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 $\theta$.
Analyze the left side of the inequality as a quadratic function of $x$, find the minimum value of this quadratic function, and let it be greater than 0, thereby determining the range of $\theta$.

Solution: Given the conditions $\cos \theta>0, \sin \theta>0$, if for all $x \in[0,1]$, we have $f(x)=$ $x^{2} \cos \theta - x(1-x) + (1-x)^{2} \sin \theta > 0$, which simplifies to $f(x) = (\cos \theta + 1 + \sin \theta) x^{2} - (1 + 2 \sin \theta) x + \sin \theta > 0$ for $x \in [0,1]$, then it must be true that $\cos \theta = f(1) > 0$ and $\sin \theta = f(0) > 0$. 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 true that $\frac{4(\cos \theta + \sin \theta + 1) \sin \theta - (1 + 2 \sin \theta)^{2}}{4(\cos \theta + \sin \theta + 1)} > 0$, which simplifies to $4 \cos \theta \sin \theta - 1 > 0$, or $\sin 2 \theta > \frac{1}{2}$. Since $\cos \theta > 0$ and $\sin \theta > 0$, it follows that $2 k \pi + \frac{\pi}{12} < \theta < 2 k \pi + \frac{5 \pi}{12}, k \in \mathbf{Z}$.