<s>[INST] Integers $n$ satsifying $\frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}}$ 
If $\displaystyle \frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}},n\in \mathbb{Z}$, then number of $n$ satisfies given equation ,is

What I tried:
Let $\displaystyle \frac{\pi}{n}=x$ and equation is $\sin 5x=\sin 3x$
$\displaystyle \sin (5x)-\sin (3x)=2\sin (4x)\cos (x)=0$
$\displaystyle 4x= m\pi$ and $\displaystyle x= 2m\pi\pm \frac{\pi}{2}$
$\displaystyle \frac{4\pi}{n}=m\pi\Rightarrow n=\frac{4}{m}\in \mathbb{Z}$ for $m=\pm 1,\pm 2\pm 3,\pm 4$
put into $\displaystyle x=2m\pi\pm \frac{\pi}{2}$
How do i solve my sum in some easy way Help me please 
 [/INST] I'm afraid you have used the wrong property: $\sin A-\sin B=2\cos\left(\dfrac{A+B}2\right)\sin\left(\dfrac{A-B}2\right)$. So you get$$2\cos(4x)\sin x=0$$Thus, $x=k\pi\vee4x=(2k+1)\pi/2$ for $k\in\Bbb Z$. For the former, $x=\pi/n=k\pi\implies nk=1$ which is true iff $n=k=1\vee n=k=-1$. In the latter case, $(2k+1)\pi/8=\pi/n\implies(2k+1)n=8$ which is true iff $n=-8,k=-1\vee n=8,k=0$.
Also, remember that $\sin 3x$ and $\sin 5x$ have to be non-zero, because they are in the denominator in the original problem statement. So we have to reject $\pm1$ to get two solutions, $n=\pm8$. 
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