Trig

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2sin^2(x) = 2 + cos(x)
interval [0,2pi)

How do I solve this? Help would be greatly appreciated!

  • Trig -

    2(1-cos^2 x) = 2 + cos x

    2 - 2 cos^2 x = 2 + cos x
    2 cos^2 x + cos x = 0

    cos x (2 cos x + 1 ) = 0
    cos x = 0 when x = pi/2 or 90 degrees or 3 pi/2 or 270 deg
    or
    cos x = -1/2 at x = 2pi/3 or 120 degrees or at 4 pi/3 or 240 degrees

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