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Maths Solve the Following Question.(4 Marks)

Trigonometric Equations · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 16 questions.

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Solve the Following Question.(4 Marks)

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16 questions · timed · auto-graded

Question 14 Marks
Find the general solutions of the following equations:
$\cos4\text{x}=\cos2\text{x}$
Answer
We have,
$\cos4\text{x}=\cos2\text{x}$
$\Rightarrow\cos4\text{x}=\cos2\text{x}=0$
$\Rightarrow2\sin\text{x}.\sin3\text{x}=0$
$\Rightarrow\text{Either}$
$\sin\text{x}=0$ or $\sin3\text{x}=0$
$\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $3\text{x}=\text{m}\pi,\text{m}\in$
Thus,
$\text{x}=\text{n}\pi$ or $\text{m}\frac{\pi}{3},\text{n,m}\in\text{z}$
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Question 24 Marks
Find the general solutions of the following equations:
$3\cos^{2}\text{x}-2\sqrt{3}\sin\text{x}\cos\text{x}-3\sin^{2}\text{x}=0$
Answer
We have,
$3\cos^{2}\text{x}-2\sqrt{3}\sin\text{x}\cos\text{x}-3\sin^{2}\text{x}=0$
$\sqrt{3}\cos^{2}\text{x}-2\sin\text{x}\cos\text{x}-\sqrt{3}\sin^{2}\text{x}=0$ $(\text{Divided by}\sqrt{3})$
$\sqrt{3}\cos^{2}\text{x}-\sin\text{x}\cos\text{x}-3\sin\text{x}\cos\text{x}\sqrt{3}\sin^{2}\text{x}=0$
$\cos\text{x}(\sqrt{3}\cos\text{x}+\sin\text{x})-\sqrt{3}\sin\text{x}(\sqrt{3}\cos\text{x}+\sin\text{x})=0$
$(\sqrt{3}\cos\text{x}+\sin\text{x})(\cos\text{x}-\sqrt{3}\sin\text{x})=0$
$\sqrt{3}\cos\text{x}+\sin\text{x}=0$ or $\cos\text{x}-\sqrt{3}\sin\text{x}=0$
$\tan\text{x}=-\sqrt{3}=-\tan\frac{\pi}{3}$ or $\tan\text{x}=\frac{1}{\sqrt{3}}=\tan\frac{\pi}{6}$
$\text{x}=\text{n}\pi-\frac{\pi}{3}$ or $\text{x}=\text{m}\pi-\frac{\pi}{6}$
$\text{n,m}\in\text{z}$
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Question 34 Marks
Solve the following equations:
$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
Answer
$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
$(\sin2\text{x}+\sin6\text{x})-\sin4\text{x}=0$
$2.\sin\Big(\frac{8\text{x}}{2}\Big).\cos\Big(\frac{4\text{x}}{2}\Big)-\sin4\text{x}=0$
$2\sin4\text{x}.\cos2\text{x}-\sin4\text{x}=0$
$\sin4\text{x}(2\cos2\text{x}-1)=0$
$\sin4\text{x}=0$ or $2\cos2\text{x}-1=0$
$4\text{x}=\text{n}(\pi)$ or $\cos2\text{x}=\frac{1}{2}$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\cos2\text{x}=\cos[\frac{\pi}{3}]$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\text{x}=\text{n}(\pi)\pm[\frac{\pi}{6}]$
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Question 44 Marks
Solve the following equation:
$2\cos^{2}\text{x}-5\cos\text{x}+2=0$
Answer
We have,
$2\cos^{2}\text{x}-5\cos\text{x}+2=0$
$\Rightarrow2\cos^{2}\text{x}-4\cos\text{x}-\cos\text{x}+2=0$ [Use factorization]
$\Rightarrow2\cos\text{x}(\cos\text{x}-2)-1(\cos\text{x}-2)=0$
$\Rightarrow(2\cos\text{x}-1)(\cos\text{x}-2)=0$
$\Rightarrow\text{Either}$
$2\cos\text{x}-1=0$ or $\cos\text{x}-2=0$
