The general solution of the equation $\frac{1-\sin x+\mathrm{...}+{(-1)}^n{\sin }^{n}x+\mathrm{....}}{1+\sin x+\mathrm{....}+{\sin }^{n}x+\mathrm{.......}}=\frac{1-\cos 2x}{1+\cos 2x},x\ne (2n+1)\frac{\pi }{2},n\in Z$ is 

$\frac{1-\sin x+\mathrm{...}+{(-1)}^n{\sin }^{n}x+\mathrm{....}}{1+\sin x+\mathrm{....}+{\sin }^{n}x+\mathrm{.......}}=\frac{1-\cos 2x}{1+\cos 2x},x\ne (2n+1)\frac{\pi }{2},n\in Z$

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

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

$(1-\sin x)(1-{\sin }^{2}x)=(1+\sin x){\sin }^{2}x$

${(1-\sin x)}^2={\sin }^{2}x(1+\sin x\ne 0)$

$1-\sin x=\sin x$

$\sin x=\frac{1}{2}$

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