8.063. $\sin 9 x=2 \sin 3 x$.

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8.063. $\sin 9 x=2 \sin 3 x$.
## Solution.

Rewriting the equation as $\sin 3(3 x)-2 \sin 3 x=0$ and using the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have

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

Then

1) $\sin 3 x=0, 3 x=\pi n, x_{1}=\frac{\pi n}{3}, n \in Z$
2) $4 \sin ^{2} 3 x-1=0, \sin 3 x= \pm \frac{1}{2}, 3 x= \pm \frac{\pi}{6}+\pi k, x_{2}= \pm \frac{\pi}{18}+\frac{\pi k}{3}, k \in Z$.

Answer: $\quad x_{1}=\frac{\pi n}{3} ; x_{2}= \pm \frac{\pi}{18}+\frac{\pi k}{3}, n, k \in Z$.