8.041. $\cos 9 x-\cos 7 x+\cos 3 x-\cos x=0$.
Solution.

Rewrite the equation as

$-2 \sin \frac{9 x+7 x}{2} \sin \frac{9 x-7 x}{2}-2 \sin \frac{3 x+x}{2} \sin \frac{3 x-x}{2}=0 \Leftrightarrow$

$\Leftrightarrow \sin 8 x \sin x+\sin 2 x \sin x=0 \Leftrightarrow \sin x(\sin 8 x+\sin 2 x)=0$.

From this:

1) $\sin x=0, \quad x_{1}=\pi n, \quad n \in Z$
2) $\sin 8 x+\sin 2 x=0 \Leftrightarrow 2 \sin \frac{8 x+2 x}{2} \cos \frac{8 x-2 x}{2}=0$, $\sin 5 x \cos 3 x=0$

Then either $\sin 5 x=0,5 x=\pi k, x_{2}=\frac{\pi k}{5}, k \in Z$, or $\cos 3 x=0$, $3 x=\frac{\pi}{2}+\pi m, x_{3}=\frac{\pi}{6}+\frac{\pi m}{3}=\frac{\pi}{6}(2 m+1), m \in Z ; x_{1}$ is included in $x_{2}$.

Answer: $\quad x_{1}=\frac{\pi k}{5} ; x_{2}=\frac{\pi}{6}(2 m+1), \quad k, m \in Z$.