sqrt{2}}{\sqrt{3}}}\right)}}\)
\(=\tan^{-1}{\sin{\left(\sin^{-1}{\frac{1}{\sqrt{3}}}\right)}}\) ➜ এখানে,
\(\text{ভূমি}=\sqrt{2} , \ \text{অতিভুজ}=\sqrt{3}\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{(\sqrt{3})^2-(\sqrt{2})^2} \)
\(=\sqrt{3-2} \)
\(=\sqrt{1}\)
\(=1\)
\(\therefore \cos^{-1}{\frac{\sqrt{2}}{\sqrt{3}}}=\sin^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\tan^{-1}{\frac{1}{\sqrt{3}}}\)
\(=\tan^{-1}{\left(\tan{\frac{\pi}{6}}\right)}\) ➜ \(\because \frac{1}{\sqrt{3}}=\tan{\frac{\pi}{6}}\)
\(=\frac{\pi}{6}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(A=\cos{\theta}, \ B=\sin{\theta}\)
এবং \(A+\sqrt{3}B=\sqrt{2}\)
\(\therefore \cos{\theta}+\sqrt{3}\sin{\theta}=\sqrt{2}\)
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{2}\) ➜ উভয় পার্শে \(\cos{x}\) ও \(\sin{x}\) এর সহগের বর্গের যোগফলের বর্গমূল
অর্থাৎ \(\sqrt{1^2+(\sqrt{3})^2}=2\) ভাগ করে,
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\)
\(\Rightarrow \cos{\theta}\frac{1}{2}+\sin{\theta}\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \cos{\theta}\cos{\frac{\pi}{3}}+\sin{\theta}\sin{\frac{\pi}{3}}=\frac{1}{\sqrt{2}}\) ➜ \(\because \frac{1}{2}=\cos{\frac{\pi}{3}}, \ \frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}\)
\(\Rightarrow \cos{\left(\theta-\frac{\pi}{3}\right)}=\cos{\frac{\pi}{4}}\) ➜ \(\because \cos{A}\cos{B}+\sin{A}\sin{B}=\cos{(A-B)}\)
এবং \(\frac{1}{\sqrt{2}}=\cos{\frac{\pi}{4}}\)
\(\Rightarrow \theta-\frac{\pi}{3}=2n\pi\pm\frac{\pi}{4}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow \theta=2n\pi+\frac{\pi}{4}+\frac{\pi}{3}, \ 2n\pi-\frac{\pi}{4}+\frac{\pi}{3}\)
\(\Rightarrow \theta=2n\pi+\frac{3\pi+4\pi}{12}, \ 2n\pi+\frac{4\pi-3\pi}{12}\)
\(\therefore \theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\)
\(\therefore\) নির্ণেয় সমাধান, \(\theta=2n\pi+\frac{7\pi}{12}, \ 2n\pi+\frac{\pi}{12}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
দেওয়া আছে,
\(A=\cos{\theta}, \ B=\sin{\theta}, \ C=\cos{2\theta}, \ D=\sin{2\theta}\)
এবং \(A+B=C+D\)
\(\therefore \cos{\theta}+\sin{\theta}=\cos{2\theta}+\sin{2\theta}\)
\(\Rightarrow \cos{\theta}-\cos{2\theta}=\sin{2\theta}-\sin{\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+2\theta}{2}}\sin{\frac{2\theta-\theta}{2}}=2\cos{\frac{2\theta+\theta}{2}}\sin{\frac{2\theta-\theta}{2}}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
এবং \(\sin{C}-\sin{D}=2\cos{\frac{C+D}{2}}\sin{\frac{C-D}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}\)
\(\Rightarrow \sin{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}-\cos{\frac{3\theta}{2}}\sin{\frac{\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}\left(\sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}\right)=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}-\cos{\frac{3\theta}{2}}=0\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \sin{\frac{3\theta}{2}}=\cos{\frac{3\theta}{2}}\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \frac{\sin{\frac{3\theta}{2}}}{\cos{\frac{3\theta}{2}}}=1\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\cos{\frac{3\theta}{2}}\) ভাগ করে,
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=1\)
\(\Rightarrow \sin{\frac{\theta}{2}}=0, \ \tan{\frac{3\theta}{2}}=\tan{\frac{\pi}{4}}\) ➜ \(\because 1=\tan{\frac{\pi}{4}}\)
\(\Rightarrow \frac{\theta}{2}=n\pi, \ \frac{3\theta}{2}=n\pi+\frac{\pi}{4}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\tan{A}=\tan{\alpha}\)
\(\Rightarrow A=n\pi+\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=2n\pi, \ \theta=\frac{2n\pi}{3}+\frac{\pi}{6}\) ➜ দ্বিতীয় সমীকরণে,
উভয় পার্শে \(\frac{2}{3}\) গুণ করে,
\(\Rightarrow \theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=2n\pi, \ \frac{2n\pi}{3}+\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=0, \ \frac{\pi}{6}\) মান দুইটি গ্রহনযোগ্য।\(\therefore \theta=0, \ \frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=2\pi, \ \frac{2\pi}{3}+\frac{\pi}{6}\)
\(\Rightarrow \theta=2\pi, \ \frac{5\pi}{6}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \left[0,\frac{\pi}{2}\right]\) ব্যাবধিতে নির্ণেয় সমাধান, \(\theta=0, \ \frac{\pi}{6}\)
\(Q.5.(xv)\) \(f(x)=\tan{\theta}\) এবং \(\sin{\theta}=\frac{12}{13}\)
\((a)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{1}{\sqrt{x}}}=\frac{1}{2}\sec^{-1}{\frac{1+x}{1-x}}\)
\((b)\) সমাধান করঃ \(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\((c)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{3}{4}}+\frac{\theta}{2}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
উত্তরঃ \((b) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) এবং \(\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\((a)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{1}{\sqrt{x}}}=\frac{1}{2}\sec^{-1}{\frac{1+x}{1-x}}\)
\((b)\) সমাধান করঃ \(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\((c)\) প্রমাণ কর যে, \(\cot^{-1}{\frac{3}{4}}+\frac{\theta}{2}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
উত্তরঃ \((b) \ \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) এবং \(\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
সমাধানঃ
\((a)\)
\(L.S=\cot^{-1}{\frac{1}{\sqrt{x}}}\)
\(=\tan^{-1}{\sqrt{x}}\) ➜ \(\because \cot^{-1}{\frac{1}{A}}=\tan^{-1}{A}\)
\(=\frac{1}{2}\times2\tan^{-1}{\sqrt{x}}\)
\(=\frac{1}{2}\cos^{-1}{\frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2}}\) ➜ \(\because 2\tan^{-1}{A}=\cos^{-1}{\frac{1-A^2}{1+A^2}}\)
\(=\frac{1}{2}\cos^{-1}{\frac{1-x}{1+x}}\)
\(=\frac{1}{2}\sec^{-1}{\frac{1+x}{1-x}}\) ➜ \(\because \cos^{-1}{A}=\sec^{-1}{\frac{1}{A}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
প্রদত্ত সমীকরণ,
\(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\tan{\left\{\tan^{-1}{\left(\frac{1}{2}\right)}\right\}}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\((c)\)
দেওয়া আছে,
\(\sin{\theta}=\frac{12}{13}\)
\(\therefore \theta=\sin^{-1}{\frac{12}{13}}\)
এবং \(\cot^{-1}{\frac{3}{4}}+\frac{\theta}{2}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
\(\Rightarrow \cot^{-1}{\frac{3}{4}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
\(\therefore \cot^{-1}{\frac{3}{4}}+\frac{1}{2}\sin^{-1}{\frac{12}{13}}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\) ➜ \(\because \theta=\sin^{-1}{\frac{12}{13}}\)
এখন,
\(L.S=\cot^{-1}{\frac{4}{3}}+\frac{1}{2}\sin^{-1}{\frac{12}{13}}-\sin^{-1}{\frac{1}{\sqrt{5}}}\)
\(=\tan^{-1}{\frac{3}{4}}+\frac{1}{2}\sin^{-1}{\frac{2\times\frac{2}{3}}{1+\left(\frac{2}{3}\right)^2}}-\sin^{-1}{\frac{1}{\sqrt{5}}}\) ➜ \(\because \cot^{-1}{\frac{1}{A}}=\tan^{-1}{A}\)
\(=\tan^{-1}{\frac{3}{4}}+\frac{1}{2}\times2\tan^{-1}{\frac{2}{3}}-\tan^{-1}{\frac{1}{2}}\) ➜ \(\because \sin^{-1}{\frac{2A}{1+A^2}}=\tan^{-1}{A}\)
শেষ পদে,
\(\text{লম্ব}=1 , \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\frac{3}{4}}+\tan^{-1}{\frac{2}{3}}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times\frac{2}{3}}\right)}-\tan^{-1}{\frac{1}{2}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{9+8}{12}}{1-\frac{1}{2}}\right)}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{12}}{1-\frac{1}{2}}\right)}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{17}{12-6}\right)}-\tan^{-1}{\frac{1}{2}}\) ➜ প্রথম পদের,
লব ও হরকে \(12\) দ্বারা গুণ করে,
\(=\tan^{-1}{\frac{17}{6}}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{6}-\frac{1}{2}}{1+\frac{17}{6}\times\frac{1}{2}}\right)}\) ➜ \(\because \tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left(\frac{x-y}{1+xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{6}-\frac{1}{2}}{1+\frac{17}{6}\times\frac{1}{2}}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{6}-\frac{1}{2}}{1+\frac{17}{12}}\right)}\)
\(=\tan^{-1}{\left(\frac{34-6}{12+17}\right)}\) ➜ লব ও হরকে \(12\) দ্বারা গুণ করে,
\(=\tan^{-1}{\left(\frac{28}{29}\right)}\)
\(=\cot^{-1}{\frac{29}{28}}\) ➜ \(\because \tan^{-1}{A}=\cot^{-1}{\frac{1}{A}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\(L.S=\cot^{-1}{\frac{1}{\sqrt{x}}}\)
\(=\tan^{-1}{\sqrt{x}}\) ➜ \(\because \cot^{-1}{\frac{1}{A}}=\tan^{-1}{A}\)
\(=\frac{1}{2}\times2\tan^{-1}{\sqrt{x}}\)
\(=\frac{1}{2}\cos^{-1}{\frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2}}\) ➜ \(\because 2\tan^{-1}{A}=\cos^{-1}{\frac{1-A^2}{1+A^2}}\)
\(=\frac{1}{2}\cos^{-1}{\frac{1-x}{1+x}}\)
\(=\frac{1}{2}\sec^{-1}{\frac{1+x}{1-x}}\) ➜ \(\because \cos^{-1}{A}=\sec^{-1}{\frac{1}{A}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
প্রদত্ত সমীকরণ,
\(\tan^2{x}+\sec^2{x}=3\tan{x}, \ 0\le{x}\le{2\pi}\)
\(\Rightarrow \tan^2{x}+1+\tan^2{x}=3\tan{x}\) ➜ \(\because \sec^2{A}=1+\tan^2{A}\)
\(\Rightarrow 2\tan^2{x}-3\tan{x}+1=0\)
