8.130. $\sin 3 x-\sin 7 x=\sqrt{3} \sin 2 x$.
## Solution.

Let's rewrite the given equation as

$2 \sin \frac{3 x-7 x}{2} \cos \frac{3 x+7 x}{2}-\sqrt{3} \sin 2 x=0, -2 \sin 2 x \cos 5 x-\sqrt{3} \sin 2 x=0$, $-\sin 2 x(2 \cos 5 x+\sqrt{3})=0$

From this:

1) $\sin 2 x=0, 2 x=\pi k, x_{1}=\frac{\pi k}{2}, k \in Z$
2) $2 \cos 5 x+\sqrt{3}=0, \quad \cos 5 x=-\frac{\sqrt{3}}{2}, \quad 5 x= \pm \frac{5}{6} \pi+2 \pi n, \quad x_{2}= \pm \frac{\pi}{6}+\frac{2}{5} \pi n$, $n \in Z$

Answer: $\quad x_{1}=\frac{\pi k}{2} ; x_{2}= \pm \frac{\pi}{6}+\frac{2}{5} \pi n, \quad k, n \in Z$