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 $$2k\pi +\frac{\pi }{4}<\theta <2k\pi +\frac{5\pi }{4}(k\in \mathbb{Z}).$$ But $\theta \in [0,2\pi )$, so the range of $\theta$ is $(\frac{\pi }{4},\frac{5\pi }{4})$. $\square$