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PRE CALC PLEASE HELP!!!!!!!!

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find the solutions of the equation that are in the interval [0,2pi)

2-cos^2x=4sin^2(1/2x)

  • PRE CALC PLEASE HELP!!!!!!!! -

    2 - (1 - 2sin^2 (x/2) = 4sin^2 (x/2)
    1 = 2 sin^2 (x/2)
    sin^2 (x/2) = 1/2
    sin (x/2) = 1/√2
    x/2 = π/4 or x/2 = 3π/4

    then x = π/2 or x = 3π/2

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