Question

Solve the following equations.
$sin\frac{3x}{2}cos\frac{x}{2}=\frac{sin2x}{2}.$

Answer 1

$\sin A\cos B=\frac{1}{2}(\sin (A+B)-\sin (A-B))$

$\Rightarrow \sin \frac{3x}{2}\cos \frac{x}{2}=\frac{1}{2}[\sin 2x-\sin x]$

$\Rightarrow \frac{\sin 2x}{2}-\frac{\sin x}{2}=\frac{\sin 2x}{2}$ (given )

$\Rightarrow \frac{\sin (-x)}{2}=0$

$\Rightarrow \sin (-x)=2n\pi$

$x=-2n\pi$