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Math Help - Solving an equaion.

  1. #1
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    Solving an equaion.

    Find the zeros of the equation 2cos^2x+cos2x-1 = 0 between 0 and 2\pi.
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  2. #2
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     2\cos^2x+\cos2x-1 = 0

     2\cos^2x+(2\cos^2x-1)-1 = 0

     4\cos^2x-2 = 0

     4\cos^2x = 2

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

     \cos(x) = \frac{1}{\sqrt{2}}

    Should be easy from here, there are 2 solutions
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  3. #3
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    Quote Originally Posted by Captcha View Post
    Find the zeros of the equation 2cos^2x+cos2x-1 = 0 between 0 and 2\pi.
    2\cos^2{x} + \cos(2x) - 1 = 0

    note the double angle identity ... \cos(2x) = 2\cos^2{x} - 1<br />


    two ways to go from here ...

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

    \cos(2x) + \cos(2x) = 0

    2\cos(2x) = 0

    2x = \frac{\pi}{2} \, , \, \frac{3\pi}{2} \, , \, \frac{5\pi}{2} \, , \, \frac{7\pi}{2}

    x = \frac{\pi}{4} \, , \, \frac{3\pi}{4} \, , \, \frac{5\pi}{4} \, , \, \frac{7\pi}{4}


    or ...


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

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

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

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

    \cos{x} = \pm \frac{1}{\sqrt{2}}

    x = \frac{\pi}{4} \, , \, \frac{3\pi}{4} \, , \, \frac{5\pi}{4} \, , \, \frac{7\pi}{4}
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