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Trigonometric Equations · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 22 questions.

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Question 13 Marks
Find the general solutions of the following equations: $\sin2\text{x}=\cos3\text{x}$
Answer
We have, $\sin2\text{x}=\cos3\text{x}$ $\Rightarrow\cos3\text{x}=\cos\Big(\frac{\pi}{2}-2\text{x}\Big)\Big[\because\cos\Big(\frac{\pi}{2}-\text{x}\Big)=\sin\text{x}\Big]$ $\Rightarrow3\text{x}=2\text{n}\pi\pm\Big(\frac{\pi}{2}-2\text{x}\Big)\text{n}\in\text{z}$ $\Rightarrow\text{ Either}$ $5\text{x}=2\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$ or$\text{x}=2\text{n}\pi-\frac{\pi}{2},\text{n}\in\text{z}$ $\Rightarrow5\text{x}(4\text{n}+1)\frac{\pi}{2},\text{n}\in\text{z}$ or $\Rightarrow\text{x}(4\text{n}-1)\frac{\pi}{2}$ $\Rightarrow\text{x}(4\text{n}-1)\frac{\pi}{10},\text{n}\in\text{z}$or$\Rightarrow\text{x}(4\text{n}-1)\frac{\pi}{2}\text{n}\in\text{z}$
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Question 23 Marks
Solve the following equations: $(\sqrt{3}-1)\cos\text{x}+(\sqrt{3}+1)\sin\text{x}=2$
Answer
We have, $(\sqrt{3}-1)\cos\text{x}+(\sqrt{3}+1)\sin\text{x}=2$ Divide on both side by $2\sqrt{2}$ $\frac{(\sqrt{3}-1)}{2\sqrt{2}}\cos\text{x}+\frac{(\sqrt{3}+1)}{2\sqrt{2}}\sin\text{x}=\frac{1}{\sqrt{2}}$ $\sin\Big(\text{x}+\tan^{-1}\Big(\frac{\sqrt{3}-1}{\sqrt{3}+1}\Big)\Big)=\sin\frac{\pi}{4}$ $\text{x}=2\text{n}\pi+\frac{\pi}{3}$ or $2\text{n}\pi-\frac{\pi}{6}\text{n}\in\text{z}$
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Question 33 Marks
Find the general solutions of the following equations: $\sin2\text{x}+\cos\text{x}=0$
Answer
We have, $\sin2\text{x}+\cos\text{x}=0$ $2\sin\text{x}\cos\text{x}+\cos\text{x}=0$ $\cos\text{x}\text{(2sinx+1)}=0$ $\cos\text{x}=0$ or $2\sin\text{x+1}=0$ $\text{x}=\text{(4m}-1)\frac{\pi}{2}$ or $\sin\text{x}=\frac{-1}{2}$ $\text{x}=\text{(4m}-1)\frac{\pi}{2}$ or $\text{x}=\text{(4n}-1)\frac{\pi}{6},\text{m,n}\in\text{z}$
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Question 43 Marks
Find the general solutions of the following equations: $\tan\text{px}=\cot\text{qx}$
Answer
We have, $\tan\text{px}=\cot\text{qx}$ $\Rightarrow\tan\text{px}=\tan\Big(\frac{\pi}{2}-\text{qx}\Big)$ $\Rightarrow\text{px}=\text{nx}\pm\Big(\frac{\pi}{2}-\text{qx}\Big),\text{n}\in\text{z}$ $\Rightarrow\text{(p+q)}\text{x}=\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$ $\Rightarrow\text{(p+q)}\text{x}=\text{(2n+1)}\frac{\pi}{2},\text{n}\in\text{z}$ $\Rightarrow\text{x}=\frac{\text{(2n+1)}}{\text{(p+q)}}\frac{\pi}{2},\text{n}\in\text{z}$
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Question 53 Marks
Solve the following equations: $\cos\text{x}+\cos3\text{x}-\cos2\text{x}=0$
Answer
