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March 25, 2017

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Solve for x in the interval [0,2pi) sin^2x+2cosx=2

  • Trig - ,

    sin^2x+2cosx=2
    1 - cos^2x + 2cosx - 2 = 0
    cos^2x - 2cosx + 1 = 0
    (cosx - 1)^2 = 0
    .
    .
    .
    cosx = 1

    Can you take it from there?

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