Question #88cb2

1 Answer
Feb 16, 2018

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

Explanation:

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