Example 7 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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Analysis Here we notice the relationship between $x^{2}, (1-x)^{2}$, and $x(1-x)$, and think of using the AM-GM inequality to handle the problem.
Solution If $f(0)>0$, then $\sin \theta>0, f(1)>0$, then $\cos \theta>0$, so $\theta$ should be an angle in the first quadrant.
If $\sin \theta \cos \theta \leqslant \frac{1}{4}$, then by the AM-GM inequality
$$
x^{2} \cos \theta + (1-x)^{2} \sin \theta \geqslant 2 \sqrt{x^{2}(1-x)^{2} \sin \theta \cos \theta} = 2 x(1-x) \sqrt{\sin \theta \cos \theta}.
$$

When $x^{2} \cos \theta = (1-x)^{2} \sin \theta$, i.e., $x = \frac{\sqrt{\sin \theta}}{\sqrt{\cos \theta} + \sqrt{\sin \theta}}$, equality holds. Clearly, at this point $x \in [0,1]$.
Let $x_{0} = \frac{\sqrt{\sin \theta}}{\sqrt{\cos \theta} + \sqrt{\sin \theta}}$, then the left side of the original inequality $= 2 x_{0}(1-x_{0}) \sqrt{\sin \theta \cos \theta} - x_{0}(1-x_{0}) = x(1-x)[\sqrt{4 \sin \theta \cos \theta} - 1] \leqslant 0$, i.e., there exists $x \in [0,1]$ such that the original inequality does not hold, which is a contradiction. Therefore, $\sin \theta \cos \theta > \frac{1}{4}$. At this point, clearly when $x \in (0,1)$, the left side of the original inequality $= x(1-x)[\sqrt{4 \sin \theta \cos \theta} - 1] > 0$; and when $x = 0, 1$, it is easy to see that the inequality holds. Therefore, the necessary and sufficient condition for the inequality to hold is $\left\{\begin{array}{l}\sin \theta > 0 \\ \cos \theta > 0 \\ \sin \theta \cos \theta > \frac{1}{4}\end{array}\right.$, solving gives: $2 k \pi + \frac{\pi}{12} < \theta < 2 k \pi + \frac{5}{12} \pi, k \in \mathbf{Z}$. /Comment Here the AM-GM inequality is used, avoiding the discussion of quadratic functions. It reminds us to pay attention to observation and association.