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Math Help - Solving trigonometric equations help

  1. #1
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    Solving trigonometric equations help

    Hey guys.

    Here's the problem. We did not cover a problem like this in class on Thursday, so that's why I'm confused.

    3\tan^22x-1=0

    I began to solve the trigonometric equation, and here's how I went about it.

    3\tan^22x = 1

    tan^22x = \frac{1}{3}

    Here's where I'm stuck. I'm thinking of taking the square root of both sides, but then I'm left with:

    \tan2x = \pm\frac{\sqrt{3}}{3}

    And then once I divide two out, I'm left with a weird fraction that doesn't look like anything that can be found on the unit circle.

    Am I doing anything wrong here?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathgeek777 View Post
    Hey guys.

    Here's the problem. We did not cover a problem like this in class on Thursday, so that's why I'm confused.

    3\tan^22x-1=0

    I began to solve the trigonometric equation, and here's how I went about it.

    3\tan^22x = 1

    tan^22x = \frac{1}{3}

    Here's where I'm stuck. I'm thinking of taking the square root of both sides, but then I'm left with:

    \tan2x = \pm\frac{\sqrt{3}}{3}

    And then once I divide two out, I'm left with a weird fraction that doesn't look like anything that can be found on the unit circle.

    Am I doing anything wrong here?
    Tsk tsk.

    tan^{-1} \left ( \frac{\sqrt{3}}{3} \right ) = \frac{\pi}{6}

    Now, this is a reference angle, and tangent is positive. Thus your possible values for 2x are \frac{\pi}{6}, \frac{7\pi}{6}.

    So
    tan(2x) = \frac{\sqrt{3}}{2} \implies 2x = \frac{\pi}{6}, \frac{7\pi}{6}

    Now you do it for tan^{-1} \left ( -\frac{\sqrt{3}}{3} \right ).

    -Dan
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  3. #3
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    The weird thing is that I'm supposed to use this equation to verify that  x = \frac{\pi}{12} and x=\frac{5\pi}{12} are solutions of the equation.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathgeek777 View Post
    The weird thing is that I'm supposed to use this equation to verify that  x = \frac{\pi}{12} and x=\frac{5\pi}{12} are solutions of the equation.
    Why is that weird? \frac{\pi}{12} is a solution using tan(2x) = +\frac{\sqrt{3}}{3} and \frac{5\pi}{12} is a solution using tan(2x) = -\frac{\sqrt{3}}{3}.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Why is that weird? \frac{\pi}{12} is a solution using tan(2x) = +\frac{\sqrt{3}}{3} and \frac{5\pi}{12} is a solution using tan(2x) = -\frac{\sqrt{3}}{3}.

    -Dan
    I'm just trying to figure out how you arrived at the answer.

    EDIT: Nevermind, I just realized it.
    Last edited by mathgeek777; January 12th 2008 at 02:52 PM.
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