8.371. $\sin 3 x=a \sin x$.

8.371. $\sin 3 x=a \sin x$.
## Solution.

From the formula $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$, we have

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

From this,

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

$$
\left\{\begin{array} { l } 
{ 2 x = \pm \operatorname { a r c c o s } \frac { a - 1 } { 2 } + 2 \pi n , } \\
{ - 1 \leq \frac { a - 1 } { 2 } \leq 1 , }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{2}= \pm \frac{1}{2} \arccos \frac{a-1}{2}+\pi n, n \in Z \\
-1 \leq a \leq 3
\end{array}\right.\right.
$$

Answer: $x_{1}=\pi k$, where $a \in R ; x_{2}= \pm \frac{1}{2} \arccos \frac{a-1}{2}+\pi n$, where $a \in[-1 ; 3]$; $k$ and $n \in Z$.