How do you solve $sin2x = cosx$ from [-2pi, pi/2]#?

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Answer 1 · 2015-08-13T04:48:24

Solve sin 2x = cos x

sin 2x - cos x = 0
2sin x.cos x - cos x = 0
cos x(2sin x - 1) = 0

a. cos x = 0 --> $x = pi/2$ and $x = (3pi)/2$

b. $sin x = 1/2$ --> $x = pi/6$ and $x = pi - pi/6 = (5pi)/6$

Answers: $pi/6, pi/2, (5pi)/6, (3pi)/2$

How do you solve sin2x = cosxsin2x = cosx from [-2pi, pi/2]#?