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Answer 1 · 2018-02-16T07:07:01

The answers are $x=pi/3,pi,(5pi)/3$ .

$sin^2x=cos^2(x/2)$

$sin^2x=(cos(x/2))^2$

$sin^2x=(sqrt((1+cosx)/2))^2$

$sin^2x=(1+cosx)/2$

$2sin^2x=1+cosx$

$2(1-cos^2x)-cosx-1=0$

$2-2cos^2x-cosx-1=0$

$-2cosx^2x-cosx+1=0$

Let $u = cosx$ :

$-2u^2-u+1=0$

$2u^2+u-1=0$

$(2u-1)(u+1)=0$

$u=-1,1/2$

Substitute $cosx$ back in for $u$ :

$cosx=-1,cosx=1/2$

The solution for $cosx=-1$ is $pi$ , and solutions for $cosx=1/2$ are $pi/3$ and $(5pi)/3$ . The final solution set is:

$x=pi/3,pi,(5pi)/3$