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=pi/6, pi/2, (5pi)/6, (3pi)/2#

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#cosx-sin2x=0#

#cosx-2sinxcosx=0#

#cosx(1-2sinx)=0#

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

#1-2sinx=0#

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