8.052. $\sin 3 x=2 \cos \left(\frac{\pi}{2}-x\right)$
## Solution.

## Rewrite the equation as

$3 \sin x-4 \sin ^{3} x=2 \sin x, 4 \sin ^{3} x-\sin x=0, \quad \sin x\left(4 \sin ^{2} x-1\right)=0$, from which:

1) $\sin x=0, x_{1}=\pi n, \quad n \in Z$
2) $4 \sin ^{2} x-1=0, \quad \sin x= \pm \frac{1}{2}, \quad x_{2,3}= \pm \frac{\pi}{6}+\pi k, \quad k \in Z$.

Answer: $\quad x_{1}=\pi n ; x_{2,3}= \pm \frac{\pi}{6}+\pi k, \quad n, k \in Z$.