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Math Help - Solving a trig equation

  1. #1
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    Question Solving a trig equation

    Find exact values for all solutions to the following equation on the interval
    0 ≤ t ≤ 2pi

    2 cos 2t = 1 - 4 cos t
    2(2cos^2 -1) = 1 - 4 cos t
    4 cos ^2 t -2 = 1 - 4 cost t
    4 cos^2 t + 4 cos t - 3 = 0
    (2cos t - 1)(2cos t + 3) = 0

    2cos t - 1 = 0
    2cos t = 1
    cos t = 1/2
    t = pi/3, 5pi/3

    2cos t = -3
    cos t = -3/2
    t = ?????


    So for this problem, would we only have two solutions? It seems as though there are no solutions for the inverse of cos -3/2. Please confirm this. Thank you so much.
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  2. #2
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    Quote Originally Posted by florx View Post
    Find exact values for all solutions to the following equation on the interval
    0 ≤ t ≤ 2pi

    2 cos 2t = 1 - 4 cos t
    2(2cos^2 -1) = 1 - 4 cos t
    4 cos ^2 t -2 = 1 - 4 cost t
    4 cos^2 t + 4 cos t - 3 = 0
    (2cos t - 1)(2cos t + 3) = 0

    2cos t - 1 = 0
    2cos t = 1
    cos t = 1/2
    t = pi/3, 5pi/3

    2cos t = -3
    cos t = -3/2
    t = ?????


    So for this problem, would we only have two solutions? It seems as though there are no solutions for the inverse of cos -3/2. Please confirm this. Thank you so much.
    Dear florx,

    You are correct. Actually 2cost-3\neq{0} since cost\neq{\frac{-3}{2}}
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