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Math Help - trig help

  1. #1
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    Nov 2008
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    trig help

    This is what the problem says:

    solve the equation for solutions in the interval [0,2pi]

    showing all steps would be appreciated.

    2sin²(x) - cos(x) - 1 = 0
    Last edited by ThePerfectHacker; November 30th 2008 at 05:07 PM.
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  2. #2
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    Smile

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  3. #3
    Junior Member
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    Nov 2008
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    I solved it this way:
    Since \sin^2(x)+\cos^2(x)=1, then \sin^2(x)=1-\cos^2(x)

    substitute in your equation and you get
    2(1-\cos^2(x))-\cos(x)-1=0

    using distributive and reordering terms you get
    -2 \cos^2(x)-\cos(x)+1=0

    Now you have to make a change of variables, so let u=\cos(x), then x=\cos^{-1}(u) (you will need this later)

    and substituting in this last equation you get
    -2u^2-u+1=0
    which has roots u_1=-1 and u_2=1/2

    Then x_1=\cos^{-1}(-1)=\pi and x_2=\cos^{-1}(1/2)=\pi/3 or 5\pi/3\

    Those are the 3 solutions
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  4. #4
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    thank you guys SO much!
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