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Thread: Equation help

  1. September 26th 2010, 12:02 PM #1
    pauline
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    Equation help

    Hello i have problems solving this one



    Equation help-trig.gif

    answer
    x= π/4 + k π/2

    thanks
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  2. September 26th 2010, 03:07 PM #2
    skeeter
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    note that \displaystyle \sin^2{u} = \frac{1 - \cos(2u)}{2}


    \displaystyle \frac{1 - \cos{x}}{1 - \cos{2x}} = 1 - \cos{x}<br />

    since \sin{x} \ne 0 , \cos{x} \ne 1 ...

    \displaystyle \frac{1}{1 - \cos(2x)} = 1

    \cos(2x) = 0

    \displaystyle 2x = \frac{\pi}{2} + k\pi \, ; \, k \in \mathbb{Z}

    \displaystyle x = \frac{\pi}{4} + \frac{k\pi}{2}
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  3. September 27th 2010, 10:01 AM #3
    pauline
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    Hi i understand but at the end why

    Equation help-64_00f20ef9f2d0846358e8d3edcfa2f637.png

    and not
    2x= π/2 + 2 k π

    i can't understand that part ...
    thank you
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  4. September 27th 2010, 03:17 PM #4
    skeeter
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    Quote Originally Posted by pauline View Post
    Hi i understand but at the end why

    Click image for larger version.  Name: 64_00f20ef9f2d0846358e8d3edcfa2f637.png  Views: 32  Size: 1.7 KB  ID: 19078

    and not
    2x= π/2 + 2 k π

    i can't understand that part ...
    thank you
    because \cos{u} = 0 at \displaystyle u = \pm \frac{\pi}{2} \, , \, \pm \frac{3\pi}{2} \, , \, \pm \frac{5\pi}{2} \, , \, ...
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