A trigonometric substitution

$\begin{aligned} \dfrac{1}{\sin{\frac{\pi}{n}}} & = \dfrac{1}{\sin{\frac{2\pi}{n}}} + \dfrac{1}{\sin{\frac{3\pi}{n}}} = \dfrac{1}{2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}} + \dfrac{1}{\sin{\frac{\pi}{n}}\cos{\frac{2\pi}{n}}+\sin{\frac{2\pi}{n}}\cos{\frac{\pi}{n}}} \\ & = \dfrac{1}{2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}} + \dfrac{1}{\sin{\frac{\pi}{n}}\cos{\frac{2\pi}{n}}+2\sin{\frac{\pi}{n}}\cos^2{\frac{\pi}{n}}} \end{aligned}$

$\begin{aligned} \Rightarrow \dfrac{1}{2\cos{\frac{\pi}{n}}} + \dfrac{1}{\cos{\frac{2\pi}{n}}+2\cos^2{\frac{\pi}{n}}} & = 1 \\ \dfrac{1}{2\cos{\frac{\pi}{n}}} + \dfrac{1}{4\cos^2{\frac{\pi}{n}}-1} & = 1 \\ 4\cos^2{\frac{\pi}{n}} + 2\cos{\frac{\pi}{n}} -1 & = 2\cos{\frac{\pi}{n}} \left( 4\cos^2{\frac{\pi}{n}}-1\right) \\ 8\cos^3 {\frac{\pi}{n}} - 4\cos^2{\frac{\pi}{n}} - 4\cos{\frac{\pi}{n}} +1 & = 0 \\ 4\cos^3 {\frac{\pi}{n}} - 2\cos^2{\frac{\pi}{n}} - 2\cos{\frac{\pi}{n}} +\frac{1}{2} & = 0 \\ (4\cos^3 {\frac{\pi}{n}} - 3\cos{\frac{\pi}{n}} ) - (2\cos^2{\frac{\pi}{n}} - 1) + \cos{\frac{\pi}{n}} & = \frac{1}{2} \\ \cos {\frac{3 \pi}{n}} - \cos{\frac{2\pi}{n}} + \cos{\frac{\pi}{n}} & = \frac{1}{2} \\ \Rightarrow \cos {\frac{3 \pi}{7}} - \cos{\frac{2\pi}{7}} + \cos{\frac{\pi}{7}} & = \frac{1}{2} \\ \Rightarrow n & = \boxed{7} \end{aligned}$

The solution $\cos {\frac{3 \pi}{7}} - \cos{\frac{2\pi}{7}} + \cos{\frac{\pi}{7}} = \frac{1}{2}$ is a property of $n^{th}$ root of $1$ , when $n>1$ is odd. The $\cos {\frac{k\pi}{n}}$ terms are the real parts of the complex roots except $1$ and from Argand diagram is obvious that the following is true.

\(\begin{array} {} \cos{\frac{\pi}{3}} = \frac{1}{2} \\ \cos{\frac{\pi}{5}} - \cos{\frac{2\pi}{5}} = \frac{1}{2} \\ \cos{\frac{\pi}{7}} - \cos{\frac{2\pi}{7}} + \cos{\frac{3\pi}{7}}= \frac{1}{2} \\ \cos{\frac{\pi}{9}} - \cos{\frac{2\pi}{9}} + \cos{\frac{3\pi}{9}} - \cos{\frac{4\pi}{9}}= \frac{1}{2} \\ ... \end{array}\)

The solution is therefore unique.