\(\Rightarrow 2\tan^2{x}-2\tan{x}-\tan{x}+1=0\)
\(\Rightarrow 2\tan{x}(\tan{x}-1)-1(\tan{x}-1)=0\)
\(\Rightarrow (\tan{x}-1)(2\tan{x}-1)=0\)
\(\Rightarrow \tan{x}-1=0, \ 2\tan{x}-1=0\)
\(\Rightarrow \tan{x}=1, \ 2\tan{x}=1\)
\(\Rightarrow \tan{x}=1, \ \tan{x}=\frac{1}{2}\)
\(\Rightarrow \tan{x}=\tan{\frac{\pi}{4}}, \ \tan{x}=\tan{\left\{\tan^{-1}{\left(\frac{1}{2}\right)}\right\}}\)
\(\Rightarrow x=n\pi+\frac{\pi}{4}, \ x=n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) ➜ \(\because \tan{A}=tan{\theta}\)
\(\Rightarrow A=n\pi+\theta\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\therefore\) সাধারণ সমাধান, \(x=n\pi+\frac{\pi}{4}, \ n\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{x}\le{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=1,\) \(x=\pi+\frac{\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{4}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
যখন, \(n=-1,\) \(x=-\pi+\frac{\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\(\Rightarrow x=-\frac{3\pi}{4}, \ -\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=2\pi+\frac{\pi}{4}, \ 2\pi+\tan^{-1}{\left(\frac{1}{2}\right)}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\le{x}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{4}, \ \frac{5\pi}{4}, \ \tan^{-1}{\left(\frac{1}{2}\right)}, \ \pi+\tan^{-1}{\left(\frac{1}{2}\right)}\)
\((c)\)
দেওয়া আছে,
\(\sin{\theta}=\frac{12}{13}\)
\(\therefore \theta=\sin^{-1}{\frac{12}{13}}\)
এবং \(\cot^{-1}{\frac{3}{4}}+\frac{\theta}{2}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
\(\Rightarrow \cot^{-1}{\frac{3}{4}}+\frac{1}{2}\theta-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\)
\(\therefore \cot^{-1}{\frac{3}{4}}+\frac{1}{2}\sin^{-1}{\frac{12}{13}}-\sin^{-1}{\frac{1}{\sqrt{5}}}=\cot^{-1}{\frac{29}{28}}\) ➜ \(\because \theta=\sin^{-1}{\frac{12}{13}}\)
এখন,
\(L.S=\cot^{-1}{\frac{4}{3}}+\frac{1}{2}\sin^{-1}{\frac{12}{13}}-\sin^{-1}{\frac{1}{\sqrt{5}}}\)
\(=\tan^{-1}{\frac{3}{4}}+\frac{1}{2}\sin^{-1}{\frac{2\times\frac{2}{3}}{1+\left(\frac{2}{3}\right)^2}}-\sin^{-1}{\frac{1}{\sqrt{5}}}\) ➜ \(\because \cot^{-1}{\frac{1}{A}}=\tan^{-1}{A}\)
\(=\tan^{-1}{\frac{3}{4}}+\frac{1}{2}\times2\tan^{-1}{\frac{2}{3}}-\tan^{-1}{\frac{1}{2}}\) ➜ \(\because \sin^{-1}{\frac{2A}{1+A^2}}=\tan^{-1}{A}\)
শেষ পদে,
\(\text{লম্ব}=1 , \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\frac{3}{4}}+\tan^{-1}{\frac{2}{3}}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times\frac{2}{3}}\right)}-\tan^{-1}{\frac{1}{2}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{9+8}{12}}{1-\frac{1}{2}}\right)}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{12}}{1-\frac{1}{2}}\right)}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{17}{12-6}\right)}-\tan^{-1}{\frac{1}{2}}\) ➜ প্রথম পদের,
লব ও হরকে \(12\) দ্বারা গুণ করে,
\(=\tan^{-1}{\frac{17}{6}}-\tan^{-1}{\frac{1}{2}}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{6}-\frac{1}{2}}{1+\frac{17}{6}\times\frac{1}{2}}\right)}\) ➜ \(\because \tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left(\frac{x-y}{1+xy}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{6}-\frac{1}{2}}{1+\frac{17}{6}\times\frac{1}{2}}\right)}\)
\(=\tan^{-1}{\left(\frac{\frac{17}{6}-\frac{1}{2}}{1+\frac{17}{12}}\right)}\)
\(=\tan^{-1}{\left(\frac{34-6}{12+17}\right)}\) ➜ লব ও হরকে \(12\) দ্বারা গুণ করে,
\(=\tan^{-1}{\left(\frac{28}{29}\right)}\)
\(=\cot^{-1}{\frac{29}{28}}\) ➜ \(\because \tan^{-1}{A}=\cot^{-1}{\frac{1}{A}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\(Q.5.(xvi)\) \(f(x)=\cos{x}\) এবং \(g(x)=\sin{x}\)
\((a)\) \(f\left(\cot^{-1}{\frac{3}{4}}\right)\) এর মান কত?
\((b)\) \(f^{-1}\left(\frac{x}{a}\right)+f^{-1}\left(\frac{y}{b}\right)=\theta\) হলে, দেখাও যে, \(\frac{x^2}{a^2}-\frac{2xy}{ab}f(\theta)+\frac{y^2}{b^2}=1-\left\{f(x)\right\}^2\)
\((c)\) সমাধান করঃ \(f(\theta)-f(7\theta)=g(4\theta)\)
উত্তরঃ \((a) \ \frac{3}{5}\)
\((c) \ \frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((a)\) \(f\left(\cot^{-1}{\frac{3}{4}}\right)\) এর মান কত?
\((b)\) \(f^{-1}\left(\frac{x}{a}\right)+f^{-1}\left(\frac{y}{b}\right)=\theta\) হলে, দেখাও যে, \(\frac{x^2}{a^2}-\frac{2xy}{ab}f(\theta)+\frac{y^2}{b^2}=1-\left\{f(x)\right\}^2\)
\((c)\) সমাধান করঃ \(f(\theta)-f(7\theta)=g(4\theta)\)
উত্তরঃ \((a) \ \frac{3}{5}\)
\((c) \ \frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
সমাধানঃ
\((a)\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
\(\therefore f\left(\cot^{-1}{\frac{3}{4}}\right)=\cos{\left(\cot^{-1}{\frac{3}{4}}\right)}\)
প্রদত্ত রাশি,
\(f\left(\cot^{-1}{\frac{3}{4}}\right)\)
\(=\cos{\left(\cot^{-1}{\frac{3}{4}}\right)}\) ➜ \(\because f\left(\cot^{-1}{\frac{3}{4}}\right)=\cos{\left(\cot^{-1}{\frac{3}{4}}\right)}\)
\(=\cos{\left(\cos^{-1}{\frac{3}{5}}\right)}\) ➜ এখানে,
\(\text{ভূমি}=3 , \ \text{লম্ব}=4\)
\(\therefore \text{অতিভুজ}=\sqrt{(\text{লম্ব})^2+(\text{ভূমি})^2} \)
\(=\sqrt{4^2+3^2} \)
\(=\sqrt{16+9} \)
\(=\sqrt{25}\)
\(=5\)
\(\therefore \cot^{-1}{\frac{3}{4}}=\cos^{-1}{\frac{3}{5}}\)
\(=\frac{3}{5}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
ধরি, \(f(x)=\cos{x}=y\)
\(\Rightarrow f(x)=y, \ \cos{x}=y\)
\(\Rightarrow x=f^{-1}(y), \ x=\cos^{-1}{y}\)
\(\Rightarrow f^{-1}(y)=\cos^{-1}{y}\)
\(\therefore f^{-1}\left(\frac{x}{a}\right)=\cos^{-1}{\left(\frac{x}{a}\right)}\)
\(f^{-1}\left(\frac{y}{b}\right)=\cos^{-1}{\left(\frac{y}{b}\right)}\)
এবং \(f(\theta)=\cos{\theta}\)
প্রদত্ত সমীকরণ,
\(f^{-1}\left(\frac{x}{a}\right)+f^{-1}\left(\frac{y}{b}\right)=\theta\)
\(\Rightarrow \cos^{-1}{\left(\frac{x}{a}\right)}+\cos^{-1}{\left(\frac{y}{b}\right)}=\theta\) ➜ \(\because f^{-1}\left(\frac{x}{a}\right)=\cos^{-1}{\left(\frac{x}{a}\right)}\)
এবং \(f^{-1}\left(\frac{y}{b}\right)=\cos^{-1}{\left(\frac{y}{b}\right)}\)
\(\Rightarrow \cos^{-1}{\left\{\frac{x}{a}\times\frac{y}{b}-\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}\right\}}=\theta\) ➜ \(\because \cos^{-1}{A}+\cos^{-1}{B}=\cos^{-1}{\{AB-\sqrt{(1-A^2)(1-B^2)}\}}\)
\(\Rightarrow \frac{xy}{ab}-\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}=\cos{\theta}\)
\(\Rightarrow \frac{xy}{ab}-\cos{\theta}=\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}\)
\(\Rightarrow \left(\frac{xy}{ab}-\cos{\theta}\right)^2=\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{x^2y^2}{a^2b^2}-2\frac{xy}{ab}\cos{\theta}+\cos^2{\theta}=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2y^2}{a^2b^2}\)
\(\Rightarrow \frac{x^2y^2}{a^2b^2}-2\frac{xy}{ab}\cos{\theta}+\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{x^2y^2}{a^2b^2}=1-\cos^2{\theta}\)
\(\Rightarrow -2\frac{xy}{ab}\cos{\theta}+\frac{x^2}{a^2}+\frac{y^2}{b^2}=1-\{\cos{\theta}\}^2\)
\(\therefore \frac{x^2}{a^2}-2\frac{xy}{ab}\cos{\theta}+\frac{y^2}{b^2}=1-\{f{\theta}\}^2\) ➜ \(\because f(\theta)=\cos{\theta}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(f(x)=\cos{x}\) এবং \(g(x)=\sin{x}\)
\(\therefore f(\theta)=\cos{\theta}, \ f(7\theta)=\cos{7\theta}\) এবং \(g(4\theta)=\sin{4\theta}\)
প্রদত্ত সমীকরণ,
\(f(\theta)-f(7\theta)=g(4\theta)\)
\(\Rightarrow \cos{\theta}-\cos{7\theta}=\sin{4\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+7\theta}{2}}\sin{\frac{7\theta-\theta}{2}}=\sin{4\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{8\theta}{2}}\sin{\frac{6\theta}{2}}=\sin{4\theta}\)
\(\Rightarrow 2\sin{4\theta}\sin{3\theta}-\sin{4\theta}=0\)
\(\Rightarrow \sin{4\theta}(2\sin{3\theta}-1)=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}-1=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}=1\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 4\theta=n\pi, \ 3\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}, \ \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
\(\therefore f\left(\cot^{-1}{\frac{3}{4}}\right)=\cos{\left(\cot^{-1}{\frac{3}{4}}\right)}\)
প্রদত্ত রাশি,
\(f\left(\cot^{-1}{\frac{3}{4}}\right)\)
\(=\cos{\left(\cot^{-1}{\frac{3}{4}}\right)}\) ➜ \(\because f\left(\cot^{-1}{\frac{3}{4}}\right)=\cos{\left(\cot^{-1}{\frac{3}{4}}\right)}\)
\(=\cos{\left(\cos^{-1}{\frac{3}{5}}\right)}\) ➜ এখানে,
\(\text{ভূমি}=3 , \ \text{লম্ব}=4\)
\(\therefore \text{অতিভুজ}=\sqrt{(\text{লম্ব})^2+(\text{ভূমি})^2} \)
\(=\sqrt{4^2+3^2} \)
\(=\sqrt{16+9} \)
\(=\sqrt{25}\)
\(=5\)
\(\therefore \cot^{-1}{\frac{3}{4}}=\cos^{-1}{\frac{3}{5}}\)
\(=\frac{3}{5}\)
ইহাই নির্ণেয় মান।
\((b)\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
ধরি, \(f(x)=\cos{x}=y\)
\(\Rightarrow f(x)=y, \ \cos{x}=y\)
\(\Rightarrow x=f^{-1}(y), \ x=\cos^{-1}{y}\)
\(\Rightarrow f^{-1}(y)=\cos^{-1}{y}\)