$\cos\text{x}+\cos3\text{x}-\cos2\text{x}=0$ $\Rightarrow2\cos2\text{x}.\cos\text{x}-\cos2\text{x}=0$ $\Rightarrow\cos2\theta(2\cos\theta-1)=1$ Either $\cos2\theta=0\ \text{or}\ 2\cos\theta=1$ $\Rightarrow2\theta=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{ z}$or $\cos\theta=\frac{1}{2}=\cos\frac{\pi}{3}$ $\Rightarrow\theta=(2\text{n}+1)\frac{\pi}{4},\text{n }\in\ \text{z}$ or $\theta=2\text{m}\pi\pm\frac{\pi}{3},\text{m }\in\ \text{z}$
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Question 63 Marks
Solve the following equations: $\sec\text{x}\cos5\text{x}+1=0,0<\text{x}<\frac{\pi}{2}$
Answer
$\sec\text{x}\cos5\text{x}+1=0$ $\frac{\cos5\text{x}+\cos\text{x}}{\cos\text{x}}=0\Rightarrow\cos\text{x}\neq0$ $2\cos3\text{x}\cos2\text{x}=0$ $\cos3\text{x}=0$ or $\cos2\text{x}=0$ $3\text{x}=\frac{\pi}{2}$ or $2\text{x}=\frac{\pi}{2}$ $\text{x}=\frac{\pi}{4},\frac{\pi}{6}$
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Question 73 Marks
Solve the following equations: $\cos\text{x}+\cos2\text{x}+\cos3\text{x}=0$
Answer
$\cos\text{x}+\cos2\text{x}+\cos3\text{x}=0$ $\Rightarrow\cos2\text{x}+2\cos2\text{x}.\cos\text{x}=0$ $[\because\cos\text{x}+\cos3\text{x}=2\cos2\text{x}.\cos\text{x}]$ $\Rightarrow\cos2\text{x}(1+2\cos\text{x})=0$ Either $\cos2\text{x}=0$ or $1+2\cos\text{x}=0$ $\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=-\frac{1}{2}$ $\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=+\cos\Big(\pi-\frac{\pi}{3}\Big)$ or $\cos\text{x}=\cos2\frac{\pi}{3}$ or $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{z}$ Thus, $\text{x}=(2\text{n}+1)\frac{\pi}{4}$ or $\Big(2\text{n}\pi\pm\frac{2\pi}{3}\Big),\text{n}\in\text{z}$
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Question 83 Marks
Solve the following equations: $\sin3\text{x}-\sin\text{x}=4\cos^{2}\text{x}-2$
Answer
$\sin3\text{x}-\sin\text{x}=4\cos^{2}\text{x}-2$ $\Rightarrow2\cos2\text{x},\sin\text{x}=2(2\cos^{2}\text{x}-1)$ $\Rightarrow2\cos2\text{x},\sin\text{x}=2\cos2\text{x}$ $[\because\cos2\text{x}=2\cos^{2}\text{x}-1]$ $\Rightarrow2\cos2\text{x}(\sin\text{x}-1)=0$ Either $\cos2\text{x}=0$ or $\sin\text{x}-1=0$ $\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{z}$or $\sin\text{x}=1=\sin\frac{\pi}{2}$ $\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\text{x}=\text{m}\pi+(-1)^\text{m}\frac{\pi}{2},\text{m}\in\text{z}$ Thus, $\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\text{m}\pi+(-1)^\text{m}\frac{\pi}{2},\text{m}\in\text{z}$
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Question 93 Marks
Find the general solutions of the following equations: $\tan\text{mx}+\cot\text{nx}=0$
Answer
We have, $\tan\text{mx}+\cot\text{nx}=0$ $\sin\text{mx}\sin\text{nx}+\cos\text{mx}\cos\text{nx}=0$ $\cos\text{(m-n)}\text{x}=0$ $\text{(m-n)}\text{x}=\Big(\frac{2\text{k}+1}{2}\Big)\pi$ $\text{x}=\Big(\frac{2\text{k}+1}{2\text{(m-n)}}\Big)\pi,\text{k}\in\text{z}$
