General solution of $(\sin x-3\sin 2x+\sin 3x)=\cos x-3\cos 2x+\cos 3x$ is

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

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

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

$\sin 2x(2\cos x-3)=\cos 2x(2\cos x-3)$

$(\sin 2x-\cos 2x)(2\cos x-3)=0$

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

$\Rightarrow \tan 2x=1\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8}$

$2\cos x-3=0\Rightarrow \cos x=\frac{3}{2}\notin [-1,1]$

$\therefore$ No solution