8.080. $\cos x - \sqrt{3} \sin x = \cos 3x$.
## Solution.

From the condition

$$
(\cos x-\cos 3 x)-\sqrt{3} \sin x=0 \Leftrightarrow-2 \sin \frac{x+3 x}{2} \sin \frac{x-3 x}{2}-\sqrt{3} \sin x=0
$$

$2 \sin 2 x \sin x-\sqrt{3} \sin x=0, \sin x(2 \sin 2 x-\sqrt{3})=0$.

From this:

1) $\sin x=0, x_{1}=\pi n, \quad n \in Z$
2) $2 \sin 2 x-\sqrt{3}=0,2 \sin 2 x=\frac{\sqrt{3}}{2}, 2 x=(-1)^{k} \frac{\pi}{3}+\pi k$,

$$
x_{2}=(-1)^{k} \frac{\pi}{6}+\frac{\pi k}{2}, \quad k \in Z
$$

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