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Math Help - Solving

  1. #1
    Junior Member Hasan1's Avatar
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    Solving

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

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

    You get:

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

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

    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 \cos x.

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

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

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

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

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

    Simplifies to:

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

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

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

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

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

    Apply the above identity again:

    1 - 2 \sin 4x = 0.

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