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Math Help - Solving for X, and General solutions?

  1. #1
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    Solving for X, and General solutions?

    I have 2cos^2(x)+cos(x)=1 and I have to solve for x within a standard circle(0-2pi).

    Then I have to find the general solution for the above equation and 2tan(x) +SqRt(12)=0(which I previously solved for.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Blahdkm View Post
    I have 2cos^2(x)+cos(x)=1 and I have to solve for x within a standard circle(0-2pi).

    [snip]
    Simply subtract one from both sides and the factor the quadratic

    2cos^2x+cosx-1=0

    (2cosx-1)(cosx+1)=0

    now see that cosx=1/2 or -1

    what angles do these correspond to?

    Quote Originally Posted by Blahdkm View Post
    [snip]

    Then I have to find the general solution for the above equation and 2tan(x) +SqRt(12)=0(which I previously solved for.
    2tanx=-\sqrt{12}
    2tanx=-2\sqrt{3}
    tanx=-\sqrt{3}
    i only know of two angles.......
    any questions?

    thanx mr. fantastic
    Last edited by VonNemo19; June 16th 2009 at 08:22 PM. Reason: Merged posts and some snipping of quote
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  3. #3
    MHF Contributor alexmahone's Avatar
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    2tanx=-\sqrt{12}
    2tanx=-2\sqrt{3}
    tanx=-\sqrt{3}
    i only know of two angles.......
    x=-\frac{\pi}{3}+n\pi
    where n is an integer.
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  4. #4
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    Thank you two SO MUCH for the help! Any chance of assistance finding the general solution of 2cos^2(x) + cos(x)=1?
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Blahdkm View Post
    Thank you two SO MUCH for the help! Any chance of assistance finding the general solution of 2cos^2(x) + cos(x)=1?
    cos x=\frac{1}{2} \implies x=2n\pi \pm \frac{\pi}{3}
    cos x=-1 \implies x=2n\pi \pm \pi
    Last edited by alexmahone; June 17th 2009 at 04:23 AM.
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