(Each question 2 marks)

Question 12 Marks

Find the general solutions of the following equations: $\tan\text{x}=-\frac{-1}{\sqrt{3}}$

Answer

we have, $\tan\text{x}=-\frac{-1}{\sqrt{3}}$ $\Rightarrow\tan\theta=\tan\Big(\frac{\pi}{6}\Big)$ $\Rightarrow\tan\theta=\tan\Big(-\frac{\pi}{6}\Big)$$\big[\because\tan(-\theta)=-\tan\theta\big]$ $\Rightarrow\theta=\text{n}\pi+\Big(-\frac{\pi}{6}\Big),\text{n}\in\text{z}$ or $\theta=\text{n}\pi-\frac{\pi}{6},\text{n}\in\text{z}$

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Question 22 Marks

Find the general solutions of the following equations: $\sqrt{3}\sec\text{x}=2$

Answer

we have, $\sqrt{3}\sec\text{x}=2$ $\Rightarrow\frac{1}{\cos\theta}=\frac{2}{\sqrt{3}}$ $\Rightarrow\cos\theta=\frac{\sqrt{3}}{2}$ $\Rightarrow\cos\theta=\cos\Big(\frac{\pi}{6}\Big)$ $\Rightarrow\theta=2\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{z}$

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Question 32 Marks

Find the general solutions of the following equations: $\cos3\text{x}=\frac{1}{2}$

Answer

we have, $\cos3\text{x}=\frac{1}{2}$ $\Rightarrow\cos\text{x}=\cos\Big(\frac{\pi}{3}\Big)$ $\Rightarrow3\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$ $\Rightarrow\text{x}=2\text{n}\frac{\pi}{3}\pm\frac{\pi}{9},\text{n}\in\text{Z}$

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Question 42 Marks

Find the general solutions of the following equations:$\text{cosec }\text{x}=-\sqrt{2}$

Answer

$\text{cosec }\text{x}=-\sqrt{2}$$\Rightarrow\frac{1}{\sin\theta}=-\sqrt{2}$
$\Rightarrow\sin\theta=-\frac{1}{\sqrt2}$
$\Rightarrow\sin\theta=\sin\Big(\pi+\frac{\pi}{4}\Big)$
$\Rightarrow\sin\theta=\sin\frac{5\pi}{4}\text{or}\sin\theta=\sin\Big(-\frac{\pi}{4}\Big)$$\because\sin(-\theta)=-\sin\theta$
$\therefore\theta=\text{n}\pi+(-1)^{\text{n+1}}\frac{\pi}{4},\text{n}\in\text{z}$

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Question 52 Marks

Find the general solutions of the following equations: $\sec\text{x}=\sqrt{2}$

Answer

we have, $\sec\text{x}=\sqrt{2}$ $\Rightarrow\frac{1}{\cos\theta}=\sqrt{2}$ $\Rightarrow\cos\theta=\frac{1}{\sqrt{2}}\Rightarrow\cos\theta=\cos\frac{\pi}{4}$ $\Rightarrow\theta=2\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{z}$

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Question 62 Marks

Find the general solutions of the following equations: $\sin2\text{x}=\frac{\sqrt{3}}{2}$

Answer

We have, $\sin2\text{x}=\frac{\sqrt{3}}{2}$ $\Rightarrow\sin2\theta=\sin\Big(\frac{\pi}{3}\Big)$ $\Rightarrow2\theta=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{3},\text{n}\in\text{z}$ $\theta=\frac{\text{n}\pi}{2}+(-1)^{n}\frac{\pi}{6},\text{n}\in\text{z}$

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Question 72 Marks

Find the general solutions of the following equations: $\cos\text{x}=-\frac{\sqrt3}{2}$

Answer

we have, $\cos\text{x}=-\frac{\sqrt3}{2}$ $\Rightarrow\cos\theta=\cos\Big(\pi+\frac{\pi}{6}\Big)$ $\Rightarrow\cos\theta=\cos\frac{7\pi}{6}\Big[\because\cos\frac{7\pi}{6}=-\frac{\sqrt3}{2}\Big]$ $\therefore$ the general solution is $\theta=2\text{n}\pi\pm\frac{7\pi}{6},\text{n}\in\text{z}$

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Question 82 Marks

Find the general solutions of the following equations:$\sin\text{x}=\frac{1}{2}$

Answer

we have, $\sin\theta=\frac{1}{2}$ $\Rightarrow\sin\theta=\sin\frac{\pi}{6}\Big[\because\sin\frac{\pi}{6}=\frac{1}{2}\Big]$ ⇒ The general solution is $\theta=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{6};\text{n}\in\text{z}$ $\Big[\because\text{if }\sin\theta=\sin\alpha\Rightarrow\theta=\text{n}\pi+(-1)^\text{n}\alpha\Big]$

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MATHS (Each question 2 marks)

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