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Question 103 Marks
Solve the following equations: $\tan3\text{x}+\tan\text{x}=2\tan2\text{x}$
Answer
$\tan3\text{x}+\tan\text{x}=2\tan2\text{x}$ $\Rightarrow\tan3\text{x}-\tan2\text{x}=\tan2\text{x}-\tan\text{x}$ $\Rightarrow\tan3\text{x}-\tan2\text{x}=\tan2\text{x}-\tan\text{x}$ $\Rightarrow2\sin^{2}\text{x}\sin2\text{x}=0$ $\Rightarrow\text{Either}$ $\sin\text{x}=0$ or $\sin2\text{x}=0$ $\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $2\text{x}=\text{m}\pi,\text{m}\in\text{z}$ $\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $\text{x}=\text{m}\frac{\pi}{2},\text{m}\in\text{z}$
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Question 113 Marks
Solve the following equation: $4\sin^{2}\text{x}-8\cos\text{x}+1=0$
Answer
We have, $4\sin^{2}\text{x}-8\cos\text{x}+1=0$ $\Rightarrow4(1-\cos^{2}\text{x})-8\cos\text{x}+1=0$ $\Rightarrow4\cos^{2}\text{x}+8\cos\text{x}-5=0$ factorise it,we get, $\Rightarrow4\cos^{2}\text{x}+10\cos\text{x}-2\cos\text{x}-5=0$ $\Rightarrow2\cos\text{x}(2\cos\text{x}+5)-1(2\cos\text{x}+5)=0$ $\Rightarrow(2\cos\text{x}-1)(2\cos\text{x}+5)=0$Either $2\cos\text{x}-1=0$ or $2\cos\text{x}+5=0$
$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-\frac{5}{2}$
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
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Question 123 Marks
Solve the following equations: $\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$
Answer
We have, $\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$
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Question 133 Marks
Solve the following equations: $\sin\theta+\sin5\theta=\sin3\theta$
Answer
$\sin\text{x}+\sin5\text{x}=\sin3\text{x}$ $\Rightarrow2\sin3\text{x},\cos2\text{x}-\sin3\text{x}=0$ $\Big[\because\sin\text{C}+\sin\text{D}=2\sin\frac{\text{C}+\text{D}}{2}.\cos\frac{\text{C}-\text{D}}{2}\Big] $ Either $\Rightarrow\sin3\text{x}=0$ or $2\cos2\text{x}-1=0 $ $\Rightarrow3\text{x}=\text{n}\pi,\text{n }\in\ \text{z}$ or $\cos2\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$ $\Rightarrow\text{x}=\frac{\text{n}\pi}{3},\text{n }\in\ \text{z}$ or $2\text{x}=2\text{m}\pi\pm\frac{\pi}{3},\text{m }\in\ \text{z}$ or $\text{x}=\text{m}\pi\pm\frac{\pi}{6}$ Thus, $\text{x}=\frac{\text{n}\pi}{3}$ or $\text{m}\pi\pm\frac{\pi}{6},\text{n, m }\in\ \text{z}$
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Question 143 Marks
Find the general solutions of the following equations: $\tan3\text{x}=\cot\text{x}$
Answer
