8.110. $\cos 3 x \cos 6 x=\cos 4 x \cos 7 x$.

8.110. $\cos 3 x \cos 6 x=\cos 4 x \cos 7 x$.
## Solution.

We have

$$
\begin{aligned}
& \frac{1}{2}(\cos (3 x-6 x)+\cos (3 x+6 x))=\frac{1}{2}(\cos (4 x-7 x)+\cos (4 x+7 x)) \\
& \cos 3 x+\cos 9 x-\cos 3 x-\cos 11 x=0, \cos 9 x-\cos 11 x=0 \Leftrightarrow \\
& \Leftrightarrow-2 \sin \frac{9 x+11 x}{2} \sin \frac{9 x-11 x}{2}=0, \quad \sin 10 x \sin x=0
\end{aligned}
$$

From this

1) $\sin 10 x=0, \quad 10 x=\pi n, \quad x_{1}=\frac{\pi n}{10}, n \in Z$
2) $\sin x=0, x_{2}=\pi k, k \in Z ; x_{2}$ is included in $x_{1}$.

Answer: $\quad x=\frac{\pi n}{10}, \quad n \in Z$.