9. 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 values for $\theta$.

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9. When $x=0$, we have $\sin \theta>0$; when $x=1$, we have $\cos \theta>0$; when $x \in(0,1)$, the original expression can be transformed into $\frac{x}{1-x} \cos \theta+\frac{1-x}{x} \sin \theta>1$. By the important inequality $\frac{x}{1-x} \cos \theta+\frac{1-x}{x} \sin \theta \geqslant 2 \sqrt{\sin \theta \cos \theta}$, it is sufficient that $2 \sqrt{\sin \theta \cos \theta}>1, \sin 2 \theta>\frac{1}{2}$, solving this gives $\theta \in\left(k \pi+\frac{\pi}{12}, k \pi+\frac{5}{12} \pi\right)$. Also, since $\sin \theta>0, \cos \theta>0, \theta \in$ $\left(2 k \pi+\frac{\pi}{12}, 2 k \pi+\frac{5}{12} \pi\right)(k \in \mathbf{Z})$.