We have, $\tan3\text{x}=\cot\text{x}$ $\Rightarrow\tan3\text{x}=\tan\Big(\frac{\pi}{2}-\text{x}\Big)\Big[\because\tan\Big(\frac{\pi}{2}-\text{x}\Big)=\cot\text{x}\Big]$ $\Rightarrow3\text{x}=\text{n}\pi+\frac{\pi}{2}-\text{x},\text{n}\in\text{z}$ $\Rightarrow4\text{x}=\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$ $\Rightarrow\text{x}=\frac{\text{n}\pi}{4}+\frac{\pi}{8},\text{n}\in\text{z}$
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Question 153 Marks
Find the general solutions of the following equations: $\sin9\text{x}=\sin\text{x}$
Answer
$\sin9\text{x}=\sin\text{x}$ $\sin9\text{x}-\sin\text{x}=0$ Apply $\sin\text{A}-\sin\text{B}$ Formula $\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A-B}}{2}\Big)$ $\sin9\text{x}-\sin\text{x}2\cos5\text{x}\sin4\text{x}=0$ $\cos5\text{x}\sin4\text{x}=0$ $\Rightarrow\cos5\text{x}=0$ or $\sin4\text{x}=0$ $5\text{x}=\frac{(2\text{n}+1)\pi}{2}$ (or) $4\text{x}=\text{n}\pi$ $\text{x}=\Big\{\frac{(2\text{n}+1)\pi}{10}\Big\}$ {or}$\text{x}=\Big\{\frac{\text{n}\pi}{4}\Big\}$where$\text{n}\in\text{z}$
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Question 163 Marks
Solve the following equations: $5\cos^{2}\text{x}+7\sin^{2}\text{x}-6=0$
Answer
$2\sin^{2}\text{x}+5-6=0$ $\sin^{2}\text{x}=\frac{1}{2}$ $\sin\text{x}=\pm\frac{1}{\sqrt{2}}$ $\text{x}=\text{n}\pi\pm\frac{\pi}{4}$
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Question 173 Marks
Find the general solutions of the following equations: $\sin\text{x}=\tan\text{x}$
Answer
We have, $\sin\text{x}=\tan\text{x}$ $\Rightarrow\sin\text{x}=\frac{\sin\text{x}}{\cos\text{x}}$ $\Rightarrow\sin\text{x}=\frac{\sin\text{x}}{\cos\text{x}}=0$ $\Rightarrow\sin\text{x}(\cos\text{x}-1)=0$ $\Rightarrow$ Either $\sin\text{x}=0$ or $\cos\text{x}-1=0$Now,
$\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $\cos\text{x}=1$$\Rightarrow\cos\text{x}=\cos0^\circ$
$\text{x}=2\text{m}\pi,\text{m}\in\text{z}$Thus,
$\text{x}=\text{n}\pi\text{n}\in\text{z}$ or $\text{x}=2\text{m}\pi,\text{m}\in\text{z}$
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Question 183 Marks
Solve the following equation: $2\sin^{2}\text{x}+\sqrt{3}\cos\text{x}+1=0$
Answer
We have, $2\sin^{2}\text{x}+\sqrt{3}\cos\text{x}+1=0$ $\Rightarrow2(1-\cos^{2}\text{x})+\sqrt{3}\cos\text{x}+1=0$ $\Rightarrow2\cos^{2}\text{x}-\sqrt{3}\cos\text{x}-3=0$ factorise it,we get, $\Rightarrow2\cos^{2}\text{x}-2\sqrt{3}\cos\text{x}+\sqrt{3}\cos\text{x}-3=0$ $\Rightarrow2\cos\text{x}(\cos\text{x}-\sqrt{3})+\sqrt{3}(\cos\text{x}-\sqrt{3})=0$ $\Rightarrow(2\cos\text{x}+\sqrt{3})(\cos\text{x}-\sqrt{3})=0$ $\Rightarrow\text{Either}$ $\cos\text{x}=-\frac{\sqrt{3}}{2}$ or $\cos\text{x}=\sqrt{3}$ [This is not possible as$-1<\cos\text{x}<1$] $\Rightarrow\cos\text{x}=\cos\Big(\pi-\frac{\pi}{6}\Big)$ $\Rightarrow\cos\text{x}=\cos\frac{5\pi}{6}$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{5\pi}{6},\text{n}\in\text{z}$
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Question 193 Marks
