How do you solve $sin 2 x = cos x$ $sin2x = cosx$ 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 --> $x = \frac{π}{2}$ $x = pi/2$ and $x = \frac{3 π}{2}$ $x = (3pi)/2$

b. $sin x = \frac{1}{2}$ $sin x = 1/2$ --> $x = \frac{π}{6}$ $x = pi/6$ and $x = π - \frac{π}{6} = \frac{5 π}{6}$ $x = pi - pi/6 = (5pi)/6$

Answers: $\frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$ $pi/6, pi/2, (5pi)/6, (3pi)/2$

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