6.1. Find all solutions to the inequality

$$
\sin ^{2018} x+\cos ^{-2019} x \geqslant \cos ^{2018} x+\sin ^{-2019} x
$$

belonging to the interval $\left[-\frac{\pi}{4} ; \frac{7 \pi}{4}\right]$.
Answer: $\left[-\frac{\pi}{4} ; 0\right) \cup\left[\frac{\pi}{4} ; \frac{\pi}{2}\right) \cup\left(\pi ; \frac{5 \pi}{4}\right] \cup\left(\frac{3 \pi}{2} ; \frac{7 \pi}{4}\right]$.

Solution. The function $f(t)=t^{2018}-t^{-2019}$ is increasing on each of the half-intervals $[-1,0)$ and $(0,1]$. Indeed, the derivative $f^{\prime}(t)=$ $t^{-2020}\left(2018 t^{4037}+2019\right)=0$ at $t=-\sqrt[4037]{\frac{2019}{2018}}$ 0 on each of the half-intervals $[-1,0)$ and $(0,1]$. In addition, $f(t) \leqslant 0$ if $00 \\
0>\sin x \geqslant \cos x \\
\left\{\begin{array}{l}
\sin x0
\end{array}\right.
\end{array}\right.
$$

Therefore, the answer within the period from $\frac{\pi}{4}$ to $\frac{7 \pi}{4}$ is: $\left[-\frac{\pi}{4} ; 0\right) \cup\left[\frac{\pi}{4} ; \frac{\pi}{2}\right) \cup$ $\left(\pi ; \frac{5 \pi}{4}\right] \cup\left(\frac{3 \pi}{2} ; \frac{7 \pi}{4}\right]$