Question 1

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

Question 2

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.

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

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

$\sin x=1,\sin x=-(1)/(2)$

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

$\sin x=\sin ((\pi)/(2)),\sin x=\sin ((7\pi)/(6))$

$x=(\pi)/(2),(7\pi)/(6)$

Hence the solution is

$x=(\pi)/(2),(7\pi)/(6)$

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