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Math Help - Solving equations

  1. #1
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    Solving equations

    Hi
    Need help solving cot(x)+3tan(x)=5cosec(x), i have tried to do the following:

    \frac{1}{tan(x)+3tan(x)}=\frac{5}{sin(x)}

    \frac{cos(x)}{sin(x)}+\frac{3sin(x)}{cos(x)}=\frac  {5}{sin(x)}

    \frac{cos^2(x)+3sin^2(x)}{sin(x)cos(x)}=\frac{5cos  (x)}{sin(x)cos(x)}

    after this i contiue and i couldn't find the solution.

    P.S
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  2. #2
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    \cot(x)+3\tan(x)\neq \frac{1}{\tan(x)+3\tan(x)}

    Try

     <br />
\cot(x)+3\tan(x)= \frac{1}{\tan(x)}+3\tan(x)
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  3. #3
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    <br />
\frac{cos^2(x)+3sin^2(x)=5cos(x)}{sin(x)cos(x)}<br />

    1-sin^2(x)+3sin^2(x)=5cos(x)

    2sin^2(x)=5cos(x)-1

    1-cos^2(x)=5cos(x)-1

    2=5cos(x)+cos^2(x)

    2=cos(x)(5+cos(x))

    so cos(x)=2 and 5+cos(x)=2

    Someone tell me where i have gone wrong.
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  4. #4
    Super Member Bacterius's Avatar
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    There is something wrong with your first equation. You cannot divide an equation. You need to put the fraction on both sides :

    \frac{cos^2(x)+3sin^2(x)}{sin(x)cos(x)}=\frac{5cos  (x)}{sin(x)cos(x)}

    This simplifies trivially to :

    cos^2(x)+3sin^2(x) = 5cos(x)

    Now you can use the trigonometric identity : cos^2(x) + sin^2(x) = 1

    (1 - sin^2(x) )+3sin^2(x) = 5cos(x)

    1 + 2sin^2(x) = 5cos(x)

    1 + 2sin^2(x) = 5cos(x)

    Use this again : cos^2(x) + sin^2(x) = 1

    1 + 2(1 - cos^2(x)) = 5cos(x)

    1 + 2 - 2 cos^2(x)) = 5cos(x)

    3 - 2 cos^2(x)) = 5cos(x)

    2 cos^2(x) + 5cos(x) - 3 = 0

    And you are left with a quadratic equation, which can be easily solved for cos(x), and thus for x.
    ____________________________________

    You got it wrong between the third and the fourth line : 2 sin^2 (x) = 2 - 2 cos^2 (x) \neq 1 - cos^2(x)
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