Find all values of x in the interval [0,2pi] that satisfy the equation.
2sin^2x=1 and
sin2x = cosx

Find all values of x in the interval [0,2pi] that satisfy the equation.
2sin^2x=1 and
sin2x = cosx

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Would you do that by \[\sqrt{\sin^2x}=\sqrt{1}\] square rooting both sides?

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