Find the general solutions of the following equations: $\tan2\text{x}\tan\text{x}=1$
Answer
We have, $\tan2\text{x}\tan\text{x}=1$ $\Rightarrow\tan2\text{x}=\frac{1}{\tan\text{x}}$ $\Rightarrow\tan2\text{x}=\cot\text{x}$ $\Rightarrow\tan2\text{x}=\tan\Big(\frac{\pi}{2}-\text{x}\Big)$ $\Rightarrow2\text{x}=\text{n}\pi+\frac{\pi}{2}-\text{x},\text{n}\in\text{z}$ $\Rightarrow3\text{x}=\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$ $\Rightarrow\text{x}=\frac{\text{n}\pi}{3}+\frac{\pi}{6},\text{n}\in\text{z}$
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Question 203 Marks
Solve the following equations: $\text{cosec }\text{x}=1+\cot\text{x}$
Answer
We have, $\text{cosec }\text{x}=1+\cot\text{x}$ $\Rightarrow\frac{1}{\sin\text{x}}=1+\frac{\cos\text{x}}{\sin\text{x}}$ $\Rightarrow1=\sin\text{x}+\cos\text{x}$ Divide both side by $\sqrt{2}$, We get, $\Rightarrow\frac{1}{\sqrt2}\sin\text{x}+\frac{1}{\sqrt2}\cos\text{x}=\frac{1}{\sqrt2}$ $\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\frac{1}{\sqrt2}$ $\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$ $\Rightarrow\text{x}=\frac{\pi}{4}=2\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{z}$ $\therefore\text{x}\Big(2\text{n}\pi+\frac{\pi}{2}\Big)$ Or $2\text{n }\pi,\text{n}\in\text{z}$
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Question 213 Marks
Find the general solutions of the following equations: $\tan^{2}\text{x}+(1-\sqrt{3})\tan\text{x}-\sqrt{3}=0$
Answer
We have, $\tan^{2}\text{x}+(1-\sqrt{3})\tan\text{x}-\sqrt{3}=0$ $\Rightarrow\tan^{2}\text{x}+\tan\text{x}-\sqrt{3}\tan\text{x}-\sqrt{3}=0$ $\Rightarrow\tan\text{x}(\tan\text{x}+1)-\sqrt{3}(\tan\text{x}+1)=0$ $\Rightarrow(\tan\text{x}-\sqrt{3})(\tan\text{x}+1)=0$ $\Rightarrow\text{Either}$ $\tan\text{x}=\sqrt{3}$ or $\tan\text{x}=-1$ $\Rightarrow\tan\text{x}=\tan\frac{\pi}{3}$ or $\tan\text{x} =-\tan\frac{\pi}{4}$ $\Rightarrow\text{x}=\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{z}$ or $\text{x}=\text{m}\pi-\frac{\pi}{4},\text{m}\in\text{z}$ $\therefore\text{x}=\text{x}\pi+\frac{\pi}{3}$ or $\text{m}\pi-\frac{\pi}{4},\text{n},\text{m}\in\text{z}$
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Question 223 Marks
Find the general solutions of the following equations: $\tan\text{x}+\cot2\text{x}=0$
Answer
We have, $\tan\text{x}+\cot2\text{x}=0$ $\tan\text{x}=-\cot2\text{x}$ $\Rightarrow\cot2\text{x}=-\tan\text{x}$ $\Rightarrow\tan2\text{x}=-\cot\text{x}$ $\Rightarrow\tan2\text{x}=-\tan\Big(\frac{\pi}{2}-\text{x}\Big)$ $\Rightarrow\tan2\text{x}=\tan\Big(\text{x}-\frac{\pi}{2}\Big)$ $\Rightarrow2\text{x}=\text{n}\pi+\Big(\text{x}-\frac{\pi}{2}\Big),\text{n}\in\text{z}$ $\text{x}=\text{n}\pi-\frac{\pi}{2},\text{n}\in\text{z}$
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