8.037. $\sin 3 x \cos 3 x=\sin 2 x$.

8.037. $\sin 3 x \cos 3 x=\sin 2 x$.
Solution.

We have

$2 \sin 3 x \cos 3 x-2 \sin 2 x=0, \Leftrightarrow \sin 6 x-2 \sin 2 x=0$,

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

## From this:

1) $\sin 2 x=0,2 x=\pi n, \quad x_{1}=\frac{\pi n}{2}, \quad n \in Z$
2) $4 \sin ^{2} 2 x-1=0, \sin ^{2} 2 x=\frac{1}{4}, \sin 2 x= \pm \frac{1}{2}, 2 x= \pm \frac{\pi}{6}+\pi k$, $x_{2,3}= \pm \frac{\pi}{12}+\frac{\pi k}{2}=\frac{\pi}{12}(6 k \pm 1), k \in Z$.

Answer: $\quad x_{1}=\frac{\pi n}{2} ; x_{2,3}=\frac{\pi}{12}(6 k \pm 1), n, k \in Z$.