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Math Help - solving trigonometry quadradic

  1. #1
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    Exclamation solving trigonometry quadradic

    solve the following equations if 0 <(less than or equal to) a <(less than) 2pi

    2sin^2[a] = 1 - sin [a]

    please answer clearly showing all working
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  2. #2
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    Quote Originally Posted by T_o_M View Post
    solve the following equations if 0 <(less than or equal to) a <(less than) 2pi

    2sin^2[a] = 1 - sin [a]

    please answer clearly showing all working
    2sin^2(a) + sin(a) - 1 = 0

    Let x = sin(a). Then
    2x^2 + x - 1 = 0

    This factors:
    (2x - 1)(x + 1) = 0

    So
    2x - 1 = 0 ==> x = 1/2 ==> sin(a) = 1/2
    or
    x + 1 = 0 ==> x = -1 ==> sin(a) = -1

    So
    sin(a) = 1/2 ==> a = (pi)/6 rad, 5(pi)/6 rad
    or
    sin(a) = -1 ==> a = 3(pi)/2 rad

    Thus
    a = (pi)/6, 5(pi)/6, or 3(pi)/2.

    -Dan
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