Example 2 If $\cos ^{5} \theta-\sin ^{5} \theta<7\left(\sin ^{3} \theta-\cos ^{3} \theta\right)$, where $\theta \in[0,2 \pi)$, then the range of values for $\theta$ is $\qquad$ [2]
(2011, National High School Mathematics Joint Competition)
Solve the original inequality, which is equivalent to
$$
\begin{array}{l}
7 \sin ^{3} \theta+\sin ^{5} \theta>7 \cos ^{3} \theta+\cos ^{5} \theta . \\
\text { Let } f(x)=7 x^{3}+x^{5} .
\end{array}
$$

Notice that, $f^{\prime}(x)=21 x^{2}+5 x^{4} \geqslant 0$ for all $x \in(-\infty,+\infty)$, hence $f(x)$ is an increasing function on $\mathbf{R}$.
The original inequality is thus $f(\sin \theta)>f(\cos \theta)$.
Therefore, $\sin \theta>\cos \theta$
$$
\Rightarrow 2 k \pi+\frac{\pi}{4}<\theta<2 k \pi+\frac{5 \pi}{4}(k \in \mathbf{Z}) \text {. }
$$

Given $\theta \in[0,2 \pi)$, the range of $\theta$ is
$$
\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)
$$