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Math Help - Where's the mistake

  1. #1
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    Where's the mistake

    I've made this exercise below. My broblem is that my solution doesn't match the solution in the book.Where did I make the mistake?

    sin ^{4} \frac{x}{3} +cos ^{4} \frac{x}{3}= \frac{5}{8}

     \left(1-cos ^{2} \frac{x}{3}   \right) ^{2}+cos ^{4} \frac{x}{3}= \frac{5}{8}

    1-2cos ^{2} \frac{x}{3}+cos ^{4} \frac{x}{3}+cos ^{4} \frac{x}{3}= \frac{5}{8}

    2cos ^{4} \frac{x}{3}-2cos ^{2} \frac{x}{3}=- \frac{3}{8}

    cos ^{4} \frac{x}{3}-cos ^{2} \frac{x}{3}=- \frac{3}{16}

    cos ^{4} \frac{x}{3}-cos ^{2} \frac{x}{3}+ \frac{3}{16}=0

    \Delta=1- \frac{3}{4}= \frac{1}{4}

     \sqrt{\Delta}= \frac{1}{2}

    cos ^{2} \frac{x}{3}= \frac{3}{4} \vee cos ^{2} \frac{x}{3}= \frac{1}{4}

    cos \frac{x}{3}= \frac{ \sqrt{3} }{2} \vee cos \frac{x}{3}=- \frac{ \sqrt{3} }{2} \vee cos \frac{x}{3}= \frac{1}{2} \vee cos \frac{x}{3}=- \frac{1}{2}

     \frac{x}{3}= \frac{\Pi}{3}+2k\Pi \vee  \frac{x}{3}=- \frac{\Pi}{3}+2k\Pi \vee  \frac{x}{3}= \frac{\Pi}{6}+2k\Pi \vee  \frac{x}{3}=- \frac{\Pi}{6}+2k\Pi

    x=\Pi+6k\Pi \vee x=-\Pi+6k\Pi \vee x= \frac{\Pi}{2}+6k\Pi \vee x=- \frac{\Pi}{6k\Pi}

    This is my result.

    Below it's the correct result:

    x=\Pi+3k\Pi \vee x=-\Pi+3k\Pi \vee x= \frac{\Pi}{2}+3k\Pi \vee x=- \frac{\Pi}{2}+3k\Pi
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  2. #2
    Moo
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    Hello,

    You're almost right.

    The mistake is here :

    \frac{x}{3}= \frac{\Pi}{3}+2k\Pi \vee \frac{x}{3}=- \frac{\Pi}{3}+2k\Pi \vee \frac{x}{3}= \frac{\Pi}{6}+2k\Pi \vee \frac{x}{3}=- \frac{\Pi}{6}+2k\Pi<br />
    Remember that \cos(x)=\cos(-x)
    so you should have been warned there's a problem here =)

    \cos \frac x3=-\frac{\sqrt{3}}{2}=\cos \frac{5 \pi}{6}

    Don't forget the \cos(x)=\cos(-x) formula, which should give twice more solutions.



    Now, you can shorten it.
    In fact, you can notice that \cos(x)=\cos(a) \implies x=a +2k \pi \vee x=-a+2k \pi

    \cos^2(x)=\cos^2(a) \implies \cos(x)=\cos(a) \vee \cos(x)=-\cos(a)=\cos(a+\pi)


    Try to gather all of these, you'll get that :

    \cos^2(x)=\cos^2(a) \implies x=a+ k \pi \vee x=-a+k \pi
    Last edited by Moo; October 19th 2008 at 10:11 AM. Reason: messed up with wedge and vee
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