Question 550916
Solve:

sin2x = cos x ,{{{-2pi <= x <= pi}}}
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sin(2x) = cos(x)
2sin(x)cos(x) = cos(x)
2sin(x)cos(x) - cos(x) = 0
cos(x)*(2sin(x) - 1) = 0
cos(x) = 0
x = pi, -pi
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2sin(x) = 1
sin(x) = 1/2
x = pi/6, 5pi/6, -11pi/6, -7pi/6