$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=2$
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$ $\big[$This is not possible as $-1<\cos\text{x}<1\big]$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
Thus,
$\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
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Question 54 Marks
Solve the following equations:
$\cot\text{x}+\tan\text{x}=2$
Answer
$\cot\text{x}+\tan\text{x}=2$
$\Rightarrow\frac{1}{\tan\text{x}}+\tan\text{x}=2$
$\Rightarrow\tan^{2}\text{x}+1=2\tan\text{x}$
$\Rightarrow\tan^{2}\text{x}-2\tan\text{x}+1=0$
$\Rightarrow(\tan\text{x}-1)^{2}=0$
$\Rightarrow\tan\text{x}=1=\tan\frac{\pi}{4}$
$\Rightarrow\text{x}=\text{n}\pi+\frac{\pi}{4},\text{n}\in\text{Z}$ $(\tan\text{x}=\tan\alpha\Rightarrow\text{x}=\text{n}\pi+\alpha,\text{n}\in\text{z)}$
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Question 64 Marks
Solve the following equations:
$\sin\text{x}+\cos\text{x}=1$
Answer
We have,
$\sin\text{x}+\cos\text{x}=1$
divide both side by $\sqrt{2},$ we get,
$\Rightarrow\frac{1}{\sqrt{2}}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=\frac{1}{\sqrt{2}}$
$\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\frac{1}{\sqrt{2}}$
$\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$
$\Rightarrow\text{x}=\frac{\pi}{4}=2\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{z}$
$\Rightarrow\text{x}=2\text{n}\pi+\frac{\pi}{2}$ or $2\text{n}\pi,\text{n}\in\text{z}$
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Question 74 Marks
Find the general solutions of the following equations:
$\sin3\text{x}+\cos2\text{x}=0$
Answer
$\cos(2\text{x})=-\sin(3\text{x})$$=-\cos(\frac{\pi}{2}-3\text{x})$
$=\cos(\frac{\pi}{2}+3\text{x})$
$\Rightarrow2\text{n}\pi+2\text{x}=\frac{\pi}{2}+3\text{x}$
$\text{x}=(4\text{m}-1)\frac{\pi}{2},\text{m}\in\text{z}$
or
$\Rightarrow2\text{n}\pi-2\text{x}=\frac{\pi}{2}+3\text{x}$
$\text{x}=(4\text{n}-1)\frac{\pi}{10},\text{n}\in\text{z}$
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Question 84 Marks
Solve the following equations:
$\sin\text{x}+\sin2\text{x}+\sin3=0$
Answer
We have,
$\sin\text{x}+\sin2\text{x}+\sin3=0$
$\Rightarrow\sin2\text{x}+2\sin2\text{x}.\cos\text{x}=0$
$\Rightarrow\sin2\text{x}+(1+2\cos\text{x})=0$
$\Rightarrow\text{Either}$
$\sin2\text{x}=0$ or $1+2\cos\text{x}=0$
$\Rightarrow2\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $\cos\text{x}=-\frac{1}{2}=\cos\Big(\pi-\frac{\pi}{3}\Big)$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{2},\text{n}\in\text{z}$ or $\text{x}=2\text{m}\pi\pm\frac{2\pi}{3},\text{m}\in\text{z}$
Thus,
$\text{x}=\frac{\text{n}\pi}{2},\text{n}\in\text{z}$ or $\text{x}=2\text{m}\pi\pm\frac{2\pi}{3},\text{m}\in\text{z} $
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Question 94 Marks
Solve the following equations:
$\cos\text{x}\cos2\text{x}\cos3\text{x}=\frac{1}{4}$
Answer
We have,
$\cos\text{x}\cos2\text{x}\cos3\text{x}=\frac{1}{4}$
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Question 104 Marks
