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Math Help - Please solve this equation for me!

  1. #1
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    Please solve this equation for me!

    I know this isn't very hard, but I seem to keep on messing up somewhere

    Using an extra variable A for simplicity,
    A= 8cosX *(sqrt(10) / (sqrt(10)+2)

    Solve for cosX (or (cosX)^2, whichever is found easier) in:
    5A/(40-A^2)=cosX


    Thanks very much in advance!
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  2. #2
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    Quote Originally Posted by machack View Post
    I know this isn't very hard, but I seem to keep on messing up somewhere

    Using an extra variable A for simplicity,
    A= 8cosX *(sqrt(10) / (sqrt(10)+2)

    Solve for cosX (or (cosX)^2, whichever is found easier) in:
    5A/(40-A^2)=cosX


    Thanks very much in advance!
    Is this your equation?

    A= 8\cos x \times \frac{\sqrt{10}}{\sqrt{10}+2}

    then

    \frac{A}{ \frac{8\sqrt{10}}{\sqrt{10}+2}}= \cos x

     \cos x=\frac{A}{ \frac{8\sqrt{10}}{\sqrt{10}+2} }

     \cos x=\frac{A(\sqrt{10}+2)}{8\sqrt{10}}
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  3. #3
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    No thats not the equation. A was an arbitrary variable made by me to make the real equation (the 2nd one) more managable to type. You gotta plug in A into the 2nd equation and solve for cos x.
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  4. #4
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    Quote Originally Posted by machack View Post
    I know this isn't very hard, but I seem to keep on messing up somewhere

    Using an extra variable A for simplicity,
    A= 8cosX *(sqrt(10) / (sqrt(10)+2)

    Solve for cosX (or (cosX)^2, whichever is found easier) in:
    5A/(40-A^2)=cosX


    Thanks very much in advance!
    Ok then,

    \frac{A}{(40-A^2)}=\cos x

    A= \frac{8\sqrt{10}\cos x}{\sqrt{10}+2}

    \frac{\frac{8\sqrt{10}\cos x}{\sqrt{10}+2}}{(40-(\frac{8\sqrt{10}\cos x}{\sqrt{10}+2})^2)}=\cos x

    \frac{\frac{8\sqrt{10}\cos x}{\sqrt{10}+2}}{(40-(\frac{8\sqrt{10}\cos x}{\sqrt{10}+2})^2)}=\cos x

    \frac{\frac{8\sqrt{10}\cos x}{\sqrt{10}+2}}{(40-(\frac{640\cos^2 x}{(\sqrt{10}+2)^2}))}=\cos x

    You then need to make a common denominator in the 'overall' denominator.

    Then divide out the fraction by multiplying the reciprocal, you should be able to then cancel out some of the \cos x

    Post what you try. I'll have a look.
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  5. #5
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    Ah i finally managed to solve this. It turned out your helped since solving for A first then plugging in the value of A to solve for cos^2(x) turned out to be simpler than solving for cos^2(x) right away.
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