8.031. $\sin 2 x+\sin (\pi-8 x)=\sqrt{2} \cos 3 x$.

8.031. $\sin 2 x+\sin (\pi-8 x)=\sqrt{2} \cos 3 x$.
Solution.

From the condition

$\sin 2 x+\sin 8 x-\sqrt{2} \cos 3 x=0 \Leftrightarrow 2 \sin \frac{2 x+8 x}{2} \cos \frac{2 x-8 x}{2}-\sqrt{2} \cos 3 x=0$, $2 \sin 5 x \cos 3 x-\sqrt{2} \cos 3 x=0, \cos 3 x(2 \sin 5 x-\sqrt{2})=0$.

From this

1) $\cos 3 x=0, \quad 3 x=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{6}+\frac{\pi k}{3}=\frac{\pi}{6}(2 k+1) \quad k \in Z$;
2) $2 \sin 5 x-\sqrt{2}=0, \sin 5 x=\frac{\sqrt{2}}{2}, 5 x=(-1)^{n} \frac{\pi}{4}+\pi n$,

$$
x_{2}=(-1)^{n} \frac{\pi}{20}+\frac{\pi n}{5}, \quad n \in Z
$$

Answer: $\quad x_{1}=\frac{\pi}{6}(2 k+1) x_{2}=(-1)^{n} \frac{\pi}{20}+\frac{\pi n}{5}, \quad k, n \in Z$.