Solve the following equations: $\tan\text{x}+\tan2\text{x}=\tan3\text{x}$
Answer
$\tan\text{x}+\tan2\text{x}=\tan(\text{x}+2\text{x})$ $\tan\text{x}+\tan2\text{x}-\frac{\tan\text{x}+\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}=0$ $[\tan\text{x}+\tan2\text{x}]\Big[1-\frac{1}{1-\tan\text{x}\tan2\text{x}}\Big]=0$ $[\tan\text{x}+\tan2\text{x}]\Big[\frac{1-\tan\text{x}\tan2\text{x}-1}{1-\tan\text{x}\tan2\text{x}}\Big]=0$ $[\tan\text{x}+\tan2\text{x}]\Big[\frac{-\tan\text{x}\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}\Big]=0$ $\tan\text{x}=0$ or $\tan2\text{x}=0$ or $\tan\text{x}+\tan2\text{x}=0$ $\text{x}=\text{n}\pi$ or $\frac{\text{n}\pi}{2}$ or $\tan\text{x}\Big[\frac{1-\tan^{2}\text{x}+2}{1-\tan^{2}\text{x}}\Big]=0$ $\text{x}=\text{n}\pi$ or $\frac{\text{n}\pi}{2}$ or $\tan\text{x}=\pm\sqrt{3}$$\text{x}=\text{m}\pi$ or $\frac{\text{n}\pi}{3}\text{m,n}\in\text{Z}$
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Question 114 Marks
Solve the following equations:
$\sin\text{x}+\sin2\text{x}+\sin3\text{x}+\sin4\text{x}=0$
Answer
Given, $\sin\text{x}+\sin2\text{x}+\sin3\text{x}+\sin4\text{x}=0$$(\sin4\text{x}+\sin2\text{x})+(\sin3\text{x}+\sin\text{x})=0$
Using, $(\sin\text{A}+\sin\text{B})\text{Formula}\Rightarrow$
$2\sin\Big[\frac{(4\text{x}+2\text{x})}{2}\Big]\cos\Big[\frac{4\text{x}-2\text{x}}{2}\Big]+2\sin\Big[\frac{( 3\text{x}+\text{x})}{2}\Big]\cos\Big[\frac{( 3\text{x}-\text{x})}{2}\Big]=0$
$2\sin3\text{x}\cos\text{x}+2\sin2\text{x}\cos\text{x}=0$
$2\cos\text{x}(\sin3\text{x}+\sin2\text{x})=0$
$2\cos\text{x}(2\sin)\Big[\frac{(3\text{x}+2\text{x})}{2}\Big]\cos\Big[\frac{(3\text{x}-2\text{x})}{2}\Big]=0$
$4\cos\text{x}\sin\frac{5\text{x}}{2}\cos\frac{\text{x}}{2}=0$
$\cos\text{x}=0;\sin\frac{5\text{x}}{2}=0;\cos\frac{\text{x}}{2}=0$
$\text{x}=\frac{(2\text{n}+1)\pi}{2};\frac{5\text{x}}{2}=\text{m}\pi;\frac{\text{x}}{2}=\frac{(2\text{x}+1)\pi}{2}$
$\text{x}=\frac{(2\text{n}+1)\pi}{2};\text{x}=\frac{2\text{m}\pi}{5};\text{x}=(2\text{x}+1)\pi,\text{m},\text{r},\text{n}\in\text{Z}$
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Question 124 Marks
Solve the following equations:
$2\sin^{2}\text{x}=3\cos\text{x},0\leq\text{x}\leq2\pi$
Answer
$2\sin^{2}\text{x}=3\cos\text{x}$$\Rightarrow2(1-\cos^{2}\text{x})=3\cos\text{x}$
$\Rightarrow2\cos^{2}\text{x}+3\cos\text{x}-2=0$
$\Rightarrow(2\cos\text{x}-1)(\cos\text{x}+2)=0$
$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-2$
But, $\cos\text{x}=-2$ is not possible. $(-1\leq\cos\text{x}\leq1)$
$\therefore\cos\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
Putting n=0 and n=1, We get
$\text{x}=\frac{\pi}{3},\frac{5\text{x}}{3}$ $(0\leq\text{x}\leq2\pi)$
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Question 134 Marks
Solve the following equations:
$\sqrt{3}\cos\text{x}+\sin\text{x}=1$
Answer
