SOLUTION: Solve: sin2x = cos x ,{{{-2pi <= x <= pi}}}

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Question 550916: Solve:
sin2x = cos x ,-2pi+%3C=+x+%3C=+pi
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
sin2x = cos x ,-2pi+%3C=+x+%3C=+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
------------
2sin(x) = 1
sin(x) = 1/2
x = pi/6, 5pi/6, -11pi/6, -7pi/6