8.095. $\sin \frac{7 x}{2} \cos \frac{3 x}{2}+\sin \frac{x}{2} \cos \frac{5 x}{2}+\sin 2 x \cos 7 x=0$.
## Solution.

Applying the formula $\sin \alpha \cos \beta=\frac{1}{2}(\sin (\alpha-\beta)+\sin (\alpha+\beta))$, we write the equation as

$$
\begin{aligned}
& \frac{1}{2}\left(\sin \left(\frac{7 x}{2}-\frac{3 x}{2}\right)+\sin \left(\frac{7 x}{2}+\frac{3 x}{2}\right)\right)+\frac{1}{2}\left(\sin \left(\frac{x}{2}-\frac{5 x}{2}\right)+\sin \left(\frac{x}{2}+\frac{5 x}{2}\right)\right)+ \\
& +\frac{1}{2}(\sin (2 x-7 x)+\sin (2 x+7 x))=0 \\
& \sin 2 x+\sin 5 x-\sin 2 x+\sin 3 x-\sin 5 x+\sin 9 x=0 \\
& \sin 9 x+\sin 3 x=0 \Leftrightarrow 2 \sin \frac{9 x+3 x}{2} \cos \frac{9 x-3 x}{2}=0, \sin 6 x \cos 3 x=0
\end{aligned}
$$

From this,

1) $\sin 6 x=0, 6 x=\pi n, x_{1}=\frac{\pi n}{6}, n \in Z$
2) $\cos 3 x=0, 3 x=\frac{\pi}{2}+\pi k, x_{2}=\frac{\pi}{6}+\frac{\pi k}{3}, k \in Z ; x_{2}$ is included in $x_{1}$.

Answer: $\quad x=\frac{\pi n}{6}, n \in Z$.