8.145. $\cos x - \cos 2x = \sin 3x$.
Solution.

Rewrite the equation as

$$
\begin{aligned}
& 2 \sin \frac{x+2 x}{2} \sin \frac{2 x-x}{2}-\sin 3 x=0, \quad 2 \sin \frac{3 x}{2} \sin \frac{x}{2}-\sin 2\left(\frac{3 x}{2}\right)=0 \\
& 2 \sin \frac{3 x}{2} \sin \frac{x}{2}-2 \sin \frac{3 x}{2} \cos \frac{3 x}{2}=0, \quad 2 \sin \frac{3 x}{2}\left(\sin \frac{x}{2}-\cos \frac{3 x}{2}\right)=0
\end{aligned}
$$

From this:

1) $\sin \frac{3 x}{2}=0, \frac{3 x}{2}=\pi k, \quad x_{1}=\frac{2}{3} \pi k, \quad k \in Z$;
2) $\sin \frac{x}{2}-\cos \frac{3 x}{2}=0 \Leftrightarrow 2 \sin \frac{\frac{x}{2}-\frac{\pi}{2}+\frac{3 x}{2}}{2} \cos \frac{\frac{x}{2}+\frac{\pi}{2}-\frac{3 x}{2}}{2}=0$, $\sin \left(x-\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)=0$

## From this

a) $\sin \left(x-\frac{\pi}{4}\right)=0, \quad x-\frac{\pi}{4}=\pi l, \quad x_{2}=\frac{\pi}{4}+\pi l=\frac{\pi}{4}(4 l+1), \quad l \in Z$;

b) $\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)=0, \quad-\frac{x}{2}+\frac{\pi}{4}=\frac{\pi}{2}+\pi n, \quad x_{3}=-\frac{1}{2} \pi+2 \pi n=\frac{\pi}{2}(4 n-1)$, $n \in Z$.

Answer: $\quad x_{1}=\frac{2}{3} \pi k ; x_{2}=\frac{\pi}{4}(4 l+1) ; x_{3}=\frac{\pi}{2}(4 n-1), \quad k, l, n \in Z$.