Solution of the equation...

Question

How to find out the analytic solutions of the equation $1-4\sin^2\frac{\theta}{2}=\frac{\sin\theta(n-1)}{\sin\theta n}$ in the interval $(0, \pi)$? $n$ is an arbitrary integer constant.

Answer 1

$2\sin^2\frac{\theta}2=1-\cos\theta$

$$\implies1-2(1-\cos\theta)=\dfrac{\sin(n-1)\theta}{\sin n\theta}$$

$$\iff\sin(n-1)\theta=\sin n\theta(2\cos\theta-1)=2\sin n\theta\cos\theta-\sin n\theta$$

Using Werner Formula, $$\sin(n-1)\theta=\sin(n+1)\theta+\sin(n-1)\theta-\sin n\theta$$

$$\iff\sin(n+1)\theta=\sin n\theta$$

Now $\sin x=\sin A\implies x=m\pi+(-1)^mA$ where $m$ is any integer