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Math Help - Trig Identities and Logs, Solve x within the given interval

  1. #1
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    Trig Identities and Logs, Solve x within the given interval

    Note: I wasn't really sure which board to post this on since this involves both trig identities and logs, but I figured that since Trigonometry generally comes after Algebra that I'd post it in this board. If I put this topic in the wrong board, I apologize for my confusion.


    Solve for x in the interval [0, 2 \pi]:

    \log_{2}\cot x - 2\log_{4}\csc2x = \log_{2}\cos x


    Now, so far I've tried turning the 2log_{4} into 4\log_{2} and then getting stuck with not knowing how to break down csc 2x so I can use log identities (subtracting logs with same bases becomes division of factors, for reference). I've also tried breaking it down to the following, but I hit a dead end because it doesn't really seem to lead anywhere that I can tell:

    \log_{2}csc x = \log_{4}csc4x^2


    The answer is supposed to be x = \frac{\pi}{3} , \frac{5\pi}{3}


    So, any tips on how to progress, or which way to progress, would be appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by tacoking View Post
    Note: I wasn't really sure which board to post this on since this involves both trig identities and logs, but I figured that since Trigonometry generally comes after Algebra that I'd post it in this board. If I put this topic in the wrong board, I apologize for my confusion.


    Solve for x in the interval [0, 2 \pi]:

    \log_{2}\cot x - 2\log_{4}\csc2x = \log_{2}\cos x


    Now, so far I've tried turning the 2log_{4} into 4\log_{2}
    Note that \log_4(x)=\frac{1}{2}\log_2(x)

    and then getting stuck with not knowing how to break down csc 2x
    Just write the whole thing with \sin and \cos. Then collect the left side inside a single \log_2, then drop the \log_2 on both sides of the equation. Remember that \sin (2x)=2\sin(x)\cos(x).
    After you have done that you get the pretty simple equation
    2\cos(x)=1
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  3. #3
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    I forgot about that trig identity... >.< Thanks much for the help!
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