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

From the condition

$$
\begin{aligned}
& 2 \sin \frac{x+3 x}{2} \sin \frac{3 x-x}{2}-\sin 2 x=0, \quad 2 \sin 2 x \sin x-\sin 2 x=0 \\
& \sin 2 x(2 \sin x-1)=0
\end{aligned}
$$

## Hence

1) $\sin 2 x=0, 2 x=\pi k, \quad x_{1}=\frac{\pi \cdot k}{2}, k \in Z$;
2) $2 \sin x-1=0, \sin x=\frac{1}{2}, \quad x_{2}=(-1)^{n} \frac{\pi}{6}+\pi n, \quad n \in Z$.

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