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Math Help - Find general solution

  1. #1
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    Find general solution

    Find the general solution of the equation:
    10 \sin \frac{\pi x}{3}+24 \cos \frac{\pi x}{3}=13

    My attempt:
    I used the identity R (\sin \theta \cos \alpha+\cos \theta \sin \alpha)\equiv R \cos(\theta-\alpha) and I could get some angles, but when I tried to form the a general solution I got stuck. The answer is supposed to be: (6n\pm1)+\frac{3}{\pi}\arctan\frac{5}{12}. My problem is with the first part, why (6n\pm1)? I was thinking maybe there should be a \pi there?
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    Find the general solution of the equation:
    10 \sin \frac{\pi x}{3}+24 \cos \frac{\pi x}{3}=13

    My attempt:
    I used the identity R (\sin \theta \cos \alpha+\cos \theta \sin \alpha)\equiv R \cos(\theta-\alpha) and I could get some angles, but when I tried to form the a general solution I got stuck. The answer is supposed to be: (6n\pm1)+\frac{3}{\pi}\arctan\frac{5}{12}. My problem is with the first part, why (6n\pm1)? I was thinking maybe there should be a \pi there?
    Thanks
    To save people re-inventing the wheel (and hence wasting time), please post your working up to where you get the equation in the form R \cos(x - \alpha) = 13.
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  3. #3
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    10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13

    10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv R(sin\frac{\pi x}{3}cos\alpha+cos\frac{\pi x}{3}sin\alpha)\equiv Rcos(\frac{\pi x}{3}- \alpha)
    Rcos\alpha=10

    Rsin\alpha=24

    R=\sqrt{10^2+24^2}=\sqrt{676}=26

    \alpha=\arctan\frac{24}{10}

    So 10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv 26cos(\frac{\pi x}{3}-\alpha)
    26cos(\frac{\pi x}{3}-\arctan\frac{24}{10})=13
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  4. #4
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    Quote Originally Posted by arze View Post
    10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13

    10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv R(sin\frac{\pi x}{3}cos\alpha+cos\frac{\pi x}{3}sin\alpha)\equiv Rcos(\frac{\pi x}{3}- \alpha)
    Oops! That should be
    24cos\frac{\pi x}{3}+10sin\frac{\pi x}{3}\equiv R({\color{red}cos}\frac{\pi x}{3}cos\alpha+{\color{red}sin}\frac{\pi x}{3}sin\alpha)\equiv Rcos(\frac{\pi x}{3}- \alpha).
    (Remember that cos(a - b) = cos a cos b + sin a sin b.)

    So
    Rcos\alpha={\color{red}24} and
    Rsin\alpha={\color{red}10}.

    Looks like the only thing that changes is
    26cos(\frac{\pi x}{3}-\arctan{\color{red}\frac{10}{24}})=13
    26cos(\frac{\pi x}{3}-\arctan\frac{5}{12})=13
    cos(\frac{\pi x}{3}-\arctan\frac{5}{12})=\frac{1}{2}
    ...


    01
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  5. #5
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    Oops! That should be



    01
    Thanks! now i got that part.
    so \alpha=\arctan\frac{10}{24}
    Now what about the other part? I can't understand why
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  6. #6
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    Quote Originally Posted by arze View Post
    10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13

    10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv R(sin\frac{\pi x}{3}cos\alpha+cos\frac{\pi x}{3}sin\alpha)\equiv R {\color{red}cos} (\frac{\pi x}{3}- \alpha) Mr F says: That red cos should be a sin ....

    Rcos\alpha=10

    Rsin\alpha=24

    R=\sqrt{10^2+24^2}=\sqrt{676}=26

    \alpha=\arctan\frac{24}{10}

    So 10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv 26cos(\frac{\pi x}{3}-\alpha)
    26cos(\frac{\pi x}{3}-\arctan\frac{24}{10})=13
    To get the solution the book gives, you should have:


    10 \sin \frac{\pi x}{3}+24 \cos\frac{\pi x}{3} \equiv R \left({\color{red}\cos \frac{\pi x}{3}} \cos\alpha + {\color{red}\sin\frac{\pi x}{3}} \sin\alpha \right) \equiv R cos \left(\frac{\pi x}{3}- \alpha \right)

    where R = 26 and \alpha= \arctan \frac{10}{24} = \frac{5}{12}.


    Then:

    26 \cos \left(\frac{\pi x}{3}- \arctan \frac{5}{12} \right) = 13

     \Rightarrow \cos \left(\frac{\pi x}{3}- \arctan \frac{5}{12} \right) = \frac{1}{2}.


    Case 1: \frac{\pi x}{3} - \arctan \frac{5}{12} = \frac{\pi}{3} + 2 n \pi

     \Rightarrow \pi x - 3 \arctan \frac{5}{12} = \pi + 6n \pi

     \Rightarrow x - \frac{3}{\pi} \arctan \frac{5}{12} = 1 + 6n

     \Rightarrow x = 6n + 1 + \frac{3}{\pi} \arctan \frac{5}{12}


    Case 2: \frac{\pi x}{3}- \arctan \frac{5}{12} = - \frac{\pi}{3} + 2 n \pi

    etc.
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  7. #7
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    Quote Originally Posted by arze View Post
    Thanks! now i got that part.
    so \alpha=\arctan\frac{10}{24}
    Now what about the other part? I can't understand why
    Let \frac{\pi x}{3}-\arctan\frac{5}{12} = y. So

    cos y=\frac{1}{2}

    Tell me what are the solutions for y?


    01


    EDIT: Best just follow mr fantastic's response above...
    Last edited by yeongil; July 16th 2009 at 05:29 AM. Reason: beaten to it, doh!
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  8. #8
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    Ok! I got it now, thanks!
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