\(\therefore f^{-1}\left(\frac{x}{a}\right)=\cos^{-1}{\left(\frac{x}{a}\right)}\)
\(f^{-1}\left(\frac{y}{b}\right)=\cos^{-1}{\left(\frac{y}{b}\right)}\)
এবং \(f(\theta)=\cos{\theta}\)
প্রদত্ত সমীকরণ,
\(f^{-1}\left(\frac{x}{a}\right)+f^{-1}\left(\frac{y}{b}\right)=\theta\)
\(\Rightarrow \cos^{-1}{\left(\frac{x}{a}\right)}+\cos^{-1}{\left(\frac{y}{b}\right)}=\theta\) ➜ \(\because f^{-1}\left(\frac{x}{a}\right)=\cos^{-1}{\left(\frac{x}{a}\right)}\)
এবং \(f^{-1}\left(\frac{y}{b}\right)=\cos^{-1}{\left(\frac{y}{b}\right)}\)
\(\Rightarrow \cos^{-1}{\left\{\frac{x}{a}\times\frac{y}{b}-\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}\right\}}=\theta\) ➜ \(\because \cos^{-1}{A}+\cos^{-1}{B}=\cos^{-1}{\{AB-\sqrt{(1-A^2)(1-B^2)}\}}\)
\(\Rightarrow \frac{xy}{ab}-\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}=\cos{\theta}\)
\(\Rightarrow \frac{xy}{ab}-\cos{\theta}=\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}\)
\(\Rightarrow \left(\frac{xy}{ab}-\cos{\theta}\right)^2=\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{x^2y^2}{a^2b^2}-2\frac{xy}{ab}\cos{\theta}+\cos^2{\theta}=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2y^2}{a^2b^2}\)
\(\Rightarrow \frac{x^2y^2}{a^2b^2}-2\frac{xy}{ab}\cos{\theta}+\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{x^2y^2}{a^2b^2}=1-\cos^2{\theta}\)
\(\Rightarrow -2\frac{xy}{ab}\cos{\theta}+\frac{x^2}{a^2}+\frac{y^2}{b^2}=1-\{\cos{\theta}\}^2\)
\(\therefore \frac{x^2}{a^2}-2\frac{xy}{ab}\cos{\theta}+\frac{y^2}{b^2}=1-\{f{\theta}\}^2\) ➜ \(\because f(\theta)=\cos{\theta}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(f(x)=\cos{x}\) এবং \(g(x)=\sin{x}\)
\(\therefore f(\theta)=\cos{\theta}, \ f(7\theta)=\cos{7\theta}\) এবং \(g(4\theta)=\sin{4\theta}\)
প্রদত্ত সমীকরণ,
\(f(\theta)-f(7\theta)=g(4\theta)\)
\(\Rightarrow \cos{\theta}-\cos{7\theta}=\sin{4\theta}\)
\(\Rightarrow 2\sin{\frac{\theta+7\theta}{2}}\sin{\frac{7\theta-\theta}{2}}=\sin{4\theta}\) ➜ \(\because \cos{C}-\cos{D}=2\sin{\frac{C+D}{2}}\sin{\frac{D-C}{2}}\)
\(\Rightarrow 2\sin{\frac{8\theta}{2}}\sin{\frac{6\theta}{2}}=\sin{4\theta}\)
\(\Rightarrow 2\sin{4\theta}\sin{3\theta}-\sin{4\theta}=0\)
\(\Rightarrow \sin{4\theta}(2\sin{3\theta}-1)=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}-1=0\)
\(\Rightarrow \sin{4\theta}=0, \ 2\sin{3\theta}=1\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{4\theta}=0, \ \sin{3\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow 4\theta=n\pi, \ 3\theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\sin{A}=\sin{\alpha}\)
\(\Rightarrow A=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{n\pi}{4}, \ \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=\frac{n\pi}{4}, \ \frac{n\pi}{3}+(-1)^n\frac{\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(Q.5.(xvii)\) দৃশ্যকল্প-১: \(\cos^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{5}{13}}+cosec^{-1}{\frac{65}{16}}\)
দৃশ্যকল্প-২: \(f(x)=\tan{x}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\cos^{-1}{\frac{1-x^2}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\) হবে।
\((c)\) দৃশ্যকল্প-২: থেকে সমাধান করঃ \(f(x)+f(2x)+f(3x)=f(x)f(2x)f(3x)\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{3}, \ \frac{2\pi}{3}, \ \pi, \ \frac{4\pi}{3}, \ \frac{5\pi}{3}\)
দৃশ্যকল্প-২: \(f(x)=\tan{x}\)
\((a)\) দেখাও যে, \(2\tan^{-1}{x}=\cos^{-1}{\frac{1-x^2}{1+x^2}}\)
\((b)\) দেখাও যে, দৃশ্যকল্প-১ এর মান \(\frac{\pi}{2}\) হবে।
\((c)\) দৃশ্যকল্প-২: থেকে সমাধান করঃ \(f(x)+f(2x)+f(3x)=f(x)f(2x)f(3x)\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((c) \ \frac{\pi}{3}, \ \frac{2\pi}{3}, \ \pi, \ \frac{4\pi}{3}, \ \frac{5\pi}{3}\)
সমাধানঃ
\((a)\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
এখন,
\(\cos{2\theta}=\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\cos^{-1}{\left(\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\cos^{-1}{\left(\frac{1-x^2}{1+x^2}\right)}\) ➜ \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)
(দেখানো হলো)
\((b)\)
প্রদত্ত রাশি,
\(\cos^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{5}{13}}+cosec^{-1}{\frac{65}{16}}\)
\(=\tan^{-1}{\frac{4}{3}}+\tan^{-1}{\frac{5}{12}}+\tan^{-1}{\frac{16}{63}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{ভূমি}=3 , \ \text{অতিভুজ}=5 \)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{5^2-3^2} \)
\(=\sqrt{25-9} \)
\(=\sqrt{16} \)
\(=4 \)
\(\therefore \cos^{-1}{\frac{3}{5}}=\tan^{-1}{\frac{4}{3}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=5 , \ \text{অতিভুজ}=13 \)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{13^2-5^2} \)
\(=\sqrt{169-25} \)
\(=\sqrt{144} \)
\(=12 \)
\(\therefore \sin^{-1}{\frac{5}{13}}=\tan^{-1}{\frac{5}{12}}\)
এবং তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=16 , \ \text{অতিভুজ}=65\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(65)^2-(16)^2} \)
\(=\sqrt{4225-256} \)
\(=\sqrt{3969}\)
\(=63\)
\(\therefore cosec^{-1}{\frac{65}{16}}=\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}}\right\}}+\tan^{-1}{\frac{16}{63}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{20}{36}}\right\}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left(\frac{48+15}{36-20}\right)}+\tan^{-1}{\frac{16}{63}}\) ➜ লব ও হরকে \(36\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{63}{16}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-\frac{63}{16}\times\frac{16}{63}}\right\}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-1}\right\}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{0}\right\}}\)
\(=\tan^{-1}{(\infty)}\)
\(=\tan^{-1}{(\tan{\frac{\pi}{2}})}\)➜ \(\because \infty=\tan{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
\(\therefore \cos^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{5}{13}}+cosec^{-1}{\frac{65}{16}}=\frac{\pi}{2}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
\(\therefore f(2x)=\tan{2x}, \ f(3x)=\tan{3x}\)
প্রদত্ত সমীকরণ,
\(f(x)+f(2x)+f(3x)=f(x)f(2x)f(3x)\) যখন, \(0\lt{x}\lt{2\pi}\)
\(\Rightarrow \tan{x}+\tan{2x}+\tan{3x}=\tan{x}\tan{2x}\tan{3x}\) ➜ \(\because f(2x)=\tan{2x}, \ f(3x)=\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=\tan{x}\tan{2x}\tan{3x}-\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=-\tan{3x}\) ➜ উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=-\tan{3x}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}+\tan{3x}=0\)
\(\Rightarrow 2\tan{3x}=0\)
\(\Rightarrow \tan{3x}=0\)
\(\Rightarrow 3x=n\pi\) ➜ \(\because \tan{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
যখন, \(n=1,\) \(x=\frac{\pi}{3}\)মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{2\pi}{3}\)মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{2\pi}{3}\)
যখন, \(n=3,\) \(x=\frac{3\pi}{3}\)
\(\Rightarrow x=\pi\)মানটি গ্রহনযোগ্য।
\(\therefore x=\pi\)
যখন, \(n=4,\) \(x=\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{4\pi}{3}\)
যখন, \(n=5,\) \(x=\frac{5\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{3}\)
যখন, \(n=3,\) \(x=\frac{6\pi}{3}\)
\(\Rightarrow x=2\pi\)মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{3}, \ \frac{2\pi}{3}, \ \pi, \ \frac{4\pi}{3}, \ \frac{5\pi}{3}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
এখন,
\(\cos{2\theta}=\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\cos^{-1}{\left(\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\cos^{-1}{\left(\frac{1-x^2}{1+x^2}\right)}\) ➜ \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)
(দেখানো হলো)
\((b)\)
প্রদত্ত রাশি,
\(\cos^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{5}{13}}+cosec^{-1}{\frac{65}{16}}\)
\(=\tan^{-1}{\frac{4}{3}}+\tan^{-1}{\frac{5}{12}}+\tan^{-1}{\frac{16}{63}}\) ➜ প্রথম পদের ক্ষেত্রে
\(\text{ভূমি}=3 , \ \text{অতিভুজ}=5 \)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{5^2-3^2} \)
\(=\sqrt{25-9} \)
\(=\sqrt{16} \)
\(=4 \)
\(\therefore \cos^{-1}{\frac{3}{5}}=\tan^{-1}{\frac{4}{3}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=5 , \ \text{অতিভুজ}=13 \)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{13^2-5^2} \)
\(=\sqrt{169-25} \)
\(=\sqrt{144} \)
\(=12 \)
\(\therefore \sin^{-1}{\frac{5}{13}}=\tan^{-1}{\frac{5}{12}}\)
এবং তৃতীয় পদের ক্ষেত্রে,
\(\text{লম্ব}=16 , \ \text{অতিভুজ}=65\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(65)^2-(16)^2} \)
\(=\sqrt{4225-256} \)
\(=\sqrt{3969}\)
\(=63\)
\(\therefore cosec^{-1}{\frac{65}{16}}=\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}}\right\}}+\tan^{-1}{\frac{16}{63}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{20}{36}}\right\}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left(\frac{48+15}{36-20}\right)}+\tan^{-1}{\frac{16}{63}}\) ➜ লব ও হরকে \(36\) দ্বারা গুন করে,
