Use the power reduction or half-angle identities to solve the equation sin^2x=cos^2(x/2) on the interval (0,2pi).

Answer is B

 sin²(x) = cos²(x/2) 
1-cos²(x) = cos²(x/2) 
1-cos²(x) = 1/2 + 1/2cos(x) 

This is a quadratic in cos(x) 
0 = cos²(x) - (1/2) + 1/2cos(x) 
0 = 2cos²(x) + cos(x) - 1 
0 = (2cos(x)-1)(cos(x)+1) 

cos(x) = -1, cos(x) = 1/2 
so x=π, x=π/3 

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