Question

Finding solutions in an interval for an equation with sine and cosine using double-angle identities

Answer 1

Given:

$\cos (2x)+\sin x=0$$\text{Use }\cos (2x)=1-2\sin ^2x.$

$1-2\sin ^2x+\sin x=0$

Multiplying by (-1) on both sides, we get

$2\sin ^2x-\sin x-1=0$

$2\sin ^2x-2\sin x+\sin x-1=0$

$2\sin ^{}x(\sin x-1)+1(\sin x-1)=0$

Taking out common terms.

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

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

$\sin x=1,\sin x=-\frac{1}{2}$

$Use\text{ }\sin (\frac{\pi}{2})=1,\sin (\pi+\frac{\pi}{6})=-\frac{1}{2}.$$\sin x=\sin (\frac{\pi}{2}),\sin x=\sin (\pi+\frac{\pi}{6})$

$\sin x=\sin (\frac{\pi}{2}),\sin x=\sin (\frac{7\pi}{6})$

$x=\frac{\pi}{2},\frac{7\pi}{6}$

Hence the solution is

$x=\frac{\pi}{2},\frac{7\pi}{6}$