$Thegeneralsolutionoftheequation$$\frac{1-\sin x+\mathrm{...}+{\left(-1\right)}^n{\sin }^{n}x+\mathrm{......}}{1+\sin x+\mathrm{...}+{\sin }^{n}x+\mathrm{......}}=\frac{1-\cos 2x}{1+\cos 2x}$$is$

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

$\Rightarrow \frac{1}{1+\sin x}.\frac{1-\sin x}{1}=\frac{2{\sin }^{2}x}{2{\cos }^{2}x}$

$\begin{array}{l}\Rightarrow 2{\sin }^{2}x+\sin x-1=0\\ \Rightarrow \sin x=\frac{-1\pm \sqrt{1+8}}{4}=\frac{-1\pm 3}{4}\\ \Rightarrow \sin x=-1or\sin x=\frac{1}{2}\\ Since\sin x\ne -1,wehave\sin x=\frac{1}{2}=\sin \left(\frac{\pi }{6}\right)\\ \therefore x=n\pi +{\left(-1\right)}^n\frac{\pi }{6}\end{array}$