8.237. $4 \sin 2 x \sin 5 x \sin 7 x - \sin 4 x = 0$.
## Solution.

Rewrite the equation as

$4 \sin 2 x \sin 5 x \sin 7 x - 2 \sin 2 x \cos 2 x = 0$,

$$
\begin{aligned}
& 2 \sin 2 x (2 \sin 5 x \sin 7 x - \cos 2 x) = 0 \Leftrightarrow \\
& \sin 2 x (\cos 2 x - \cos 12 x - \cos 2 x) = 0, \sin 2 x \cos 12 x = 0
\end{aligned}
$$

From this,

1) $\sin 2 x = 0, 2 x = \pi k, x_{1} = \frac{\pi k}{2}, k \in Z$;
2) $\cos 12 x = 0, 12 x = \frac{\pi}{2} + \pi n, x_{2} = \frac{\pi}{24} + \frac{\pi n}{12} = \frac{\pi}{24}(2 n + 1), n \in Z$.

Answer: $x_{1} = \frac{\pi k}{2} ; x_{2} = \frac{\pi}{24}(2 n + 1)$, where $k$ and $n \in Z$.