Example 18 (1999 National High School Mathematics League Question) Given that when $x \in[0,1]$, the inequality $x^{2} \cos \theta-x(1-x)+$ $(1-x)^{2} \sin \theta>0$ always holds, try to find the range of values for $\theta$.

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Solution: Let $f(x)=x^{2} \cos \theta-x(1-x)+(1-$ $x)^{2} \sin \theta$. From the conditions, we know that $\cos \theta>0, \sin \theta>0$. If for all $x \in[0,1]$, it always holds that $f(x)=x^{2} \cos \theta-x(1-x)+$ $(1-x)^{2} \sin \theta>0$, that is,
$$
f(x)=(\cos \theta+1+\sin \theta) x^{2}-(1+2 \sin \theta) x
$$
$+\sin \theta>0$ for $x \in[0,1]$, then it must be that $\cos \theta$ $=f(1)>0, \sin \theta=f(0)>0$. On the other hand, the axis of symmetry is $x=\frac{1+2 \sin \theta}{2(\cos \theta+\sin \theta+1)} \in[0,1]$, so it must be that $\frac{4(\cos \theta+\sin \theta+1) \sin \theta-(1+2 \sin \theta)^{2}}{4(\cos \theta+\sin \theta+1)}>0$, which means $4 \cos \theta \sin \theta-1>0, \sin 2 \theta>\frac{1}{2}$. Since $\cos \theta>0$, $\sin \theta>0$, it follows that $2 k \pi+\frac{\pi}{12}<\theta<2 k \pi+\frac{5 \pi}{12}, k \in \mathbf{Z}$.