<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] No need to go with sum-to-product formulas. From $\sin5x=\sin3x$ you get either
$$
5x=3x+2k\pi
$$
or
$$
5x=\pi-3x+2k\pi
$$
The former yields $x=k\pi$, so $1=kn$, whence $n=\pm1$.
The latter yields
$$
\frac{\pi}{n}=\frac{1}{8}(2k+1)\pi
$$
that is, $(2k+1)n=8$. Hence $n=\pm8$.
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