How to solve for x on the interval [0,2π)?

Question

Solve for x on the interval [0,2π). Cos x - sin2x = 0.

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Answer 1 · 2018-05-29T01:33:30

$x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$ $x=pi/6, pi/2, (5pi)/6, (3pi)/2$

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$cos x - sin 2 x = 0$ $cosx-sin2x=0$

$cos x - 2 sin x cos x = 0$ $cosx-2sinxcosx=0$

$cos x (1 - 2 sin x) = 0$ $cosx(1-2sinx)=0$

$cosx=0, :. x=pi/2, (3pi)/2$

$1-2sinx=0$

$sinx=1/2, :. x=pi/6, (5pi)/6$