11. Solve the equation $(\sin x)^{\arctan (\sin x+\cos x)}=(\csc x)^{\arctan (\sin 2 x)+\frac{\pi}{4}}$, where $2 n \pi<x<$ $(2 n+1) \pi, n$ is any integer.
11. The equation can be transformed into $(\sin x)^{\arctan (\sin x+\cos x)+\arctan (\sin 2 x)+\frac{\pi}{4}}=1$, so $\sin x=1, x=2 n \pi+\frac{\pi}{2}(n$ is an integer), or $\arctan (\sin x+\cos x)+\arctan (\sin 2 x)+\frac{\pi}{4}=0$, i.e., $\arctan (\sin x+\cos x)+\arctan (\sin 2 x)=$ $-\frac{\pi}{4}$. Taking the tangent yields $\frac{\sin x+\cos x+\sin 2 x}{1-(\sin x+\cos x) \sin 2 x}=-1$, let $y=\sin x+\cos x$, then $\sin 2 x=2 \sin x \cos x=$ $y^{2}-1$, substituting into the above equation can solve for $y=0,-1$, 2. Noting that $2 n \pi<x<(2 n+1) \pi$, we can then get $x=2 n \pi+\frac{3 \pi}{4}(n$ is an integer). Therefore, the solutions to the original equation are $x_{1}=2 n \pi+\frac{\pi}{2}, x_{2}=2 n \pi+\frac{3 \pi}{4}$ ( $n$ is any integer).