The general solution of \(\sin x-3 \sin 2 x+\sin 3\)\( x=\cos x-3 \cos 2 x+\cos 3 x\) is

\(\begin{aligned}
& \sin x-3 \sin 2 x+\sin 3 x \\
& =\cos x-3 \cos 2 x+\cos 3 x \\
& \Rightarrow(\sin x+\sin 3 x)-3 \sin 2 x-(\cos x+\cos 3 x) \\
& +3 \cos 2 x=0 \\
& \Rightarrow 2 \sin 2 x \cos x-3 \sin 2 x-2 \cos 2 x \cos x \\
& +3 \cos 2 x=0 \\
& \Rightarrow \sin 2 x(2 \cos x-3)-\cos 2 x(2 \cos x-3)=0 \\
& \Rightarrow(\sin 2 x-\cos 2 x)(2 \cos x-3)=0 \\
& \Rightarrow \cos 2 x=\sin 2 x \quad \ldots\left[\because \cos x \neq \frac{3}{2}\right]
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow \cos 2 x=\cos \left(\frac{\pi}{2}-2 x\right) \\
& \Rightarrow 2 x=2 n \pi \pm\left(\frac{\pi}{2}-2 x\right)
\end{aligned}\)
Neglecting \((-)\) sign, we get
\(x=\frac{\mathrm{n} \pi}{2}+\frac{\pi}{8}\)