The general solution of the equation $\frac{1-\sin x+\dots +(-1{)}^n{\sin }^{n}x+\dots }{1+\sin x+\dots ..+{\sin }^{n}x+\dots ..}=\frac{1-\cos 2x}{1+\cos 2x}$  is

The equation  $\frac{1}{1+\sin x}\cdot \frac{1-\sin x}{1}=\frac{2{\sin }^{2}x}{2{\cos }^{2}x}$

$\Rightarrow \sin x=\frac{-1\pm \sqrt{1+8}}{4}=\frac{-1\pm 3}{4}$

$\Rightarrow \sin x=-1 \text{or }\sin x=\frac{1}{2}$

Since,  $\sin x\ne -1$, 

we have $\sin x=\frac{1}{2}=\sin \left(\frac{\pi }{6}\right),$ 

 $\therefore x=n\pi +(-1{)}^n\frac{\pi }{6}$