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Math Help - [SOLVED] solving a trig. equation.

  1. #1
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    [SOLVED] solving a trig. equation.

    for
    0 less than or equal to x less than 2pi.

    1+cosx=1-cosx?
    I cant figure out how to solve this, you cant really factor it/ use the quadratic equation.
    I assume it has something to do with the identity
    cosx=cos(-x)

    If anyone could steer me in the direction of solving this Id appreciate it, thank's.
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  2. #2
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    Quote Originally Posted by brentwoodbc View Post
    for
    0 less than or equal to x less than 2pi.

    1+cosx=1-cosx?
    I cant figure out how to solve this, you cant really factor it/ use the quadratic equation.
    I assume it has something to do with the identity
    cosx=cos(-x)

    If anyone could steer me in the direction of solving this Id appreciate it, thank's.
    Hi

    If I well understand your equation is 1+\cos x = 1-\cos x right ?

    This is equivalent to 2\cos x = 0 therefore x = \frac{\pi}{2} or x = \frac{3\pi}{2}
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  3. #3
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    Quote Originally Posted by brentwoodbc View Post
    for
    0 less than or equal to x less than 2pi.

    1+cosx=1-cosx?
    I cant figure out how to solve this, you cant really factor it/ use the quadratic equation.
    I assume it has something to do with the identity
    cosx=cos(-x)

    If anyone could steer me in the direction of solving this Id appreciate it, thank's.
    1 + \cos{x} = 1 - \cos{x}

    2\cos{x} = 0

    \cos{x} = 0

    x = \frac{\pi}{2}

    x = \frac{3\pi}{2}
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  4. #4
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    thanks to both of you!
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