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Math Help - Solving a Multiple angle equation using double angle formulas

  1. #1
    mathstudent101
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    Exclamation Solving a Multiple angle equation using double angle formulas

    I have a bunch of these problems to do but if someone can show me how to do one for example: 2((cos(t))^2)-cos(t)-1 = 0 so i have to find t with interval [0,2pi] can some one demonstrate?
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by mathstudent101 View Post
    I have a bunch of these problems to do but if someone can show me how to do one for example: 2((cos(t))^2)-cos(t)-1 = 0 so i have to find t with interval [0,2pi] can some one demonstrate?
    Hello,

    this equation has to be solved in 2 steps:
    1. solve for cos(t)
    2. solve for t

    to 1.:

    Substitute y = cos(t) and your equation becomes:

    2y² - y - 1 = 0

    which wil give: y = 1 or y = -0.5

    to 2.:

    Plug in the values:

    cos(t) = 1 ===> t = 0 or t = 2pi

    cos(t) = -0.5 ===> t = (2/3)pi or t = (4/3)pi

    EB
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  3. #3
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    Hello, mathstudent101!

    In this problem, the angles are the "same size".
    We don't need any identities.


    2·cos²(t) - cos(t) - 1 .= .0 . on [0, 2π]
    This is a quadratic . . . factor it.

    . . [2·cos(t) + 1][cos(t) - 1] .= .0


    Set each factor equal to zero and solve:

    . . 2·cos(t) + 1 .= .0 . . cos(t) = -½ . . t = 2π/3, 4π/3

    . . cos(t) - 1 .= .0 . . cos(t) = 1 . . t = 0

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