Question

Solve $\cos 3x+\sin 2x-\sin 4x=0$, then $x=(2n+1)\frac{\pi }{m},n\in I$ or $n\pi +{(-1)}^n\frac{\pi }{m},n\in I$. Find the value of $m$.

Answer 1

$\cos 3x+\sin 2x-\sin 4x=0$
$\Rightarrow \cos 3x+2\cos 3x.\sin (-x)=0$ $(\sin C-\sin D=2cos(\frac{C+D}{2}\left)\sin \right(\frac{C-D}{2}\left)\right)$
$\Rightarrow \cos 3x-2\cos 3x.\sin x=0$
$\Rightarrow \cos 3x(1-2\sin x)=0$
$\Rightarrow \cos 3x=0$ or $1-2\sin =0$
$\Rightarrow 3x=(2n+1)\frac{\pi }{2},n\in I$ or $\sin x=\frac{1}{2}$
$\Rightarrow x=(2n+1)\frac{\pi }{6},n\in I$ or $x=n\pi +{(-1)}^n\frac{\pi }{6},n\in I$.
$\therefore$ Solution of given equation is
$x=(2n+1)\frac{\pi }{6},n\in I$ or $n\pi +{(-1)}^n\frac{\pi }{6},n\in I$