8.027. $\sin x \sin 3 x+\sin 4 x \sin 8 x=0$.
Solution.

Rewrite the equation as

$$
\begin{aligned}
& \frac{1}{2}(\cos (x-3 x)-\cos (x+3 x))+\frac{1}{2}(\cos (4 x-8 x)-\cos (4 x+8 x))=0 \\
& \cos 2 x-\cos 12 x=0 \Leftrightarrow 2 \sin \frac{2 x+12 x}{2} \sin \frac{12 x-2 x}{2}=0, \sin 7 x \sin 5 x=0
\end{aligned}
$$

From this:

1) $\sin 7 x=0,7 x=\pi n, \quad x_{1}=\frac{\pi n}{7}, \quad n \in Z$;
2) $\sin 5 x=0, \quad 5 x=\pi k, \quad x_{2}=\frac{\pi k}{5}, \quad k \in Z$.

Answer: $\quad x_{1}=\frac{\pi n}{7} ; x_{2}=\frac{\pi k}{5}, \quad n, k \in Z$.