We have, $\sqrt{3}\cos\text{x}+\sin\text{x}=1$ Divide both side by 2, we get $\frac{\sqrt{3}}{2}\cos\text{x}+\frac{1}{2}\sin\text{x}=\frac{1}{2}$ $\Rightarrow\cos\frac{\pi}{6}\cos\text{x}+\sin\frac{\pi}{6}\sin\text{x}=\frac{1}{2}$ $\Big[\because\sin\frac{\pi}{6}=\frac{1}{2},\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\Big]$ $\Rightarrow\cos\Big(\text{x}-\frac{\pi}{6}\Big)=\cos\frac{\pi}{3}$ $\Rightarrow\text{x}=\frac{\pi}{6}=2\text{n}\pm\frac{\pi}{3},\text{n}\in\text{z}$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}+\frac{\pi}{6},\text{n}\in\text{z}$$\Rightarrow\text{x}=(4\text{n}+1)\frac{\pi}{2}$ or $(12\text{m}-1)\frac{\pi}{6},\text{n},\text{m}\in\text{z}$
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Question 144 Marks
Solve the following equation:
$\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$
Answer
We have,
$\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$
$\Rightarrow1-\cos^{2}\text{x}-\cos\text{x}=\frac{1}{4}$ $[\because\sin^{2}\text{x}=1-\cos^{2}\text{x}]$
$\Rightarrow\cos^{2}\text{x}+\cos\text{x}-\frac{3}{4}=0$
$\Rightarrow4\cos^{2}\text{x}+4\cos\text{x}-3=0$
$\Rightarrow4\cos^{2}\text{x}+6\cos\text{x}+2\cos\text{x}-3=0$ [factorize it]
$\Rightarrow2\cos\text{x}(2\cos\text{x}+3)-1(\cos\text{x}+3)=0$
$\Rightarrow(2\cos\text{x}-1)(2\cos\text{x}+3)=0$
$\Rightarrow\text{ Either}$
$2\cos\text{x}-1=0$ or $2\cos\text{x}+3=0$
$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-\frac{3}{2}$ [This is not possible as $-1<\cos\text{x}<1$]
$\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 154 Marks
Solve the following equations:
$\tan\text{x}+\tan2\text{x}+\tan3\text{x}=0$
Answer
$\tan\text{x}+\tan2\text{x}+\frac{(\tan\text{x}+\tan2\text{x})}{1-\tan\text{x}.\tan2\text{x}}=0$
$[\tan\text{x}+\tan2\text{x}]\Big[1+\frac{1}{1-\tan\text{x}.\tan2\text{x}}\Big]=0$
$\tan\text{x}+\tan2\text{x}(2-\tan\text{x}.\tan2\text{x})=0$
$\tan\text{x}=\tan(-2\text{x})$ or $\tan\text{x}.\tan2\text{x}=0$
$\text{x}=\text{n}\pi-2\text{x}$ or $\tan\text{x}.\frac{2\tan\text{x}}{1-\tan^{2}\text{x}}=2$
$3\text{x}=\text{n}\pi$ or $\frac{2\tan^{2}\text{x}}{1-\tan^{2}\text{x}}=2$
$3\text{x}=\text{n}\pi$ or $2\tan^{2}\text{x}=2-2\tan^{2}\text{x}$
$3\text{x}=\text{n}\pi$ or $4\tan^{2}\text{x}=2$
$\text{x}=\frac{\text{n}\pi}{3}$ or $\tan^{2}\text{x}=\frac{1}{2}$
$\text{x}=\frac{\text{n}\pi}{3}$ or $\text{x}=\text{m}\pi\pm\tan^{-1}(\frac{1}{\sqrt{2}}),\text{n,m}\in\text{Z}$
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Question 164 Marks
Solve the following equations:
$\sin\text{x}+\cos\text{x}=\sqrt{2}$
Answer
We have,
$\sin\text{x}+\cos\text{x}=\sqrt{2}$
$\Rightarrow\frac{1}{\sqrt{2}}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=1$
$\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=1$$\Big[\because\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{1}{\sqrt{3}}\Big]$
$\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos0^\circ$
$\Rightarrow\text{x}-\frac{\pi}{4}=2\text{n}\pi,\text{n}\in\text{z}$
$\Rightarrow\text{x}=2\text{n}\pi+\frac{\pi}{4},\text{n}\in\text{z}$
$\therefore\text{x}=(8\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$
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