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Math Help - trig equations help!

  1. #1
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    trig equations help!

    (cotx)(cos squaredx) = 2(cotx)
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  2. #2
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    I expect you're solving for \displaystyle x...


    \displaystyle \cot{x}\cos^2{x} = 2\cot{x}

    \displaystyle \cot{x}\cos^2{x} - 2\cot{x} = 0

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

    \displaystyle \cot{x} = 0 or \displaystyle \cos^2{x} - 2 = 0


    Case 1: \displaystyle \cot{x} = 0

    \displaystyle \frac{\cos{x}}{\sin{x}} = 0

    \displaystyle \cos{x} = 0

    \displaystyle x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n, n \in \mathbf{Z}

    \displaystyle x = \frac{\pi}{2} + \pi n, n \in \mathbf{Z}.


    Case 2: \displaystyle \cos^2{x} - 2 = 0

    \displaystyle \cos^2{x} = 2

    \displaystyle \cos{x} = \pm \sqrt{2}

    But since \displaystyle -1 \leq \cos{x} \leq 1 for all real \displaystyle x, there are not any real solutions.


    So the only solution to your equation is \displaystyle x = \frac{\pi}{2} + \pi n, n \in \mathbf{Z}.
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