8.019. $\cos \frac{x}{2} \cos \frac{3 x}{2}-\sin x \sin 3 x-\sin 2 x \sin 3 x=0$.
Solution.

Rewrite the equation as

$$
\begin{aligned}
& \cos x + \cos 2x - \cos 2x + \cos 4x - \cos x + \cos 5x = 0 \Leftrightarrow \\
& \Leftrightarrow \cos 4x + \cos 5x = 0, \Leftrightarrow 2 \cos \frac{9x}{2} \cos \frac{x}{2} = 0
\end{aligned}
$$

Then either

1) $\cos \frac{9x}{2} = 0, \quad \frac{9x}{2} = \frac{\pi}{2} + \pi k, \quad x_1 = \frac{\pi}{9} + \frac{2\pi k}{9} = \frac{\pi}{9}(2k + 1), \quad k \in \mathbb{Z}$,

or
2) $\cos \frac{x}{2} = 0, \quad \frac{x}{2} = \frac{\pi}{2} + \pi n, \quad x_2 = \pi + 2\pi n = \pi(2n + 1), \quad n \in \mathbb{Z}; x_2 \quad$ is included in $x_1$.

Answer: $\quad x = \frac{\pi}{9}(2k + 1) \quad k \in \mathbb{Z}$.