The general solution of the equation  $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ is

$\sin x+\sin 3x-3\sin 2x$$=\cos x+\cos 3x-3\cos 2x$

$\begin{array}{l}\Rightarrow 2\sin 2x\cos x-3\sin 2x=2\cos 2x\cos x-3\cos 2x\\ \Rightarrow (2\cos x-3)(\sin 2x-\cos 2x)=0\\ \Rightarrow 2\cos x\ne 3\therefore \tan 2x=1\\ \Rightarrow \tan 2x=\tan \frac{\pi }{4}\therefore 2x=n\pi +\frac{\pi }{4}\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8}\end{array}$