Solve: $\frac{1}{\sin(\pi/n)}=\frac{1}{\sin(2\pi/n)}+\frac{1}{\sin(3\pi/n)}$ 
The positve integer satisfying value of $n: n>3$ satisfying the equation: $\dfrac{1}{\sin(\pi/n)}=\dfrac{1}{\sin(2\pi/n)}+\dfrac{1}{\sin(3\pi/n)}$ is

This question is from a Practice Book for CBSE Term$-1$ Maths Preparation Class $10$.
What I tried:
$ \dfrac{1}{\sin ({\pi / n})}-\dfrac{1}{\sin({3\pi / n})}=\dfrac{1}{\sin({2 \pi / n})} $
$ \dfrac{\sin({3\pi / n})-\sin({\pi / n})}{\sin({\pi / n})\times \sin({3\pi / n})}=\dfrac{1}{\sin ({2 \pi / n})} $
$\dfrac{2\cos({2 \pi / n})\sin({\pi / n})}{\sin({\pi / n})\sin({3\pi / n})}=\dfrac{1}{\sin ({2 \pi / n})}$
$ 2\cos ({2 \pi / n})\sin({2 \pi / n})=\sin({3\pi / n})  $
$ \sin (4\pi/n)=\sin ({3\pi / n}) $
How to continue this solution? Also this solution is getting pretty complex? Is there any other way to solve this?
 A: An alternative to what Vasya suggests in the comments, note that $\sin x = \sin y \Rightarrow x = kπ +(-1)^k y, \ \forall k \in \mathbb{Z} $, so you have $\frac{4π}{n}= kπ +(-1)^k \left(\frac{3π}{n}\right), \ \forall k \in \mathbb{Z} \Rightarrow nk =4 -(-1)^k \cdot 3, \ \forall k\in \mathbb{Z}$

Now, if $k$ is even then write $k = 2p$ for some $p \in \mathbb{Z}$ and hence observe that $$4-(-1)^{2p}= n \cdot (2p) \Rightarrow 1 = n \cdot (2p) $$ has no solution in $n$ no matter what integer $p$ you take. (Why?)
If $k$ is odd then $$4- (-1)^k \cdot 3 =nk \Rightarrow 7= nk \Rightarrow n = \frac{7}{k}$$ then $k =1$ (Why?).
Hence $n =7$ is the unique solution.

Footnote:
$\sin x = \sin y \Rightarrow x = kπ +(-1)^k y, \ \forall k \in \mathbb{Z} $
A: Using the formula
$$\sin A-\sin B=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
we have
\begin{align*}
\sin \dfrac{4\pi}{n}&=\sin \dfrac{3\pi}{n}\\
\sin \dfrac{4\pi}{n}-\sin \dfrac{3\pi}{n}&=0\\
2\sin\left(\frac{\pi}{2n}\right)\cos\left(\frac{7\pi}{2n}\right)&=0
\end{align*}
then
$$
\begin{array}{rcl}
\sin\left(\dfrac{\pi}{2n}\right)&=&0\\[2mm]
\dfrac{\pi}{2n}&\in&\{k\pi:k\in\mathbb Z_{\neq 0}\}\\
n&\in&\left\{\dfrac{1}{2k}:k\in\mathbb Z_{\neq 0}\right\}
\end{array}\quad\text{  or }\quad
\begin{array}{rcl}
\cos\left(\dfrac{7\pi}{2n}\right)&=&0\\[2mm]
\dfrac{\pi}{2n}&\in&\left\{(2k+1)\frac{\pi}{2}:k\in\mathbb Z\right\}\\
n&\in&\left\{\dfrac{7}{2k+1}:k\in\mathbb Z\right\}
\end{array}
$$
Thus
$$n\in\left\{\dfrac{1}{2k}:k\in\mathbb Z_{\neq 0}\right\}\cup\left\{\dfrac{7}{2k+1}:k\in\mathbb Z\right\}$$
