12. Let $0<x<\frac{\pi}{2}$, then the range of the function $y=\sin x-2 \cos x+\frac{32}{125 \sin x(1-\cos x)}$ is
12. $\left[\frac{2}{5},+\infty\right) \quad y+2=\sin x+(2-2 \cos x)+\frac{32}{125 \sin x(1-\cos x)}$

Since $0<\sin x(1-\cos x) \leqslant 2$ and $\sin x(1-\cos x)>0$, by the AM-GM inequality, $y+2 \geqslant 3\left[\sin x(2-2 \cos x) \frac{32}{125 \sin x(1-\cos x)}\right]^{\frac{1}{3}}=$ $\frac{12}{5}$. Equality holds if and only if $\sin x=2-2 \cos x=\frac{32}{125 \sin x(1-\cos x)}$, i.e., $\tan \frac{x}{2}=\frac{1}{2}$. Therefore, the value should be $\left[\frac{2}{5},+\infty\right)$.