# Find general solution

• Jul 16th 2009, 12:29 AM
arze
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?(Wondering)
Thanks
• Jul 16th 2009, 04:42 AM
mr fantastic
Quote:

Originally Posted by arze
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?(Wondering)
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 $R \cos(x - \alpha) = 13$.
• Jul 16th 2009, 05:02 AM
arze
$10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13$

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

$Rsin\alpha=24$

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

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

So $10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv 26cos(\frac{\pi x}{3}-\alpha)$
$26cos(\frac{\pi x}{3}-\arctan\frac{24}{10})=13$
• Jul 16th 2009, 05:19 AM
yeongil
Quote:

Originally Posted by arze
$10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13$

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

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
• Jul 16th 2009, 05:24 AM
arze
Thanks! now i got that part.
so $\alpha=\arctan\frac{10}{24}$
Now what about the other part? I can't understand why http://www.mathhelpforum.com/math-he...b38b2ae8-1.gif
• Jul 16th 2009, 05:26 AM
mr fantastic
Quote:

Originally Posted by arze
$10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}=13$

$10sin\frac{\pi x}{3}+24cos\frac{\pi x}{3}\equiv R(sin\frac{\pi x}{3}cos\alpha+cos\frac{\pi x}{3}sin\alpha)\equiv R {\color{red}cos} (\frac{\pi x}{3}- \alpha)$ 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.
• Jul 16th 2009, 05:27 AM
yeongil
Quote:

Originally Posted by arze
Thanks! now i got that part.
so $\alpha=\arctan\frac{10}{24}$
Now what about the other part? I can't understand why http://www.mathhelpforum.com/math-he...b38b2ae8-1.gif

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... (Rofl)
• Jul 16th 2009, 04:23 PM
arze
Ok! I got it now, thanks!