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> 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
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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?