Three. (Full marks 20 points) Given that when $x \in [0,1]$, the inequality $x^{2} \cos \theta - x(1-x) + (1-x)^{2} \sin \theta > 0$ always holds, find the range of $\theta$.


$$
2 k \pi+\frac{\pi}{12}<\theta<2 k \pi+\frac{5 \pi}{12}, k \in \mathbf{Z}$, then $\cos \theta=f(1)>0, \sin \theta=f(0)>0$.
Take $x_{0}=\frac{\sqrt{\sin \theta}}{\sqrt{\cos \theta}+\sqrt{\sin \theta}} \in(0,1)$, we have $\sqrt{\cos \theta} x_{0}-\sqrt{\sin \theta}\left(1-x_{0}\right)=0$.
$$
\begin{array}{l}
\because f(x)=[\sqrt{\cos \theta} x-\sqrt{\sin \theta}(1-x)]^{2}+2\left(-\frac{1}{2}+\sqrt{\cos \theta \sin \theta}\right) x(1-x), \\
\therefore 00 .
\end{array}
$$

Conversely, when (1) and (2) hold, $f(0)=\sin \theta>0, f(1)=\cos \theta>0$, and for $x \in(0,1)$, $f(x) \geqslant 2$ $\left(-\frac{1}{2}+\sqrt{\cos \theta \sin \theta}\right) x(1-x)>0$.
First solve (1) and (2) in $[0,2 \pi]$: From $\cos \theta>0, \sin \theta>0$, we get $00 \Rightarrow \sqrt{\cos \theta \sin \theta}>\frac{1}{2}, \therefore \frac{1}{2} \sin 2 \theta>\frac{1}{4}, \sin 2 \theta>\frac{1}{2}$,
and $\because 0<2 \theta<\pi, \therefore \frac{\pi}{6}<2 \theta<\frac{5 \pi}{6}, \therefore \frac{\pi}{12}<\theta<\frac{5 \pi}{12}$.
Therefore, the range of $\theta$ is $2 k \pi+\frac{\pi}{12}<\theta<2 k \pi+\frac{5 \pi}{12}, k \in \mathbf{Z}$.