How to solve this equation $\sin 5x = \sin (x + \frac{\pi}{3})$

Question

Could you give a hint how to solve this equation $\sin 5x=\sin (x + \frac{\pi}{3})$?

I tried to change $\sin 5x$ in function of $\sin x$ and $\cos x$, but I wasn't able to go further.

score 6 · Accepted Answer · answered Sep 06 '18 at 20:33

6

hint If $$\sin(X)=\sin(a)$$ then

$$X=a+2k\pi$$

OR

$$X=\pi-a+2k\pi$$

answered Sep 06 '18 at 20:33

hamam_Abdallah

... and vice versa. – hmakholm left over Monica Sep 06 '18 at 20:38
1

@HenningMakholm You mean $a=X+2K\pi$ or $a=\pi-X+2K\pi$. – hamam_Abdallah Sep 06 '18 at 20:44
I mean if either of the two equations are true, then the sines are equal. – hmakholm left over Monica Sep 06 '18 at 20:46

Mohammad Riazi-Kermani · Answer 2 · 2018-09-06T23:00:18.647

1

you want

$$5x= 2k\pi + (x + \pi/3)$$ or

$$ 5x= 2k\pi + \pi -(x + \pi/3)$$

Therefore we have $$ x=(6k+1)\pi/{12} $$ or

$$ x= (3k+1)\pi/9$$ where $k$ is an arbitrary integer.

edited Sep 06 '18 at 23:00

answered Sep 06 '18 at 20:43

The algebra is wrong here, and the initial equation does not give equality when $k$ is odd. – hmakholm left over Monica Sep 06 '18 at 20:49
The second solution seems incorrect. – hamam_Abdallah Sep 06 '18 at 21:01
Please check the edited answer again – Mohammad Riazi-Kermani Sep 06 '18 at 23:02

Larry · Answer 3 · 2018-09-06T21:16:05.523

1

With Salahamam_Fatima's hint, suppose $a = x + \frac{\pi}{3}$

$sin(5x) = sin(\pi-(x+\frac{\pi}{3})+2k\pi)$

and

$sin(5x) = sin(x + \frac{\pi}{3}+2k\pi)$

Solve for x, and we get

$x = \frac{\pi}{9} + \frac{k\pi}{3}$ and $\frac{\pi}{12} + \frac{k\pi}{2}$

where k is an integer.

edited Sep 06 '18 at 21:16

answered Sep 06 '18 at 21:10

Larry

3 Answers3