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 
 A: 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 $n=\pm8$.
A: 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, $n=\pm8$. 
A: Since $-n$ is a solution if $n$ is a solution, it suffices to look for solutions with $n\gt1$. Since $\sin x$ is strictly increasing for $0\le x\le\pi/2$, we cannot have $\sin(3\pi/n)=\sin(5\pi/n)$ if $n\ge10$, so it suffices to consider $2\le n\le9$.
From $\sin x=\sin(\pi-x)$, we have
$$\sin\left(5\pi\over n\right)=\sin\left(\pi-{5\pi\over n}\right)=\sin\left((n-5)\pi\over n\right)$$
For $2\le n\le5$, the signs of $\sin(3\pi/n)$ and $\sin((n-5)\pi/n)$ do not agree (e.g., $\sin(3\pi/2)=-1$ while $\sin(-3\pi/2)=1$).  For $6\le n\le9$, we have $0\le{3\pi\over n},{(n-5)\pi\over n}\le{\pi\over2}$, in which case 
$$\sin\left(3\pi\over n\right)=\sin\left((n-5)\pi\over n\right)\iff{3\pi\over n}={(n-5)\pi\over n}\iff3=n-5\iff n=8$$
Thus we have two solutions, $n=8$ and $n=-8$.
Remark: The cases $n=3$ and $n=5$ could have been rejected outright, since $\sin(3\pi/n)\sin(5\pi/n)=0$ in those cases, guaranteeing a forbidden $0$ in one of the denominators in the original expression.
