Question

Solve the following equation:
$2\sin 2x\sin 4x-\cos 2x=\sin 3x$

Answer 1

$\cos (4x-2x)-\cos \left(\tan 2x\right)-\cos 2x=\sin 3x$

$-(1-{2\sin }^{2}3x)=\sin 3x$

${2\sin }^{2}3x-\sin 3x-1=0$

$(2\sin 3x+1)(\sin 3x-1)=0$

So, $\sin 3x=-\frac{1}{2},3x=n\pi +{(-1)}^n(-\frac{\pi }{6}),x=n\frac{\pi }{3}+{(-1)}^n\frac{\pi }{18}.$ and $\sin 3x=-1,3x=2n\pi -\frac{\pi }{2},x=2n\frac{\pi }{3}-\frac{\pi }{6}.$