Question

Solve the following equation:
$\tan 3x-\tan x=\sec x\sec 3x\sin x$

Answer 1

$\sin 3x\cos x-\cos 3x\sin x=\sin x$

$\sin 2x=\sin x$

$\sin 2x-\sin x=0$

$2\sin \frac{x}{2}\cos \frac{3x}{2}=0$

So, $\sin \frac{x}{2}=0,\frac{x}{2}=n\pi ,x=2n\pi .$ and $\cos \frac{3x}{2}=0,\frac{3x}{2}=(2n+1)\frac{\pi }{2},x=(2n+1)\frac{\pi }{3}.$