China Mathematical Competition

From the inequality
$$\cos^5\theta - \sin^5\theta < 7(\sin^3\theta - \cos^3\theta),$$
we have
$$\sin^3\theta + \frac{1}{7}\sin^5\theta > \cos^3\theta + \frac{1}{7}\cos^5\theta.$$
Since $f(x) = x^3 + \frac{1}{7}x^5$ is increasing over $(-\infty, +\infty)$, then $\sin \theta > \cos \theta$, and that means
$$2k\pi + \frac{\pi}{4} < \theta < 2k\pi + \frac{5\pi}{4} \quad (k \in \mathbb{Z}).$$
But $\theta \in [0, 2\pi)$, so the range of $\theta$ is $(\frac{\pi}{4}, \frac{5\pi}{4})$. $\square$