\(=\tan^{-1}{\frac{63}{16}}+\tan^{-1}{\frac{16}{63}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-\frac{63}{16}\times\frac{16}{63}}\right\}}\) ➜ \(\because \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{1-1}\right\}}\)
\(=\tan^{-1}{\left\{\frac{\frac{63}{16}+\frac{16}{63}}{0}\right\}}\)
\(=\tan^{-1}{(\infty)}\)
\(=\tan^{-1}{(\tan{\frac{\pi}{2}})}\)➜ \(\because \infty=\tan{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
\(\therefore \cos^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{5}{13}}+cosec^{-1}{\frac{65}{16}}=\frac{\pi}{2}\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(f(x)=\tan{x}\)
\(\therefore f(2x)=\tan{2x}, \ f(3x)=\tan{3x}\)
প্রদত্ত সমীকরণ,
\(f(x)+f(2x)+f(3x)=f(x)f(2x)f(3x)\) যখন, \(0\lt{x}\lt{2\pi}\)
\(\Rightarrow \tan{x}+\tan{2x}+\tan{3x}=\tan{x}\tan{2x}\tan{3x}\) ➜ \(\because f(2x)=\tan{2x}, \ f(3x)=\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=\tan{x}\tan{2x}\tan{3x}-\tan{3x}\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \tan{x}+\tan{2x}=-\tan{3x}(1-\tan{x}\tan{2x})\)
\(\Rightarrow \frac{\tan{x}+\tan{2x}}{1-\tan{x}\tan{2x}}=-\tan{3x}\) ➜ উভয় পার্শে \((1-\tan{x}\tan{2x})\) ভাগ করে,
\(\Rightarrow \tan{(x+2x)}=-\tan{3x}\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\Rightarrow \tan{3x}+\tan{3x}=0\)
\(\Rightarrow 2\tan{3x}=0\)
\(\Rightarrow \tan{3x}=0\)
\(\Rightarrow 3x=n\pi\) ➜ \(\because \tan{A}=0\)
\(\Rightarrow A=n\pi\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{n\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{n\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{2\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=0\) মানটি গ্রহনযোগ্য নয়।যখন, \(n=1,\) \(x=\frac{\pi}{3}\)মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{3}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{2\pi}{3}\)মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{2\pi}{3}\)
যখন, \(n=3,\) \(x=\frac{3\pi}{3}\)
\(\Rightarrow x=\pi\)মানটি গ্রহনযোগ্য।
\(\therefore x=\pi\)
যখন, \(n=4,\) \(x=\frac{4\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{4\pi}{3}\)
যখন, \(n=5,\) \(x=\frac{5\pi}{3}\) মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{3}\)
যখন, \(n=3,\) \(x=\frac{6\pi}{3}\)
\(\Rightarrow x=2\pi\)মানটি গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{3}, \ \frac{2\pi}{3}, \ \pi, \ \frac{4\pi}{3}, \ \frac{5\pi}{3}\)
\(Q.5.(xviii)\) \(f(x)=\cos{x}\)
\((a)\) \(\cos{3x}=-\frac{\sqrt{3}}{2}\) হলে, \(x\) এর মান নির্ণয় কর।
\((b)\) সমাধান করঃ \(f(x)+f(2x)+f(3x)=0\)
\((c)\) সমাধান করঃ \(4f(x)f(2x)f(3x)=1\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{2n\pi}{3}\pm\frac{7\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ (2n+1)\frac{\pi}{4}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
\((a)\) \(\cos{3x}=-\frac{\sqrt{3}}{2}\) হলে, \(x\) এর মান নির্ণয় কর।
\((b)\) সমাধান করঃ \(f(x)+f(2x)+f(3x)=0\)
\((c)\) সমাধান করঃ \(4f(x)f(2x)f(3x)=1\) যখন, \(0\lt{x}\lt{2\pi}\)
উত্তরঃ \((a) \ \frac{2n\pi}{3}\pm\frac{7\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b) \ (2n+1)\frac{\pi}{4}, \ 2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c) \ \frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
সমাধানঃ
\((a)\)
প্রদত্ত সমীকরণ,
\(\cos{3x}=-\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{3x}=-\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\Rightarrow \cos{3x}=\cos{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\cos{A}=\cos{(\pi+A)}\)
\(\Rightarrow \cos{3x}=\cos{\frac{7\pi}{6}}\)
\(\Rightarrow 3x=2n\pi\pm\frac{7\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm{\alpha}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{2n\pi}{3}\pm\frac{7\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{2n\pi}{3}\pm\frac{7\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b)\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
প্রদত্ত সমীকরণ,
\(f(x)+f(2x)+f(3x)=0\)
\(\Rightarrow \cos{x}+\cos{2x}+\cos{3x}=0\) ➜ \(\because f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
\(\Rightarrow \cos{3x}+\cos{x}+\cos{2x}=0\)
\(\Rightarrow 2\cos{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}+\cos{2x}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{4x}{2}}\cos{\frac{2x}{2}}+\cos{2x}=0\)
\(\Rightarrow 2\cos{2x}\cos{x}+\cos{2x}=0\)
\(\Rightarrow \cos{2x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
প্রদত্ত সমীকরণ,
\(4f(x)f(2x)f(3x)=1\) যখন, \(0\lt{x}\lt{2\pi}\)
\(\Rightarrow 4\cos{x}\cos{2x}\cos{3x}=1\) ➜ \(\because f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
\(\Rightarrow 2\cos{2x}\times2\cos{3x}\cos{x}=1\)
\(\Rightarrow 2\cos{2x}\{\cos{(3x-x)}+\cos{(3x+x)}\}=1\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow 2\cos{2x}\{\cos{2x}+\cos{4x}\}-1=0\)
\(\Rightarrow 2\cos^2{2x}+2\cos{4x}\cos{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+2\cos^2{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+\cos{4x}=0\) ➜ \(\because 2\cos^2{A}-1=\cos{2A}\)
\(\Rightarrow \cos{4x}(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 4x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{8}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{\pi}{8}, \ \frac{\pi}{3}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{8}, \ \pi+\frac{\pi}{3}, \ \pi-\frac{\pi}{3}\)
\(\Rightarrow x=\frac{3\pi}{8}, \ \frac{4\pi}{3}, \ \frac{2\pi}{3}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{8}, \ \frac{2\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{8}, \ -\pi+\frac{\pi}{3}, \ -\pi-\frac{\pi}{3}\)
\(\Rightarrow x=-\frac{\pi}{8}, \ -\frac{2\pi}{3}, \ -\frac{4\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{5\pi}{8}, \ 2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{8}\)
যখন, \(n=3,\) \(x=\frac{7\pi}{8}, \ 3\pi+\frac{\pi}{3}, \ 3\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{7\pi}{8}\)
যখন, \(n=4,\) \(x=\frac{9\pi}{8}, \ 4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
প্রদত্ত সমীকরণ,
\(\cos{3x}=-\frac{\sqrt{3}}{2}\)
\(\Rightarrow \cos{3x}=-\cos{\frac{\pi}{6}}\) ➜ \(\because \frac{\sqrt{3}}{2}=\cos{\frac{\pi}{6}}\)
\(\Rightarrow \cos{3x}=\cos{\left(\pi+\frac{\pi}{6}\right)}\) ➜ \(\because -\cos{A}=\cos{(\pi+A)}\)
\(\Rightarrow \cos{3x}=\cos{\frac{7\pi}{6}}\)
\(\Rightarrow 3x=2n\pi\pm\frac{7\pi}{6}\) ➜ \(\because \cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm{\alpha}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=\frac{2n\pi}{3}\pm\frac{7\pi}{18}\)
\(\therefore\) সাধারণ সমাধান, \(x=\frac{2n\pi}{3}\pm\frac{7\pi}{18}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((b)\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
প্রদত্ত সমীকরণ,
\(f(x)+f(2x)+f(3x)=0\)
\(\Rightarrow \cos{x}+\cos{2x}+\cos{3x}=0\) ➜ \(\because f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
\(\Rightarrow \cos{3x}+\cos{x}+\cos{2x}=0\)
\(\Rightarrow 2\cos{\frac{3x+x}{2}}\cos{\frac{3x-x}{2}}+\cos{2x}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{4x}{2}}\cos{\frac{2x}{2}}+\cos{2x}=0\)
\(\Rightarrow 2\cos{2x}\cos{x}+\cos{2x}=0\)
\(\Rightarrow \cos{2x}(2\cos{x}+1)=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}+1=0\)
\(\Rightarrow \cos{2x}=0, \ 2\cos{x}=-1\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=-\frac{1}{2}\)
\(\Rightarrow \cos{2x}=0, \ \cos{x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 2x=(2n+1)\frac{\pi}{2}, \ x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{4}, \ x=2n\pi\pm\frac{2\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\((c)\)
দেওয়া আছে,
\(f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
প্রদত্ত সমীকরণ,
\(4f(x)f(2x)f(3x)=1\) যখন, \(0\lt{x}\lt{2\pi}\)
\(\Rightarrow 4\cos{x}\cos{2x}\cos{3x}=1\) ➜ \(\because f(x)=\cos{x}\)
\(\therefore f(2x)=\cos{2x}, \ f(3x)=\cos{3x}\)
\(\Rightarrow 2\cos{2x}\times2\cos{3x}\cos{x}=1\)
\(\Rightarrow 2\cos{2x}\{\cos{(3x-x)}+\cos{(3x+x)}\}=1\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow 2\cos{2x}\{\cos{2x}+\cos{4x}\}-1=0\)
\(\Rightarrow 2\cos^2{2x}+2\cos{4x}\cos{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+2\cos^2{2x}-1=0\)
\(\Rightarrow 2\cos{4x}\cos{2x}+\cos{4x}=0\) ➜ \(\because 2\cos^2{A}-1=\cos{2A}\)
\(\Rightarrow \cos{4x}(2\cos{2x}+1)=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}+1=0\)
\(\Rightarrow \cos{4x}=0, \ 2\cos{2x}=-1\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=-\frac{1}{2}\)
\(\Rightarrow \cos{4x}=0, \ \cos{2x}=\cos{\frac{2\pi}{3}}\) ➜ \(\because -\frac{1}{2}=\cos{\frac{2\pi}{3}}\)
\(\Rightarrow 4x=(2n+1)\frac{\pi}{2}, \ 2x=2n\pi\pm\frac{2\pi}{3}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
এবং \(\cos{A}=\cos{\alpha}\)
