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Thread: Solving

  1. #1
    Junior Member Hasan1's Avatar
    Dec 2007


    $\displaystyle \tan2x = 8 \cos^2x - \cot x [0,\frac{\pi}{2}]$
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  2. #2
    Jun 2006
    San Diego
    Arrange it all on one side so it's

    $\displaystyle \tan 2x + \cot x - 8 \cos^2 x = 0$.

    Now the strategy is to factor the left side to get smaller pieces that are forced to be equal to zero.

    Start by recalling $\displaystyle \tan 2x = \frac{\sin 2x}{\cos 2x}$ and $\displaystyle \cot x = \frac{\cos x}{\sin x}$, so multiply through by $\displaystyle \cos 2x \sin x$.

    You get:

    $\displaystyle \sin 2x \sin x + \cos x \cos 2x - 8 \cos^2 x \cos 2x \sin x = 0$.

    Now remember $\displaystyle \sin 2x = 2\sin x \cos x$. So:

    $\displaystyle 2 \sin^2 x \cos x + \cos x \cos 2x - 8 \cos^2 x \cos 2x \sin x = 0$.

    And now we can make our first factor, which is $\displaystyle \cos x$.

    $\displaystyle \cos x \left(2 \sin^2 x + \cos 2x - 8 \cos x \cos 2x \sin x \right) = 0$.

    No we know $\displaystyle \cos x = 0$ when $\displaystyle x = \frac{\pi}{2}$ (in your range), so that is one solution. Let's forget about that factor now and concentrate on the rest:

    $\displaystyle 2 \sin^2 x + \cos 2x - 8 \cos x \cos 2x \sin x = 0$.

    Recall that $\displaystyle \cos 2x = \cos^2 x - \sin^2 x$, and apply this to the middle term:

    $\displaystyle 2 \sin^2 x + \cos^2 x - \sin^2 x - 8 \cos x \cos 2x \sin x = 0$

    Simplifies to:

    $\displaystyle \sin^2 x + \cos^2 x - 8 \cos x \cos 2x \sin x = 0$.

    Now, remember that $\displaystyle \sin^2 x + \cos^2 x = 1$, so we now have:

    $\displaystyle 1 - 8 \cos x \cos 2x \sin x = 0$.

    But refer again to the fact that $\displaystyle 2 \sin x \cos x = \sin 2x$. so we have:

    $\displaystyle 1 - 4 \sin 2x \cos 2x = 0$.

    Apply the above identity again:

    $\displaystyle 1 - 2 \sin 4x = 0$.

    Now solve for $\displaystyle x$ by just isolating it the old fashioned way. From this, you should get two more values in the interval $\displaystyle 0$ to $\displaystyle \frac{\pi}{2}$, making three values total because of the one we found earlier.
    Last edited by Soltras; Dec 15th 2007 at 04:19 PM.
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