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

$\frac{\left(\frac{1}{1+\sin x}\right)}{\left(\frac{1}{1-\sin x}\right)}=\frac{2{\sin }^{2}x}{2{\cos }^{2}x}$

$\Rightarrow 2{\sin }^{2}x+\sin x-1=0\Rightarrow \sin x=\frac{-1\pm 3}{4}$

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

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

$\therefore x=n\pi +{\left(-1\right)}^n\frac{\pi }{6},\forall n\in I$