Find the principal and general solutions of the following equations:
1. $\tan x =\sqrt{3}$
We know that,
$\tan \frac{\pi}{3} =\sqrt{3}$  and $\tan \left (\pi+\frac{\pi}{3} \right )=\tan \frac{4\pi}{3} =\sqrt{3}$
∴ The principal solutions are $\frac{\pi}{3}$  and $\frac{4\pi}{3}$
The general solution is,
$\tan x=\tan \frac{\pi}{3}$
⇒ $x=n\pi+\frac{\pi}{3}$ [tan x=tan y⇒x=nÏ€+y]
2. $\sec x=2$
⇒ $\cos x=\frac{1}{2}$
We know that,
$\cos \frac{\pi}{3}=\frac{1}{2}$  and $\cos \left (2\pi-\frac{\pi}{3} \right )=\cos \frac{5\pi}{3}=\frac{1}{2}$
∴ The principal solutions are $\frac{\pi}{3}$  and $\frac{5\pi}{3}$
The general solution is,
$\cos x=\cos \frac{\pi}{3}$ 
⇒ $x=2n\pi\pm \frac{\pi}{3}$
3.  $\cot x=-\sqrt{3}$
⇒ $\tan x=-\frac{1}{\sqrt{3}}$
We know that,
$-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$
⇒  $\tan \left (\pi-\frac{\pi}{6} \right )=\tan \frac{5\pi}{6}=-\frac{1}{\sqrt{3}}$  and $\tan \left (2\pi-\frac{\pi}{6} \right )=\tan \frac{11\pi}{6}=-\frac{1}{\sqrt{3}}$
∴ The principal solutions are $\frac{5\pi}{6}$  and $\frac{11\pi}{6}$
The general solution is,
$\tan x=\tan \frac{5\pi}{6}$ 
⇒ $x= n\pi+\frac{5\pi}{6}$
4. $\csc x=-2$
⇒ $\sin x=-\frac{1}{2}$
We know that,
$-\sin \frac{\pi}{6}=-\frac{1}{2}$
⇒ $\sin \left (\pi+\frac{\pi}{6} \right )=\sin \frac{7\pi}{6}=-\frac{1}{2}$  and $\sin \left (2\pi-\frac{\pi}{6} \right )=\sin \frac{11\pi}{6}=-\frac{1}{2}$
∴ The principal solutions are $\frac{7\pi}{6}$  and $\frac{11\pi}{6}$
The general solution is,
$\sin x=\sin \frac{7\pi}{6}$
⇒ $x=n\pi+(-1)^{n}\frac{7\pi}{6}$
Find the general solution for each of the following equations:
5. $\cos 4x=\cos 2x$
⇒ $4x=2n\pi\pm 2x$
⇒ $2x=n\pi\pm x$
⇒ $2x-x=n\pi$  or $2x+x=n\pi$
⇒ $x=n\pi$  or $x=\frac{n\pi}{3}$
We note that $x=n\pi$ i.e. $x=\pi ,2\pi ,3\pi ,...$  are included on the solutions $x=\frac{n\pi}{3}$ on taking x=3,6,9,...
∴ the solutions are $x=\frac{n\pi}{3}$, n∈I
6. $\cos 3x+\cos x-\cos 2x=0$ 
⇒ $2\cos (\frac{3x+x}{2})\cos (\frac{3x-x}{2})-\cos 2x=0$
⇒ $2\cos 2x\cos x-\cos 2x=0$
⇒ $\cos 2x(2\cos x-1)=0$
⇒ $\cos 2x=0$  or $2\cos x-1=0$
⇒ $2x=(2n+1)\frac{\pi}{2}$  or $\cos x=\frac{1}{2}=\cos \frac{\pi}{3}$
⇒ $x=(2n+1)\frac{\pi}{4}$  or $x=2n\pi\pm \frac{\pi}{3}$ , n∈I
7. $\sin 2x+cos x=0$
⇒ $2\sin x\cos x+cos x=0$
⇒ $\cos x(2\sin x+1)=0$
⇒ $\cos x=0$  or $2\sin x+1=0$
⇒ $x=(2n+1)\frac{\pi}{2}$  or $\sin x=-\frac{1}{2}$ $=\sin(\pi+\frac{\pi}{6})=\sin \frac{7\pi}{6}$
⇒ $x=(2n+1)\frac{\pi}{2}$ or $x=n\pi+(-1)^{n}\frac{7\pi}{6}$, n∈I
8. $\sec^{2}2x=1-\tan 2x$
⇒ $1+\tan^{2}2x=1-\tan 2x$
⇒ $\tan^{2}2x+\tan 2x=0$
⇒ $\tan 2x(\tan 2x+1)=0$
⇒ $\tan 2x=0$  or $\tan 2x+1=0$
⇒ $2x=n\pi$  or $\tan 2x=-1=\tan \frac{3\pi}{4}$
⇒ $x=\frac{n\pi}{2}$  or $2x=n\pi+\frac{3\pi}{4}$ 
⇒ $x=\frac{n\pi}{2}$ or $x=\frac{n\pi}{2}+\frac{3\pi}{8}$ , n∈I
9. $\sin x+\sin 3x+\sin 5x=0$
⇒ $\left (\sin 5x+\sin x \right )+\sin 3x=0$
⇒ $2\sin \left ( \frac{5x+x}{2} \right )\cos \left ( \frac{5x-x}{2} \right )+\sin 3x=0$
⇒ $2\sin 3x\cos 2x+\sin 3x=0$
⇒ $\sin 3x(2\cos 2x+1)=0$
⇒ $\sin 3x=0$  or $2\cos 2x+1=0$
⇒ $3x=n\pi$  or $\cos 2x=-\frac{1}{2}=\cos\frac{2\pi}{3}$
⇒ $x=\frac{n\pi}{3}$  or $2x=2n\pi\pm \frac{2\pi}{3}$
⇒ $x=\frac{n\pi}{3}$  or $x=n\pi\pm \frac{\pi}{3}$ , n∈I