8.123. $\sin 3 x-4 \sin x \cos 2 x=0$.
## Solution.

Let's rewrite the equation as

$\sin 3 x-2(\sin (x-2 x)+\sin (x+2 x))=0, \quad \sin 3 x+2 \sin x-2 \sin 3 x=0$, $\sin 3 x-2 \sin x=0$.

Since $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have $3 \sin x-4 \sin ^{3} x-2 \sin x=0$, $4 \sin ^{3} x-\sin x=0, \sin x\left(4 \sin ^{2} x-1\right)=0$.

From this:

1) $\sin x=0, \quad x_{1}=\pi n, \quad n \in Z$,
2) $4 \sin ^{2} x-1=0, \sin x= \pm \frac{1}{2}, \quad x_{2}= \pm \frac{\pi}{6}+\pi k=\frac{\pi}{6}(6 k \pm 1), k \in Z$.

Answer: $\quad x_{1}=\pi n ; x_{2}=\frac{\pi}{6}(6 k \pm 1) \quad n, k \in Z$.