\(\Rightarrow A=2n\pi\pm\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow x=(2n+1)\frac{\pi}{8}, \ x=n\pi\pm\frac{\pi}{3}\)
\(\therefore x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\)
\(\therefore\) সাধারণ সমাধান, \(x=(2n+1)\frac{\pi}{8}, \ n\pi\pm\frac{\pi}{3}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\lt{x}\lt{\pi}\) ব্যাবধিতে \(x\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ -\frac{\pi}{3}\) প্রথম মান দুইটি গ্রহনযোগ্য।\(\therefore x=\frac{\pi}{8}, \ \frac{\pi}{3}\)
যখন, \(n=1,\) \(x=\frac{3\pi}{8}, \ \pi+\frac{\pi}{3}, \ \pi-\frac{\pi}{3}\)
\(\Rightarrow x=\frac{3\pi}{8}, \ \frac{4\pi}{3}, \ \frac{2\pi}{3}\) প্রথম ও শেষ মান দুইটি গ্রহনযোগ্য।
\(\therefore x=\frac{3\pi}{8}, \ \frac{2\pi}{3}\)
যখন, \(n=-1,\) \(x=-\frac{\pi}{8}, \ -\pi+\frac{\pi}{3}, \ -\pi-\frac{\pi}{3}\)
\(\Rightarrow x=-\frac{\pi}{8}, \ -\frac{2\pi}{3}, \ -\frac{4\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(x=\frac{5\pi}{8}, \ 2\pi+\frac{\pi}{3}, \ 2\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{5\pi}{8}\)
যখন, \(n=3,\) \(x=\frac{7\pi}{8}, \ 3\pi+\frac{\pi}{3}, \ 3\pi-\frac{\pi}{3}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore x=\frac{7\pi}{8}\)
যখন, \(n=4,\) \(x=\frac{9\pi}{8}, \ 4\pi+\frac{\pi}{3}, \ 4\pi-\frac{\pi}{3}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore \ 0\lt{x}\lt{\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(x=\frac{\pi}{8}, \ \frac{\pi}{3}, \ \frac{3\pi}{8}, \ \frac{2\pi}{3}, \ \frac{5\pi}{8}\) এবং \(\frac{7\pi}{8}\)
\(Q.5.(xix)\) \(f(x)=\tan^{-1}{x}, \ h(x)=\sin^{-1}{x}, \ g(x)=\sec{x}\)
\((a)\) দেখাও যে, \(f\left(\frac{5}{6}\right)-f\left(\frac{49}{71}\right)=f\left(\frac{1}{11}\right)\)
\((b)\) দেখাও যে, \(h\left(\sqrt{2}\sin{\theta}\right)+h\left(\sqrt{\cos{2\theta}}\right)=\frac{\pi}{2}\)
\((c)\) সমাধান করঃ \(g(4\theta)-g(2\theta)=2\) যখন, \(0\lt{x}\lt{180^{o}}\)
উত্তরঃ \((c) \ 18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}\) এবং \(162^{o}\)
\((a)\) দেখাও যে, \(f\left(\frac{5}{6}\right)-f\left(\frac{49}{71}\right)=f\left(\frac{1}{11}\right)\)
\((b)\) দেখাও যে, \(h\left(\sqrt{2}\sin{\theta}\right)+h\left(\sqrt{\cos{2\theta}}\right)=\frac{\pi}{2}\)
\((c)\) সমাধান করঃ \(g(4\theta)-g(2\theta)=2\) যখন, \(0\lt{x}\lt{180^{o}}\)
উত্তরঃ \((c) \ 18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}\) এবং \(162^{o}\)
সমাধানঃ
\((a)\)
দেওয়া আছে,
\(f(x)=\tan^{-1}{x}\)
\(\therefore f\left(\frac{5}{6}\right)=\tan^{-1}{\left(\frac{5}{6}\right)}\)
এবং \(f\left(\frac{49}{71}\right)=\tan^{-1}{\left(\frac{49}{71}\right)}\)
এখন,
\(L.S=\tan^{-1}{\left(\frac{5}{6}\right)}-\tan^{-1}{\left(\frac{49}{71}\right)}\) ➜ \(\because f(x)=\tan^{-1}{x}\)
\(\therefore f\left(\frac{5}{6}\right)=\tan^{-1}{\left(\frac{5}{6}\right)}\)
এবং \(f\left(\frac{49}{71}\right)=\tan^{-1}{\left(\frac{49}{71}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{5}{6}-\frac{49}{71}}{1+\frac{5}{6}\times\frac{49}{71}}\right\}}\) ➜ \(\because \tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left\{\frac{x-y}{1+xy}\right\}}\)
\(=\tan^{-1}{\left\{\frac{\frac{5}{6}-\frac{49}{71}}{1+\frac{245}{426}}\right\}}\)
\(=\tan^{-1}{\left\{\frac{355-294}{426+245}\right\}}\) ➜ লব ও হরকে \(426\) দ্বারা গুন করে,
\(=\tan^{-1}{\left\{\frac{61}{671}\right\}}\)
\(=\tan^{-1}{\frac{1}{11}}\)
\(=f\left(\frac{1}{11}\right)\) ➜ \(\because f(x)=\tan^{-1}{x}\)
\(\therefore f\left(\frac{1}{11}\right)=\tan^{-1}{\frac{1}{11}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(h(x)=\sin^{-1}{x}\)
\(\therefore h\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}\)
এবং \(h\left(\sqrt{\cos{2\theta}}\right)=\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\)
এখন,
\(L.S=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}+\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\) ➜ \(\because h(x)=\sin^{-1}{x}\)
\(\therefore h\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}\)
এবং \(h\left(\sqrt{\cos{2\theta}}\right)=\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-(\sqrt{\cos{2\theta}})^2}+\sqrt{\cos{2\theta}}\sqrt{1-(\sqrt{2}\sin{\theta})^2}\right\}}\) ➜ \(\because \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{\left\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-\cos{2\theta}}+\sqrt{\cos{2\theta}}\sqrt{1-2\sin^2{\theta}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2\sin^2{\theta}}+\sqrt{\cos{2\theta}}\sqrt{\cos{2\theta}}\right\}}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1-2\sin^2{A}=\cos{2A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2}\sin{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{2\sin^2{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{1-\cos{2\theta}+\cos{2\theta}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\sin^{-1}{\left\{1\right\}}\)
\(=\sin^{-1}{\left\{\sin{\frac{\pi}{2}}\right\}}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(g(x)=\sec{x}\)
\(\therefore g(4\theta)=\sec{4\theta}, \ g(2\theta)=\sec{2\theta}\)
প্রদত্ত সমীকরণ,
\(g(4\theta)-g(2\theta)=2\) যখন, \(0\lt{x}\lt{180^{o}}\)
\(\Rightarrow \sec{4\theta}-\sec{2\theta}=2\) ➜ \(\because g(x)=\sec{x}\)
\(\therefore g(4\theta)=\sec{4\theta}, \ g(2\theta)=\sec{2\theta}\)
\(\Rightarrow \frac{1}{\cos{4\theta}}-\frac{1}{\cos{2\theta}}=2\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{2\theta}-\cos{4\theta}}{\cos{4\theta}\cos{2\theta}}=2\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=2\cos{4\theta}\cos{2\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{(4\theta-2\theta)}+\cos{(4\theta+2\theta)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{2\theta}+\cos{6\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}-\cos{2\theta}-\cos{6\theta}=0\)
\(\Rightarrow -\cos{4\theta}-\cos{6\theta}=0\)
\(\Rightarrow \cos{6\theta}+\cos{4\theta}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{\frac{6\theta+4\theta}{2}}\cos{\frac{6\theta-4\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{5\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{5\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 5\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\Rightarrow \theta=18^{o}, \ 90^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=18^{o}, \ 90^{o}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{10}, \ \frac{3\pi}{2}\)
\(\Rightarrow \theta=54^{o}, \ 270^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=54^{o}, \ 270^{o}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{10}, \ -\frac{\pi}{2}\)
\(\Rightarrow \theta=-18^{o}, \ -90^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{10}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=\frac{\pi}{2}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=90^{o}, \ 450^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=90^{o}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{10}, \ \frac{7\pi}{2}\)
\(\Rightarrow \theta=126^{o}, \ 630^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=126^{o}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{10}, \ \frac{9\pi}{2}\)
\(\Rightarrow \theta=162^{o}, \ 810^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=5,\) \(\theta=\frac{11\pi}{10}, \ \frac{11\pi}{2}\)
\(\Rightarrow \theta=198^{o}, \ 990^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}\) এবং \(162^{o}\)
দেওয়া আছে,
\(f(x)=\tan^{-1}{x}\)
\(\therefore f\left(\frac{5}{6}\right)=\tan^{-1}{\left(\frac{5}{6}\right)}\)
এবং \(f\left(\frac{49}{71}\right)=\tan^{-1}{\left(\frac{49}{71}\right)}\)
এখন,
\(L.S=\tan^{-1}{\left(\frac{5}{6}\right)}-\tan^{-1}{\left(\frac{49}{71}\right)}\) ➜ \(\because f(x)=\tan^{-1}{x}\)
\(\therefore f\left(\frac{5}{6}\right)=\tan^{-1}{\left(\frac{5}{6}\right)}\)
এবং \(f\left(\frac{49}{71}\right)=\tan^{-1}{\left(\frac{49}{71}\right)}\)
\(=\tan^{-1}{\left\{\frac{\frac{5}{6}-\frac{49}{71}}{1+\frac{5}{6}\times\frac{49}{71}}\right\}}\) ➜ \(\because \tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left\{\frac{x-y}{1+xy}\right\}}\)
\(=\tan^{-1}{\left\{\frac{\frac{5}{6}-\frac{49}{71}}{1+\frac{245}{426}}\right\}}\)
\(=\tan^{-1}{\left\{\frac{355-294}{426+245}\right\}}\) ➜ লব ও হরকে \(426\) দ্বারা গুন করে,
\(=\tan^{-1}{\left\{\frac{61}{671}\right\}}\)
\(=\tan^{-1}{\frac{1}{11}}\)
\(=f\left(\frac{1}{11}\right)\) ➜ \(\because f(x)=\tan^{-1}{x}\)
\(\therefore f\left(\frac{1}{11}\right)=\tan^{-1}{\frac{1}{11}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(h(x)=\sin^{-1}{x}\)
\(\therefore h\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}\)
এবং \(h\left(\sqrt{\cos{2\theta}}\right)=\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\)
এখন,
