Results 1 to 8 of 8

Thread: Find general solution

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Find general solution

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

    My attempt:
    I used the identity $\displaystyle R (\sin \theta \cos \alpha+\cos \theta \sin \alpha)\equiv R \cos(\theta-\alpha)$ 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: $\displaystyle (6n\pm1)+\frac{3}{\pi}\arctan\frac{5}{12}$. My problem is with the first part, why $\displaystyle (6n\pm1)$? I was thinking maybe there should be a $\displaystyle \pi$ there?
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by arze View Post
    Find the general solution of the equation:
    $\displaystyle 10 \sin \frac{\pi x}{3}+24 \cos \frac{\pi x}{3}=13$

    My attempt:
    I used the identity $\displaystyle R (\sin \theta \cos \alpha+\cos \theta \sin \alpha)\equiv R \cos(\theta-\alpha)$ 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: $\displaystyle (6n\pm1)+\frac{3}{\pi}\arctan\frac{5}{12}$. My problem is with the first part, why $\displaystyle (6n\pm1)$? I was thinking maybe there should be a $\displaystyle \pi$ 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 $\displaystyle R \cos(x - \alpha) = 13$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    $\displaystyle 10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13$

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

    $\displaystyle Rsin\alpha=24$

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

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

    So $\displaystyle 10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv 26cos(\frac{\pi x}{3}-\alpha)$
    $\displaystyle 26cos(\frac{\pi x}{3}-\arctan\frac{24}{10})=13$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    May 2009
    Posts
    612
    Thanks
    309
    Quote Originally Posted by arze View Post
    $\displaystyle 10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13$

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

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

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


    01
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    Oops! That should be



    01
    Thanks! now i got that part.
    so $\displaystyle \alpha=\arctan\frac{10}{24}$
    Now what about the other part? I can't understand why
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by arze View Post
    $\displaystyle 10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13$

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

    $\displaystyle Rcos\alpha=10$

    $\displaystyle Rsin\alpha=24$

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

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

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


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

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


    Then:

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

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


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

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

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

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


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

    etc.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    May 2009
    Posts
    612
    Thanks
    309
    Quote Originally Posted by arze View Post
    Thanks! now i got that part.
    so $\displaystyle \alpha=\arctan\frac{10}{24}$
    Now what about the other part? I can't understand why
    Let $\displaystyle \frac{\pi x}{3}-\arctan\frac{5}{12} = y$. So

    $\displaystyle cos y=\frac{1}{2}$

    Tell me what are the solutions for y?


    01


    EDIT: Best just follow mr fantastic's response above...
    Last edited by yeongil; Jul 16th 2009 at 05:29 AM. Reason: beaten to it, doh!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    Ok! I got it now, thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find the general solution for...
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: Nov 5th 2011, 06:00 AM
  2. PDE - find the general solution
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 12th 2009, 04:36 AM
  3. find the general solution when 1 solution is given
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: Mar 4th 2009, 09:09 PM
  4. find the general solution ODE pls help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 24th 2008, 03:24 AM
  5. find the general solution..
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 6th 2007, 09:23 AM

Search Tags


/mathhelpforum @mathhelpforum