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Math Help - [SOLVED] solving an equation

  1. #1
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    [SOLVED] solving an equation

    I seem to have partially gotten this right, but am missing some answers, so chances are i stumbled across the (partial) right answer improperly. If someone could tell me what I'm doing wrong, that would be grand

    on an interval of 0 - 2pi
    tan2x - cotx = 0
    using double angle identities and cotanget identity
    \frac{2tanx}{1-tan^2x} - \frac{1}{tanx} = 0
    \frac{2tanx}{1-tan^2x} = \frac{1}{tanx}
    multiply both sides by tanx
    \frac{2tan^2x}{1-tan^2x} = 1
    multiply both sides by 1-tan^2x
    2tan^2x = 1-tan^2x
    3tan^2x = 1
    tan^x = \frac{1}{3}
    tanx = \sqrt{\frac{1}{3}}

    so that gives an answer at \frac{pi}{6}, \frac{5pi}{6}, \frac{7pi}{6} and \frac{11pi}{6}, but the book specifies additional answers at \frac{pi}{2}, \frac{3pi}{2}. Did I make a mistake above? If not, where are those 2 additional answers from?
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  2. #2
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    I think you "lost" the two additional solutions when you multiplied by \tan{x} .


    \tan(2x) - \cot{x} = 0

    \frac{\sin(2x)}{\cos(2x)} - \frac{\cos{x}}{\sin{x}} = 0<br />

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

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

    2\sin^2{x}\cos{x} - \cos{x}(2\cos^2{x}-1) = 0<br />

    2(1-\cos^2{x})\cos{x} - \cos{x}(2\cos^2{x}-1) = 0<br />

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

    3\cos{x} - 4\cos^3{x} = 0

    \cos{x}(3 - 4\cos^2{x}) = 0

    \cos{x} = 0

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

    \cos{x} = \pm \frac{\sqrt{3}}{2}

    x = \frac{\pi}{6} \, , \, \frac{5\pi}{6} \, , \, \frac{7\pi}{6} \, , \, \frac{11\pi}{6}
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  3. #3
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    ah, that's right, you're not supposed to multiply/divide by trig functions for fear of losing a solution. Totally forgot that. Thanks for the assistance.
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