\(L.S=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}+\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\) ➜ \(\because h(x)=\sin^{-1}{x}\)
\(\therefore h\left(\sqrt{2}\sin{\theta}\right)=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}\)
এবং \(h\left(\sqrt{\cos{2\theta}}\right)=\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-(\sqrt{\cos{2\theta}})^2}+\sqrt{\cos{2\theta}}\sqrt{1-(\sqrt{2}\sin{\theta})^2}\right\}}\) ➜ \(\because \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{\left\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-\cos{2\theta}}+\sqrt{\cos{2\theta}}\sqrt{1-2\sin^2{\theta}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2\sin^2{\theta}}+\sqrt{\cos{2\theta}}\sqrt{\cos{2\theta}}\right\}}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1-2\sin^2{A}=\cos{2A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2}\sin{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{2\sin^2{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{1-\cos{2\theta}+\cos{2\theta}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\sin^{-1}{\left\{1\right\}}\)
\(=\sin^{-1}{\left\{\sin{\frac{\pi}{2}}\right\}}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((c)\)
দেওয়া আছে,
\(g(x)=\sec{x}\)
\(\therefore g(4\theta)=\sec{4\theta}, \ g(2\theta)=\sec{2\theta}\)
প্রদত্ত সমীকরণ,
\(g(4\theta)-g(2\theta)=2\) যখন, \(0\lt{x}\lt{180^{o}}\)
\(\Rightarrow \sec{4\theta}-\sec{2\theta}=2\) ➜ \(\because g(x)=\sec{x}\)
\(\therefore g(4\theta)=\sec{4\theta}, \ g(2\theta)=\sec{2\theta}\)
\(\Rightarrow \frac{1}{\cos{4\theta}}-\frac{1}{\cos{2\theta}}=2\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos{2\theta}-\cos{4\theta}}{\cos{4\theta}\cos{2\theta}}=2\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=2\cos{4\theta}\cos{2\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{(4\theta-2\theta)}+\cos{(4\theta+2\theta)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}=\cos{2\theta}+\cos{6\theta}\)
\(\Rightarrow \cos{2\theta}-\cos{4\theta}-\cos{2\theta}-\cos{6\theta}=0\)
\(\Rightarrow -\cos{4\theta}-\cos{6\theta}=0\)
\(\Rightarrow \cos{6\theta}+\cos{4\theta}=0\) ➜ উভয় পার্শে \(-1\) গুণ করে,
\(\Rightarrow 2\cos{\frac{6\theta+4\theta}{2}}\cos{\frac{6\theta-4\theta}{2}}=0\) ➜ \(\because \cos{C}+\cos{D}=2\cos{\frac{C+D}{2}}\cos{\frac{C-D}{2}}\)
\(\Rightarrow 2\cos{\frac{10\theta}{2}}\cos{\frac{2\theta}{2}}=0\)
\(\Rightarrow 2\cos{5\theta}\cos{\theta}=0\)
\(\Rightarrow \cos{5\theta}=0, \ \cos{\theta}=0\)
\(\Rightarrow 5\theta=(2n+1)\frac{\pi}{2}, \ \theta=(2n+1)\frac{\pi}{2}\) ➜ \(\because \cos{A}=0\)
\(\Rightarrow A=(2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=(2n+1)\frac{\pi}{10}, \ (2n+1)\frac{\pi}{2}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(0\le{\theta}\le{180^{o}}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{10}, \ \frac{\pi}{2}\) \(\Rightarrow \theta=18^{o}, \ 90^{o}\) সবগুলি মান গ্রহনযোগ্য।
\(\therefore \theta=18^{o}, \ 90^{o}\)
যখন, \(n=1,\) \(\theta=\frac{3\pi}{10}, \ \frac{3\pi}{2}\)
\(\Rightarrow \theta=54^{o}, \ 270^{o}\) মান দুইটি গ্রহনযোগ্য।
\(\therefore \theta=54^{o}, \ 270^{o}\)
যখন, \(n=-1,\) \(\theta=-\frac{\pi}{10}, \ -\frac{\pi}{2}\)
\(\Rightarrow \theta=-18^{o}, \ -90^{o}\) কোনো মান গ্রহনযোগ্য নয়।
যখন, \(n=2,\) \(\theta=\frac{5\pi}{10}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=\frac{\pi}{2}, \ \frac{5\pi}{2}\)
\(\Rightarrow \theta=90^{o}, \ 450^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=90^{o}\)
যখন, \(n=3,\) \(\theta=\frac{7\pi}{10}, \ \frac{7\pi}{2}\)
\(\Rightarrow \theta=126^{o}, \ 630^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=126^{o}\)
যখন, \(n=4,\) \(\theta=\frac{9\pi}{10}, \ \frac{9\pi}{2}\)
\(\Rightarrow \theta=162^{o}, \ 810^{o}\) প্রথম মানটি গ্রহনযোগ্য।
\(\therefore \theta=162^{o}\)
যখন, \(n=5,\) \(\theta=\frac{11\pi}{10}, \ \frac{11\pi}{2}\)
\(\Rightarrow \theta=198^{o}, \ 990^{o}\) কোনো মান গ্রহনযোগ্য নয়।
\(\therefore 0\le{\theta}\le{360^{o}}\) ব্যাবধিতে নির্ণেয় সমাধান, \(18^{o}, \ 54^{o}, \ 90^{o}, \ 126^{o}\) এবং \(162^{o}\)
\(Q.5.(xx)\) \(\cot{\theta}+\tan{\theta}=2\sec{\theta}\) একটি ত্রিকোণমিতিক সমীকরণ এবং \(f(x)=\sin^{-1}{\left\{\sqrt{2(1-x^2)}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\((a)\) প্রমাণ কর যে, \(\sin{\cot^{-1}{\tan{\cos^{-1}{\frac{3}{4}}}}}=\frac{3}{4}\)
\((b)\) দেখাও যে, \(f(\cos{\theta})=\frac{\pi}{2}\)
\((c)\) \(-2\pi\lt{\theta}\lt{2\pi}\) সীমার মধ্যে ত্রিকোণমিতিক সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
\((a)\) প্রমাণ কর যে, \(\sin{\cot^{-1}{\tan{\cos^{-1}{\frac{3}{4}}}}}=\frac{3}{4}\)
\((b)\) দেখাও যে, \(f(\cos{\theta})=\frac{\pi}{2}\)
\((c)\) \(-2\pi\lt{\theta}\lt{2\pi}\) সীমার মধ্যে ত্রিকোণমিতিক সমীকরণটি সমাধান কর।
উত্তরঃ \((c) \ -\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}\) এবং \(\frac{5\pi}{6}\)
সমাধানঃ
\((a)\)
\(L.S=\sin{\cot^{-1}{\tan{\cos^{-1}{\frac{3}{4}}}}}\)
\(=\sin{\cot^{-1}{\tan{\tan^{-1}{\frac{\sqrt{7}}{3}}}}}\) ➜ এখানে,
\(\text{ভূমি}=3, \ \text{অতিভুজ}=4\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{4^2-3^2} \)
\(=\sqrt{16-9} \)
\(=\sqrt{7}\)
\(\therefore \cos^{-1}{\frac{3}{4}}=\tan^{-1}{\frac{\sqrt{7}}{3}}\)
\(=\sin{\cot^{-1}{\frac{\sqrt{7}}{3}}}\)
\(=\sin{\sin^{-1}{\frac{3}{4}}}\) ➜ এখানে,
\(\text{ভূমি}=\sqrt{7}, \ \text{লম্ব}=3\)
\(\therefore \text{অতিভুজ}=\sqrt{(\text{লম্ব})^2+(\text{ভূমি})^2} \)
\(=\sqrt{3^2+(\sqrt{7})^2} \)
\(=\sqrt{9+7} \)
\(=\sqrt{16}\)
\(=4\)
\(\therefore \cot^{-1}{\frac{\sqrt{7}}{3}}=\sin^{-1}{\frac{3}{4}}\)
\(=\frac{3}{4}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(f(x)=\sin^{-1}{\left\{\sqrt{2(1-x^2)}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
এখন,
\(L.S=f(\cos{\theta})\)
\(=\sin^{-1}{\left\{\sqrt{2(1-\cos^2{\theta})}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\) ➜ \(\because f(x)=\sin^{-1}{\left\{\sqrt{2(1-x^2)}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\(\therefore f(\cos{\theta})=\sin^{-1}{\left\{\sqrt{2(1-\cos^2{\theta})}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2\sin^2{\theta}}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\) ➜ \(\because 1-\cos^2{A}=\sin^2{A}\)
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}+\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\) ➜ \(\because 1-\cos^2{A}=\sin^2{A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-(\sqrt{\cos{2\theta}})^2}+\sqrt{\cos{2\theta}}\sqrt{1-(\sqrt{2}\sin{\theta})^2}\right\}}\) ➜ \(\because \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{\left\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-\cos{2\theta}}+\sqrt{\cos{2\theta}}\sqrt{1-2\sin^2{\theta}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2\sin^2{\theta}}+\sqrt{\cos{2\theta}}\sqrt{\cos{2\theta}}\right\}}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1-2\sin^2{A}=\cos{2A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2}\sin{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{2\sin^2{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{1-\cos{2\theta}+\cos{2\theta}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\sin^{-1}{\left\{1\right\}}\)
\(=\sin^{-1}{\left\{\sin{\frac{\pi}{2}}\right\}}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((c)\)
প্রদত্ত সমীকরণ,
\(\cot{\theta}+\tan{\theta}=2\sec{\theta}, \ -2\pi\lt{\theta}\lt{2\pi}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}+\frac{\sin{\theta}}{\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
\(\Rightarrow \frac{1}{\sin{\theta}}=2, \ \because \cos{\theta}\ne{0}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{\theta}=\sin{\alpha}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore \theta=\frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{7\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{7\pi}{6}\)
যখন, \(n=2,\) \(\theta=2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{13\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(\theta=-2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{11\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{17\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন,\(n=-3,\) \(\theta=-3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{19\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
\(L.S=\sin{\cot^{-1}{\tan{\cos^{-1}{\frac{3}{4}}}}}\)
\(=\sin{\cot^{-1}{\tan{\tan^{-1}{\frac{\sqrt{7}}{3}}}}}\) ➜ এখানে,
\(\text{ভূমি}=3, \ \text{অতিভুজ}=4\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\text{ভূমি})^2} \)
\(=\sqrt{4^2-3^2} \)
\(=\sqrt{16-9} \)
\(=\sqrt{7}\)
\(\therefore \cos^{-1}{\frac{3}{4}}=\tan^{-1}{\frac{\sqrt{7}}{3}}\)
\(=\sin{\cot^{-1}{\frac{\sqrt{7}}{3}}}\)
\(=\sin{\sin^{-1}{\frac{3}{4}}}\) ➜ এখানে,
