8.169. $\sin x + \sin 3x = 4 \cos^3 x$.
## Solution.

From the condition

$$
\begin{aligned}
& 2 \sin \frac{x+3 x}{2} \cos \frac{x-3 x}{2}-4 \cos ^{3} x=0, \quad 2 \sin 2 x \cos x-4 \cos ^{3} x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos x\left(2 \sin x \cos x-2 \cos ^{2} x\right)=0, \quad 4 \cos ^{2} x(\sin x-\cos x)=0
\end{aligned}
$$

From this:

1) $\cos x=0, \quad x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), \quad k \in Z$;
2) $\sin x-\cos x=0 \Leftrightarrow \operatorname{tg} x=1, x_{2}=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), \quad n \in Z$.

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