$5.58 \sin 3 x=a \sin x$.
5.58 Transform the left side of the equation:

$\sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\cos 2 x \sin x=$

$=2 \sin x \cos ^{2} x+\cos ^{2} x \sin x-\sin ^{3} x=3 \sin x\left(1-\sin ^{2} x\right)-$ $-\sin ^{3} x=3 \sin x-4 \sin ^{3} x$.

It remains to solve the equation $3 \sin x-4 \sin ^{3} x=a \sin x$. We have:

1) $\sin x=0 ; x_{1}=\pi k$ for any $a$;
2) $\sin ^{2} x=\frac{3-a}{4} ;$ here $0 \leqslant \frac{3-a}{4} \leqslant 1 ; 0 \leqslant 3-a \leqslant 4$;

$-3 \leqslant-a \leqslant 1$, i.e., $-1 \leqslant a \leqslant 3-$ for such $a$ there exists $\sin x$.

Thus,

$$
\begin{aligned}
& 2 \sin ^{2} x=\frac{3-a}{2}, 1-\cos 2 x=\frac{3-a}{2} ; \cos 2 x=\frac{a-1}{2} \\
& x_{2}= \pm \frac{1}{2} \arccos \frac{a-1}{2}+\pi k \text{ for } -1 \leqslant a \leqslant 3
\end{aligned}
$$

Answer: $\quad x_{1}=\pi k$ for any $a ; x_{2}= \pm \frac{1}{2} \arccos \frac{a-1}{2}+\pi k \text{ for } -1 \leqslant a \leqslant 3$ $(k \in Z)$.