\(\text{ভূমি}=\sqrt{7}, \ \text{লম্ব}=3\)
\(\therefore \text{অতিভুজ}=\sqrt{(\text{লম্ব})^2+(\text{ভূমি})^2} \)
\(=\sqrt{3^2+(\sqrt{7})^2} \)
\(=\sqrt{9+7} \)
\(=\sqrt{16}\)
\(=4\)
\(\therefore \cot^{-1}{\frac{\sqrt{7}}{3}}=\sin^{-1}{\frac{3}{4}}\)
\(=\frac{3}{4}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(প্রমাণিত)
\((b)\)
দেওয়া আছে,
\(f(x)=\sin^{-1}{\left\{\sqrt{2(1-x^2)}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
এখন,
\(L.S=f(\cos{\theta})\)
\(=\sin^{-1}{\left\{\sqrt{2(1-\cos^2{\theta})}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\) ➜ \(\because f(x)=\sin^{-1}{\left\{\sqrt{2(1-x^2)}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\(\therefore f(\cos{\theta})=\sin^{-1}{\left\{\sqrt{2(1-\cos^2{\theta})}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2\sin^2{\theta}}\right\}}+\sin^{-1}{\left\{\sqrt{\cos{(2\theta)}}\right\}}\) ➜ \(\because 1-\cos^2{A}=\sin^2{A}\)
\(=\sin^{-1}{\left(\sqrt{2}\sin{\theta}\right)}+\sin^{-1}{\left(\sqrt{\cos{2\theta}}\right)}\) ➜ \(\because 1-\cos^2{A}=\sin^2{A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-(\sqrt{\cos{2\theta}})^2}+\sqrt{\cos{2\theta}}\sqrt{1-(\sqrt{2}\sin{\theta})^2}\right\}}\) ➜ \(\because \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{\left\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{1-\cos{2\theta}}+\sqrt{\cos{2\theta}}\sqrt{1-2\sin^2{\theta}}\right\}}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2\sin^2{\theta}}+\sqrt{\cos{2\theta}}\sqrt{\cos{2\theta}}\right\}}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
এবং \(1-2\sin^2{A}=\cos{2A}\)
\(=\sin^{-1}{\left\{\sqrt{2}\sin{\theta}\sqrt{2}\sin{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{2\sin^2{\theta}+\cos{2\theta}\right\}}\)
\(=\sin^{-1}{\left\{1-\cos{2\theta}+\cos{2\theta}\right\}}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\sin^{-1}{\left\{1\right\}}\)
\(=\sin^{-1}{\left\{\sin{\frac{\pi}{2}}\right\}}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((c)\)
প্রদত্ত সমীকরণ,
\(\cot{\theta}+\tan{\theta}=2\sec{\theta}, \ -2\pi\lt{\theta}\lt{2\pi}\)
\(\Rightarrow \frac{\cos{\theta}}{\sin{\theta}}+\frac{\sin{\theta}}{\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(\tan{A}=\frac{\sin{A}}{\cos{A}}\)
এবং \(\sec{A}=\frac{1}{\cos{A}}\)
\(\Rightarrow \frac{\cos^2{\theta}+\sin^2{\theta}}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\)
\(\Rightarrow \frac{1}{\sin{\theta}\cos{\theta}}=\frac{2}{\cos{\theta}}\) ➜ \(\because \cos^2{A}+\sin^2{A}=1\)
\(\Rightarrow \frac{1}{\sin{\theta}}=2, \ \because \cos{\theta}\ne{0}\)
\(\Rightarrow \sin{\theta}=\frac{1}{2}\)
\(\Rightarrow \sin{\theta}=\sin{\frac{\pi}{6}}\) ➜ \(\because \frac{1}{2}=\sin{\frac{\pi}{6}}\)
\(\Rightarrow \theta=n\pi+(-1)^n\frac{\pi}{6}\) ➜ \(\because \sin{\theta}=\sin{\alpha}\)
\(\Rightarrow \theta=n\pi+(-1)^n\alpha\) যেখানে, \(n\in{\mathbb{Z}}\)
\(\therefore\) সাধারণ সমাধান, \(\theta=n\pi+(-1)^n\frac{\pi}{6}\) যেখানে, \(n\in{\mathbb{Z}}\)
\(-2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে \(\theta\) এর মান নির্ণয়ঃ
যখন, \(n=0,\) \(\theta=\frac{\pi}{6}\) মানটি গ্রহনযোগ্য।\(\therefore \theta=\frac{\pi}{6}\)
যখন, \(n=1,\) \(\theta=\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=\frac{5\pi}{6}\)
যখন, \(n=-1,\) \(\theta=-\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{7\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{7\pi}{6}\)
যখন, \(n=2,\) \(\theta=2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{13\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন, \(n=-2,\) \(\theta=-2\pi+\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{11\pi}{6}\) মানটি গ্রহনযোগ্য।
\(\therefore \theta=-\frac{11\pi}{6}\)
যখন, \(n=3,\) \(\theta=3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{17\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
যখন,\(n=-3,\) \(\theta=-3\pi-\frac{\pi}{6}\)
\(\Rightarrow \theta=-\frac{19\pi}{6}\) মানটি গ্রহনযোগ্য নয়।
\(\therefore -2\pi\le{\theta}\le{2\pi}\) ব্যাবধিতে নির্ণেয় সমাধান, \(-\frac{11\pi}{6}, \ -\frac{7\pi}{6}, \ \frac{\pi}{6}, \ \frac{5\pi}{6}\)
\(Q.5.(xxi)\) \(A=\sin^{-1}{\frac{x}{y}}, \ B=\cos^{-1}{\frac{a}{x}}\) এবং \(C=\cos^{-1}{\frac{b}{y}}\)
\((a)\) দেখাও যে, \(\{\cos{(\sin^{-1}{x})}\}^2=\{\sin{(\cos^{-1}{x})}\}^2\)
\((b)\) দেখাও যে, \(B+C=\cos^{-1}{\left\{\frac{ab}{xy}-\frac{1}{xy}\sqrt{(x^2-a^2)(y^2-b^2)}\right\}}\)
\((c)\) \(A+B+C=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(x^4-x^2y^2-2abx^2+a^2y^2+b^2x^2=0\)
\((a)\) দেখাও যে, \(\{\cos{(\sin^{-1}{x})}\}^2=\{\sin{(\cos^{-1}{x})}\}^2\)
\((b)\) দেখাও যে, \(B+C=\cos^{-1}{\left\{\frac{ab}{xy}-\frac{1}{xy}\sqrt{(x^2-a^2)(y^2-b^2)}\right\}}\)
\((c)\) \(A+B+C=\frac{\pi}{2}\) হলে, প্রমাণ কর যে, \(x^4-x^2y^2-2abx^2+a^2y^2+b^2x^2=0\)
সমাধানঃ
\((a)\)
\(L.S=\left\{\cos{(\sin^{-1}{x})}\right\}^2\)
\(=\left\{\cos{\left(\cos^{-1}{\sqrt{1-x^2}}\right)}\right\}^2\) ➜ \(\because \sin^{-1}{A}=\cos^{-1}{\left(\sqrt{1-A^2}\right)}\)
\(=\left\{\sqrt{1-x^2}\right\}^2\)
\(=1-x^2\)
আবার,
\(R.S=\left\{\sin{(\cos^{-1}{x})}\right\}^2\)
\(=\left\{\sin{\left(\sin^{-1}{\sqrt{1-x^2}}\right)}\right\}^2\) ➜ \(\because \cos^{-1}{A}=\sin^{-1}{\left(\sqrt{1-A^2}\right)}\)
\(=\left\{\sqrt{1-x^2}\right\}^2\)
\(=1-x^2\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((b)\)
দেওয়া আছে,
\(B=\cos^{-1}{\frac{a}{x}}\) এবং \(C=\cos^{-1}{\frac{b}{y}}\)
এখন,
\(L.S=B+C\)
\(=\cos^{-1}{\frac{a}{x}}+\cos^{-1}{\frac{b}{y}}\) ➜ \(\because B=\cos^{-1}{\frac{a}{x}}\)
এবং \(C=\cos^{-1}{\frac{b}{y}}\)
\(=\cos^{-1}{\left\{\frac{a}{x}\times\frac{b}{y}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}\right\}}\) ➜ \(\because \cos^{-1}{a}+\cos^{-1}{b}=\cos^{-1}{\left\{ab-\sqrt{\left(1-a^2\right)\left(1-b^2\right)}\right\}}\)
\(=\cos^{-1}{\left\{\frac{ab}{xy}-\sqrt{\left(\frac{x^2-a^2}{x^2}\right)\left(\frac{y^2-b^2}{y^2}\right)}\right\}}\)
\(=\cos^{-1}{\left\{\frac{ab}{xy}-\sqrt{\frac{(x^2-a^2)(y^2-b^2)}{x^2y^2}}\right\}}\)
\(=\cos^{-1}{\left\{\frac{ab}{xy}-\frac{1}{xy}\sqrt{(x^2-a^2)(y^2-b^2)}\right\}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(A=\sin^{-1}{\frac{x}{y}}, \ B=\cos^{-1}{\frac{a}{x}}\) এবং \(C=\cos^{-1}{\frac{b}{y}}\)
প্রদত্ত সমীকরণ,
\(A+B+C=\frac{\pi}{2}\) \(\Rightarrow B+C=\frac{\pi}{2}-A\)
\(\Rightarrow \cos^{-1}{\frac{a}{x}}+\cos^{-1}{\frac{b}{y}}=\frac{\pi}{2}-A\) ➜ \(\because B=\cos^{-1}{\frac{a}{x}}\)
এবং \(C=\cos^{-1}{\frac{b}{y}}\)
\(\Rightarrow \cos^{-1}{\left\{\frac{a}{x}\times\frac{b}{y}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}\right\}}=\frac{\pi}{2}-A\) ➜ \(\because \cos^{-1}{a}+\cos^{-1}{b}=\cos^{-1}{\left\{ab-\sqrt{\left(1-a^2\right)\left(1-b^2\right)}\right\}}\)
\(\Rightarrow \frac{ab}{xy}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}=\cos{\left(\frac{\pi}{2}-A\right)}\)
\(\Rightarrow \frac{ab}{xy}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}=\sin{A}\)
\(\Rightarrow \frac{ab}{xy}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}=\frac{x}{y}\) ➜ \(\because A=\sin^{-1}{\frac{x}{y}}\)
\(\therefore \sin{A}=\frac{x}{y}\)
\(\Rightarrow \frac{ab}{xy}-\frac{x}{y}=\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}\)
\(\Rightarrow \left(\frac{ab}{xy}-\frac{x}{y}\right)^2=\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{a^2b^2}{x^2y^2}-2.\frac{ab}{xy}.\frac{x}{y}+\frac{x^2}{y^2}=1-\frac{a^2}{x^2}-\frac{b^2}{y^2}+\frac{a^2b^2}{x^2y^2}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(\Rightarrow \frac{a^2b^2}{x^2y^2}-\frac{2ab}{y^2}+\frac{x^2}{y^2}-1+\frac{a^2}{x^2}+\frac{b^2}{y^2}-\frac{a^2b^2}{x^2y^2}=0\)
\(\Rightarrow \frac{a^2}{x^2}-\frac{2ab}{y^2}+\frac{x^2}{y^2}+\frac{b^2}{y^2}-1=0\)
\(\Rightarrow \frac{x^2}{y^2}-\frac{2ab}{y^2}+\frac{a^2}{x^2}+\frac{b^2}{y^2}-1=0\)
\(\Rightarrow x^4-2abx^2+a^2y^2+b^2x^2-x^2y^2=0\) ➜ উভয় পার্শে \(x^2y^2\) গুণ করে,
\(\therefore x^4-x^2y^2-2abx^2+a^2y^2+b^2x^2=0\)
(প্রমাণিত)
\(L.S=\left\{\cos{(\sin^{-1}{x})}\right\}^2\)
\(=\left\{\cos{\left(\cos^{-1}{\sqrt{1-x^2}}\right)}\right\}^2\) ➜ \(\because \sin^{-1}{A}=\cos^{-1}{\left(\sqrt{1-A^2}\right)}\)
\(=\left\{\sqrt{1-x^2}\right\}^2\)
\(=1-x^2\)
আবার,
\(R.S=\left\{\sin{(\cos^{-1}{x})}\right\}^2\)
\(=\left\{\sin{\left(\sin^{-1}{\sqrt{1-x^2}}\right)}\right\}^2\) ➜ \(\because \cos^{-1}{A}=\sin^{-1}{\left(\sqrt{1-A^2}\right)}\)
\(=\left\{\sqrt{1-x^2}\right\}^2\)
\(=1-x^2\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((b)\)
দেওয়া আছে,
\(B=\cos^{-1}{\frac{a}{x}}\) এবং \(C=\cos^{-1}{\frac{b}{y}}\)
এখন,
\(L.S=B+C\)
\(=\cos^{-1}{\frac{a}{x}}+\cos^{-1}{\frac{b}{y}}\) ➜ \(\because B=\cos^{-1}{\frac{a}{x}}\)
এবং \(C=\cos^{-1}{\frac{b}{y}}\)
