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

Question

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

Answer 1

sin2x = 2sin x cos x

Solve 2sin x cos x = cos x for x,

cos x = 0, x = pi/2, 3pi/2

or

2sin x = 1

=> sin x = 1/2, x = pi/6, 5pi/6

Answer: x = pi/6, pi/2, 5pi/6, 3pi/2.