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Find all solutions in the interval [0,2pi): sin2x=cosx

This is for my trig class. Thanks for the help!
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1 Answer

sin2x = 2sin x cos x
Solve 2sin x cos x = cos x for x,
cos x = 0, x = pi/2, 3pi/2
2sin x = 1
=> sin x = 1/2, x = pi/6, 5pi/6
Answer: x = pi/6, pi/2, 5pi/6, 3pi/2.