\(=\cos^{-1}{\left\{\frac{a}{x}\times\frac{b}{y}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}\right\}}\) ➜ \(\because \cos^{-1}{a}+\cos^{-1}{b}=\cos^{-1}{\left\{ab-\sqrt{\left(1-a^2\right)\left(1-b^2\right)}\right\}}\)
\(=\cos^{-1}{\left\{\frac{ab}{xy}-\sqrt{\left(\frac{x^2-a^2}{x^2}\right)\left(\frac{y^2-b^2}{y^2}\right)}\right\}}\)
\(=\cos^{-1}{\left\{\frac{ab}{xy}-\sqrt{\frac{(x^2-a^2)(y^2-b^2)}{x^2y^2}}\right\}}\)
\(=\cos^{-1}{\left\{\frac{ab}{xy}-\frac{1}{xy}\sqrt{(x^2-a^2)(y^2-b^2)}\right\}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(A=\sin^{-1}{\frac{x}{y}}, \ B=\cos^{-1}{\frac{a}{x}}\) এবং \(C=\cos^{-1}{\frac{b}{y}}\)
প্রদত্ত সমীকরণ,
\(A+B+C=\frac{\pi}{2}\) \(\Rightarrow B+C=\frac{\pi}{2}-A\)
\(\Rightarrow \cos^{-1}{\frac{a}{x}}+\cos^{-1}{\frac{b}{y}}=\frac{\pi}{2}-A\) ➜ \(\because B=\cos^{-1}{\frac{a}{x}}\)
এবং \(C=\cos^{-1}{\frac{b}{y}}\)
\(\Rightarrow \cos^{-1}{\left\{\frac{a}{x}\times\frac{b}{y}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}\right\}}=\frac{\pi}{2}-A\) ➜ \(\because \cos^{-1}{a}+\cos^{-1}{b}=\cos^{-1}{\left\{ab-\sqrt{\left(1-a^2\right)\left(1-b^2\right)}\right\}}\)
\(\Rightarrow \frac{ab}{xy}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}=\cos{\left(\frac{\pi}{2}-A\right)}\)
\(\Rightarrow \frac{ab}{xy}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}=\sin{A}\)
\(\Rightarrow \frac{ab}{xy}-\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}=\frac{x}{y}\) ➜ \(\because A=\sin^{-1}{\frac{x}{y}}\)
\(\therefore \sin{A}=\frac{x}{y}\)
\(\Rightarrow \frac{ab}{xy}-\frac{x}{y}=\sqrt{\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)}\)
\(\Rightarrow \left(\frac{ab}{xy}-\frac{x}{y}\right)^2=\left(1-\frac{a^2}{x^2}\right)\left(1-\frac{b^2}{y^2}\right)\) ➜ উভয় পার্শে বর্গ করে,
\(\Rightarrow \frac{a^2b^2}{x^2y^2}-2.\frac{ab}{xy}.\frac{x}{y}+\frac{x^2}{y^2}=1-\frac{a^2}{x^2}-\frac{b^2}{y^2}+\frac{a^2b^2}{x^2y^2}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(\Rightarrow \frac{a^2b^2}{x^2y^2}-\frac{2ab}{y^2}+\frac{x^2}{y^2}-1+\frac{a^2}{x^2}+\frac{b^2}{y^2}-\frac{a^2b^2}{x^2y^2}=0\)
\(\Rightarrow \frac{a^2}{x^2}-\frac{2ab}{y^2}+\frac{x^2}{y^2}+\frac{b^2}{y^2}-1=0\)
\(\Rightarrow \frac{x^2}{y^2}-\frac{2ab}{y^2}+\frac{a^2}{x^2}+\frac{b^2}{y^2}-1=0\)
\(\Rightarrow x^4-2abx^2+a^2y^2+b^2x^2-x^2y^2=0\) ➜ উভয় পার্শে \(x^2y^2\) গুণ করে,
\(\therefore x^4-x^2y^2-2abx^2+a^2y^2+b^2x^2=0\)
(প্রমাণিত)
\(Q.5.(xxii)\) \(P=\sin^{-1}{\frac{1}{\sqrt{5}}}-\frac{1}{2}\cos^{-1}{\frac{4}{5}}+\tan^{-1}{\frac{1}{3}}\) এবং \(\tan^{-1}{x}=A\)
\((a)\) \(\frac{1}{2}\sin^{-1}{x}\) কে \(\tan^{-1}{y}\) এ রুপান্তর কর, যেখানে \(y=\frac{x}{1+\sqrt{1-x^2}}\)
\((b)\) দেখাও যে, \(5A=\tan^{-1}{\left\{\frac{x^5-10x^3+5x}{5x^4-10x^2+1}\right\}}\)
\((c)\) প্রমাণ কর যে, \(P=\tan^{-1}{\frac{1}{2}}\)
\((a)\) \(\frac{1}{2}\sin^{-1}{x}\) কে \(\tan^{-1}{y}\) এ রুপান্তর কর, যেখানে \(y=\frac{x}{1+\sqrt{1-x^2}}\)
\((b)\) দেখাও যে, \(5A=\tan^{-1}{\left\{\frac{x^5-10x^3+5x}{5x^4-10x^2+1}\right\}}\)
\((c)\) প্রমাণ কর যে, \(P=\tan^{-1}{\frac{1}{2}}\)
সমাধানঃ
\((a)\)
ধরি,
\(\sin^{-1}{x}=\theta\)
\(\therefore x=\sin{\theta}\)
এখন,
\(\tan{\frac{\theta}{2}}=\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{2\cos^2{\frac{\theta}{2}}}\) ➜ লব ও হরকে \(2\cos{\frac{\theta}{2}}\) দ্বারা গুণ করে,
\(=\frac{\sin{\theta}}{1+\cos{\theta}}\) ➜ \(\because 2\sin{\frac{A}{2}}\cos{\frac{A}{2}}=\sin{A}\)
এবং \(2\cos^2{\frac{A}{2}}=1+\cos{A}\)
\(=\frac{\sin{\theta}}{1+\sqrt{1-\sin^2{\theta}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
\(=\frac{x}{1+\sqrt{1-x^2}}\) ➜ \(\because \sin{\theta}=x\)
\(\therefore \tan{\frac{\theta}{2}}=\frac{x}{1+\sqrt{1-x^2}}\)
\(\Rightarrow \frac{\theta}{2}=\tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}\)
\(\Rightarrow \frac{1}{2}\theta=\tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}\)
\(\Rightarrow \frac{1}{2}\sin^{-1}{x}=\tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}=\tan^{-1}{y}\)
\(\Rightarrow \tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}=\tan^{-1}{y}\)
\(\Rightarrow \frac{x}{1+\sqrt{1-x^2}}=y\)
\(\therefore y=\frac{x}{1+\sqrt{1-x^2}}\)
(দেখানো হলো)
\((b)\)
দেওয়া আছে,
\(\tan^{-1}{x}=A\)
এখন,
\(L.S=5A\)
\(=5\tan^{-1}{x}\) ➜ \(\because A=\tan^{-1}{x}\)
\(=\tan^{-1}{x}+4\tan^{-1}{x}\)
\(=\tan^{-1}{x}+2\times2\tan^{-1}{x}\)
\(=\tan^{-1}{x}+2\tan^{-1}{\frac{2x}{1-x^2}}\) ➜ \(\because 2\tan^{-1}{a}=\tan^{-1}{\frac{2a}{1-a^2}}\)
\(=\tan^{-1}{x}+\tan^{-1}{\left(\frac{2\frac{2x}{1-x^2}}{1-\frac{4x^2}{(1-x^2)^2}}\right)}\) ➜ \(\because 2\tan^{-1}{a}=\tan^{-1}{\frac{2a}{1-a^2}}\)
\(=\tan^{-1}{x}+\tan^{-1}{\left\{\frac{4x(1-x^2)}{(1-x^2)^2-4x^2}\right\}}\) ➜ লব ও হরকে \((1-a^2)^2\) দ্বারা গুণ করে,
\(=\tan^{-1}{x}+\tan^{-1}{\left\{\frac{4x-4x^3}{1-2x^2+x^4-4x^2}\right\}}\)
\(=\tan^{-1}{x}+\tan^{-1}{\left\{\frac{4x-4x^3}{1-6x^2+x^4}\right\}}\)
\(=\tan^{-1}{\left\{\frac{x+\frac{4x-4x^3}{1-6x^2+x^4}}{1-x\times\frac{4x-4x^3}{1-6x^2+x^4}}\right\}}\) ➜ \(\because \tan^{-1}{a}+\tan^{-1}{b}=\tan^{-1}{\frac{a+b}{1-ab}}\)
\(=\tan^{-1}{\left\{\frac{x+\frac{4x-4x^3}{1-6x^2+x^4}}{1-\frac{4x^2-4x^4}{1-6x^2+x^4}}\right\}}\)
\(=\tan^{-1}{\left\{\frac{x-6x^3+x^5+4x-4x^3}{1-6x^2+x^4-4x^2+4x^4}\right\}}\) ➜ লব ও হরকে \((1-6x^2+x^4)\) দ্বারা গুণ করে,
\(=\tan^{-1}{\left\{\frac{x^5-10x^3+5x}{5x^4-10x^2+1}\right\}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(P=\sin^{-1}{\frac{1}{\sqrt{5}}}-\frac{1}{2}\cos^{-1}{\frac{4}{5}}+\tan^{-1}{\frac{1}{3}}\)
\(=\tan^{-1}{\frac{1}{2}}-\frac{1}{2}\tan^{-1}{\frac{3}{4}}+\frac{1}{2}\times2\tan^{-1}{\frac{1}{3}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=4 , \ \text{অতিভুজ}=5\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\te
ধরি,
\(\sin^{-1}{x}=\theta\)
\(\therefore x=\sin{\theta}\)
এখন,
\(\tan{\frac{\theta}{2}}=\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{2\cos^2{\frac{\theta}{2}}}\) ➜ লব ও হরকে \(2\cos{\frac{\theta}{2}}\) দ্বারা গুণ করে,
\(=\frac{\sin{\theta}}{1+\cos{\theta}}\) ➜ \(\because 2\sin{\frac{A}{2}}\cos{\frac{A}{2}}=\sin{A}\)
এবং \(2\cos^2{\frac{A}{2}}=1+\cos{A}\)
\(=\frac{\sin{\theta}}{1+\sqrt{1-\sin^2{\theta}}}\) ➜ \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
\(=\frac{x}{1+\sqrt{1-x^2}}\) ➜ \(\because \sin{\theta}=x\)
\(\therefore \tan{\frac{\theta}{2}}=\frac{x}{1+\sqrt{1-x^2}}\)
\(\Rightarrow \frac{\theta}{2}=\tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}\)
\(\Rightarrow \frac{1}{2}\theta=\tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}\)
\(\Rightarrow \frac{1}{2}\sin^{-1}{x}=\tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}=\tan^{-1}{y}\)
\(\Rightarrow \tan^{-1}{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}=\tan^{-1}{y}\)
\(\Rightarrow \frac{x}{1+\sqrt{1-x^2}}=y\)
\(\therefore y=\frac{x}{1+\sqrt{1-x^2}}\)
(দেখানো হলো)
\((b)\)
দেওয়া আছে,
\(\tan^{-1}{x}=A\)
এখন,
\(L.S=5A\)
\(=5\tan^{-1}{x}\) ➜ \(\because A=\tan^{-1}{x}\)
\(=\tan^{-1}{x}+4\tan^{-1}{x}\)
\(=\tan^{-1}{x}+2\times2\tan^{-1}{x}\)
\(=\tan^{-1}{x}+2\tan^{-1}{\frac{2x}{1-x^2}}\) ➜ \(\because 2\tan^{-1}{a}=\tan^{-1}{\frac{2a}{1-a^2}}\)
\(=\tan^{-1}{x}+\tan^{-1}{\left(\frac{2\frac{2x}{1-x^2}}{1-\frac{4x^2}{(1-x^2)^2}}\right)}\) ➜ \(\because 2\tan^{-1}{a}=\tan^{-1}{\frac{2a}{1-a^2}}\)
\(=\tan^{-1}{x}+\tan^{-1}{\left\{\frac{4x(1-x^2)}{(1-x^2)^2-4x^2}\right\}}\) ➜ লব ও হরকে \((1-a^2)^2\) দ্বারা গুণ করে,
\(=\tan^{-1}{x}+\tan^{-1}{\left\{\frac{4x-4x^3}{1-2x^2+x^4-4x^2}\right\}}\)
\(=\tan^{-1}{x}+\tan^{-1}{\left\{\frac{4x-4x^3}{1-6x^2+x^4}\right\}}\)
\(=\tan^{-1}{\left\{\frac{x+\frac{4x-4x^3}{1-6x^2+x^4}}{1-x\times\frac{4x-4x^3}{1-6x^2+x^4}}\right\}}\) ➜ \(\because \tan^{-1}{a}+\tan^{-1}{b}=\tan^{-1}{\frac{a+b}{1-ab}}\)
\(=\tan^{-1}{\left\{\frac{x+\frac{4x-4x^3}{1-6x^2+x^4}}{1-\frac{4x^2-4x^4}{1-6x^2+x^4}}\right\}}\)
\(=\tan^{-1}{\left\{\frac{x-6x^3+x^5+4x-4x^3}{1-6x^2+x^4-4x^2+4x^4}\right\}}\) ➜ লব ও হরকে \((1-6x^2+x^4)\) দ্বারা গুণ করে,
\(=\tan^{-1}{\left\{\frac{x^5-10x^3+5x}{5x^4-10x^2+1}\right\}}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(দেখানো হলো)
\((c)\)
দেওয়া আছে,
\(P=\sin^{-1}{\frac{1}{\sqrt{5}}}-\frac{1}{2}\cos^{-1}{\frac{4}{5}}+\tan^{-1}{\frac{1}{3}}\)
\(=\tan^{-1}{\frac{1}{2}}-\frac{1}{2}\tan^{-1}{\frac{3}{4}}+\frac{1}{2}\times2\tan^{-1}{\frac{1}{3}}\) ➜ প্রথম পদের ক্ষেত্রে,
\(\text{লম্ব}=1, \ \text{অতিভুজ}=\sqrt{5}\)
\(\therefore \text{ভূমি}=\sqrt{(\text{অতিভুজ})^2-(\text{লম্ব})^2} \)
\(=\sqrt{(\sqrt{5})^2-1^2} \)
\(=\sqrt{5-1} \)
\(=\sqrt{4}\)
\(=2\)
\(\therefore \sin^{-1}{\frac{1}{\sqrt{5}}}=\tan^{-1}{\frac{1}{2}}\)
দ্বিতীয় পদের ক্ষেত্রে,
\(\text{ভূমি}=4 , \ \text{অতিভুজ}=5\)
\(\therefore \text{লম্ব}=\sqrt{(\text{অতিভুজ})^2-(\te