8.001. $\cos 3 x-\sin x=\sqrt{3}(\cos x-\sin 3 x)$

8.001. $\cos 3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$
## Solution.

$\cos 3 x-\sqrt{3} \cos x=\sin x-\sqrt{3} \sin 3 x, \Leftrightarrow \frac{1}{2} \cos 3 x+\frac{\sqrt{3}}{2} \sin 3 x=$ $=\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x, \Leftrightarrow \cos 3 x \cos \frac{\pi}{3}+\sin 3 x \sin \frac{\pi}{3}=$ $=\cos \frac{\pi}{6} \cos x+\sin \frac{\pi}{6} \sin x, \Leftrightarrow \cos \left(3 x-\frac{\pi}{3}\right)=\cos \left(x-\frac{\pi}{6}\right) \Leftrightarrow$

$$
\Leftrightarrow \cos \left(3 x-\frac{\pi}{3}\right)-\cos \left(x-\frac{\pi}{6}\right)=0, \Leftrightarrow-2 \sin \frac{3 x-\frac{\pi}{3}+x-\frac{\pi}{6}}{2} \times
$$

$$
\times \sin \frac{3 x-\frac{\pi}{3}-x+\frac{\pi}{6}}{2}=0 \Leftrightarrow \sin \left(2 x-\frac{\pi}{4}\right) \sin \left(x-\frac{\pi}{12}\right)=0 .
$$

Then:

1) $\sin \left(2 x-\frac{\pi}{4}\right)=0,2 x_{1}-\frac{\pi}{4}=\pi k, \quad x_{1}=\frac{\pi}{8}+\frac{\pi k}{2}=\frac{\pi}{8}(4 k+1), k \in Z$;
2) $\sin \left(x-\frac{\pi}{12}\right)=0, \quad x_{2}-\frac{\pi}{12}=\pi n, \quad x_{2}=\frac{\pi}{12}+\pi n=\frac{\pi}{12}(12 n+1), \quad n \in Z$.

Answer: $\quad x_{1}=\frac{\pi}{8}(4 k+1) x_{2}=\frac{\pi}{12}(12 n+1) \quad k, n \in Z$.