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Math Help - [SOLVED] Trig equation

  1. #1
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    [SOLVED] Trig equation

    Hello

    Im having problems to solve the following trig equation.

    cosx-\frac{\frac{1}{2}}{cosx}=\frac{1}{2}

    Im given the following information

    x_{1}=2n\pi

    x_{2}=\frac{a\pi }{b}+2n\pi

    x_{3}=\frac{c\pi }{d}+2n\pi

    0 < \frac{a\pi }{b}+2n\pi < \frac{c\pi }{d}+2n\pi < 2\pi

    where n are integers

    im not allowed to use a calculator, so im unsure how to eaven start...
    Last edited by JannetheSwede; July 4th 2009 at 06:49 AM.
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  2. #2
    Super Member malaygoel's Avatar
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    What we have to solve or find here?
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  3. #3
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    I have to find a, b, c and d

    x_{2}=\frac{a\pi }{b}+2n\pi

    x_{3}=\frac{c\pi }{d}+2n\pi
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  4. #4
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    Quote Originally Posted by JannetheSwede View Post

    Im having problems to solve the following trig equation.

    cos x-(1/2)/cos x=1/2

    \frac{\cos{x} - \frac{1}{2}}{\cos{x}} = \frac{1}{2}

    \frac{2\cos{x} - 1}{\cos{x}} = 1

    2\cos{x} - 1 = \cos{x}

    \cos{x} = 1

    x = 2k\pi , k \in \mathbb{Z}
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  5. #5
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    Quote Originally Posted by skeeter View Post
    \frac{\cos{x} - \frac{1}{2}}{\cos{x}} = \frac{1}{2}

    \frac{2\cos{x} - 1}{\cos{x}} = 1

    2\cos{x} - 1 = \cos{x}

    \cos{x} = 1

    x = 2k\pi , k \in \mathbb{Z}
    Yea thats one of three awnsers, but what i'm looking for are the other two... :S

    btw, just to clearify cosx-\frac{\frac{1}{2}}{cosx}=\frac{1}{2}
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  6. #6
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    Quote Originally Posted by JannetheSwede View Post
    Yea thats one of three awnsers, but what i'm looking for are the other two... :S
    what makes you think there are more solutions?
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  7. #7
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    ok, i have changed the initial question to clearify what it was
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  8. #8
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    Quote Originally Posted by JannetheSwede View Post
    ok, i have changed the initial question to clearify what it was
    \cos{x} - \frac{\frac{1}{2}}{\cos{x}} = \frac{1}{2}<br />

    multiply every term by 2\cos{x} ...

    2\cos^2{x} - 1 = \cos{x}

    2\cos^2{x} - \cos{x} - 1 = 0

    (2\cos{x} + 1)(\cos{x} - 1) = 0

    \cos{x} = -\frac{1}{2} ... \cos{x} = 1

    you finish up.
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  9. #9
    Super Member malaygoel's Avatar
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    2cos^2x-cosx-1=0

    cosx=1,\frac{-1}{2}

    Spoiler:

    cosx=1 gives x=2n\pi

    cosx=\frac{-1}{2} gives x=\frac{2\pi}{3},\frac{4\pi}{3}
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  10. #10
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    Ummh, or no... I still don't get it

    I tried to apply this on another problem of the same type

    cosx+\frac{\frac{1}{2}}{cosx}=\frac{3}{2}

    but I was unable to solve it...

    I did like this

    2cos^{2}x+cos x=3cos x
    2cos^{2}x-2cos x=0

    and thats how far I got... what do I do from there?
    Last edited by JannetheSwede; July 4th 2009 at 07:43 AM.
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  11. #11
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    Quote Originally Posted by JannetheSwede View Post
    Ummh, or no... I still don't get it

    I tried to apply this on another problem of the same type

    cosx+\frac{\frac{1}{2}}{cosx}=\frac{3}{2}

    but I was unable to solve it...

    I did like this

    2cos^{2}x+cos x=3cos x
    2cos^{2}x-2cos x=0

    and thats how far I got... what do I do from there?

    you are wrong .

    It simplifies to

    2cos^2x-3cosx+1=0

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

    cosx=\frac{1}{2} , cosx=1
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