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

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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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