Find the general solutions of the following equations:
Question:

Find the general solutions of the following equations:

(i) $\sin x=\frac{1}{2}$

(ii) $\cos x=-\frac{\sqrt{3}}{2}$

(iii) $\operatorname{cosec} x=-\sqrt{2}$

(iv) $\sec x=\sqrt{2}$

(v) $\tan x=-\frac{1}{\sqrt{3}}$

(vi) $\sqrt{3} \sec x=2$

Solution:

We have:

(i) $\sin x=\frac{1}{2}$

The value of $x$ satisfying $\sin x=\frac{1}{2}$ is $\frac{\pi}{6}$.

$\therefore \sin x=\frac{1}{2}$

$\Rightarrow \sin x=\sin \frac{\pi}{6}$

$\Rightarrow x=n \pi+(-1)^{n} \frac{\pi}{6}, n \in Z$

(ii) $\cos x=-\frac{\sqrt{3}}{2}$

The value of $x$ satisfying $\cos x=-\frac{\sqrt{3}}{2}$ is $\frac{7 \pi}{6}$.

$\therefore \cos x=-\frac{\sqrt{3}}{2}$

$\Rightarrow \cos x=\cos \frac{7 \pi}{6}$

$\Rightarrow x=2 n \pi \pm \frac{7 \pi}{6}, n \in Z$

(iii) $\operatorname{cosec} x=-\sqrt{2}$ (or) $\sin x=-\frac{1}{\sqrt{2}}$

The value of $x$ satisfying $\sin x=-\frac{1}{\sqrt{2}}$ is $-\frac{\pi}{4}$.

$\therefore \sin x=-\frac{1}{\sqrt{2}}$

$\Rightarrow \sin x=\sin \left(-\frac{\pi}{4}\right)$

$\Rightarrow x=n \pi+(-1)^{n}\left(-\frac{\pi}{4}\right), n \in Z$

$\Rightarrow x=n \pi+(-1)^{n+1} \frac{\pi}{4}, \mathrm{n} \in Z$

(iv) $\sec x=\sqrt{2}$ (or) $\cos x=\frac{1}{\sqrt{2}}$

The value of $x$ satisfying $\cos x=\frac{1}{\sqrt{2}}$ is $\frac{\pi}{4}$.

$\therefore \cos x=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos x=\cos \frac{\pi}{4}$

$\Rightarrow x=2 n \pi \pm \frac{\pi}{4}, n \in Z$

(v) $\tan x=-\frac{1}{\sqrt{3}}$

The value of $x$ satisfying $\tan x=-\frac{1}{\sqrt{3}}$ is $-\frac{\pi}{6}$.

$\therefore \tan x=-\frac{1}{\sqrt{3}}$

$\Rightarrow \tan x=\tan \left(-\frac{\pi}{6}\right)$

$\Rightarrow x=n \pi-\frac{\pi}{6}, n \in Z$

(vi) $\sqrt{3} \sec x=2$

$\Rightarrow \sec x=\frac{2}{\sqrt{3}}$ (or) $\cos x=\frac{\sqrt{3}}{2}$

The value of $x$ satisfying $\cos x=\frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$.

$\therefore \cos x=\frac{\sqrt{3}}{2}$

$\Rightarrow \cos x=\cos \frac{\pi}{6}$

$\Rightarrow x=2 n \pi \pm \frac{\pi}{6}, n \in Z$