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Math Help - Trigonometric Equation Help

  1. #1
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    Jan 2010
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    Trigonometric Equation Help

    For Negative Pi Less Than or Equal to Theta which is less than or equal to Pi
    find all solutions


    2(sin^2 x - cos^2 x ) = 8sinx-5

    x is theta, and it is not in the exponent.

    I have issues solving trigonometric equations of this style because I seem to be unsure how to handle the exponents.

    I seek assistance here as usual because the lack of my educator's teaching skills.
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  2. #2
    Senior Member
    Joined
    Nov 2009
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    Smile

    Quote Originally Posted by jd3005 View Post
    For Negative Pi Less Than or Equal to Theta which is less than or equal to Pi
    find all solutions


    2(sin^2 x - cos^2 x ) = 8sinx-5

    x is theta, and it is not in the exponent.

    I have issues solving trigonometric equations of this style because I seem to be unsure how to handle the exponents.

    I seek assistance here as usual because the lack of my educator's teaching skills.
    \cos^2 x = 1 - \sin^2 x

    so

    2(sin^2 x - cos^2 x ) = 8sinx-5

    2(\sin^2 x - (1 - \sin^2 x )) = 8 \sin x - 5

    4\sin^2 x -8 \sin x +4 = 0

    \sin^2 x -2 \sin x + 1 = 0

    let t = \sin x

    solve this quadratic equation
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  3. #3
    Junior Member
    Joined
    Jun 2009
    Posts
    61
    you need to first use the identity:

    sin^2(x)-cos^2(x) = cos(2x)

    then the equation you get after this, use the identity:

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

    after that you will receive a square equation with

    \alpha sin^2(x) + \beta sin(x) + \gamma = 0

    and you need to solve that equation by calling sin(x) = t

    and sin^2(x) = t^2.

    from the answers you'll get, you need to figure which one is suitable for

    the condition of the question.
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