Question

Solve: $\frac{\sin x}{\sin 3x}$

Answer 1

$\frac{\sin x}{\sin 3x}=\frac{\sin x}{3\sin x-4{\sin }^{3}x}$

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