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=-(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)`