Question

Prove that the general solution of $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ is $n\pi /2+\pi /8$.

Answer 1

The given equation can be written as
$\sin 3x+\sin x-3\sin 2x$
$=\cos 3x+\cos x-3\cos 2x$
$2\sin 2x\cos x-3\sin 2x$
$=2\cos 2x\cos x-3{\cos }^{2}2x$
$\sin 2x(2\cos x-3)=cos2x(2\cos x-3)=0$
$\therefore \sin 2x=\cos 2x$ as $2\cos x-3\ne 0$
$\because \cos x\ne 3/2$
$\therefore \tan 2x=1=\tan \left(\pi /4\right)$
$\therefore 2x=n\pi +\frac{\pi }{4}$ or $x=\frac{n\pi }{2}+\frac{\pi }{8}$.