The general solution of $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ is

$\sin 3\alpha =4\sin \alpha \left({\sin }^{2}x-{\sin }^2\alpha \right)$
$\begin{array}{l}\Rightarrow 3\sin \alpha -4{\sin }^3\alpha =4\sin \alpha {\sin }^{2}x-4{\sin }^3\alpha \\ \Rightarrow 3\sin \cdot \alpha =4\sin \alpha {\sin }^{2}x\text{ or }\sin \alpha =0\end{array}$
$if \sin \alpha \ne 0,{\sin }^{2}x=3/4=(\sqrt{3}/2{)}^2={\sin }^2(\pi /3)$ $\text{ therefore }x=n\pi \pm \pi /3,\mathrm{\forall }n\in Z$
if $\sin \alpha =0,\text{ i.e., }\alpha =n\pi ,\text{ equation becomes an identity. }$