5. Solve the inequality $\arcsin \left(\frac{5}{2 \pi} \arccos x\right)>\arccos \left(\frac{10}{3 \pi} \arcsin x\right)$.
Answer: $\left[\cos \frac{2 \pi}{5} ; \cos \frac{8 \pi}{25}\right) \cup\left(\cos \frac{8 \pi}{25} ; \cos \frac{\pi}{5}\right]$.

Solution. Let $u=\frac{5}{2 \pi} \arccos x, v=\frac{10}{3 \pi} \arcsin x$. Then the relation $4 u+3 v=5$ holds. The solution to the inequality $\arcsin u>\arccos v$ is the set

$$
0 \leqslant u \leqslant 1,0 \leqslant v \leqslant 1, u^{2}+v^{2}>1
$$

on the diagram, it is shaded gray. The line $4 u+3 v=5$ has common points with this set - this is the segment with the point $u=4 / 5, v=3 / 5$ removed. Thus,

$$
u \in\left[\frac{1}{2} ; \frac{4}{5}\right) \cup\left(\frac{4}{5} ; 1\right] \Leftrightarrow \arccos x \in\left[\frac{\pi}{5} ; \frac{8 \pi}{25}\right) \cup\left(\frac{8 \pi}{25} ; \frac{2 \pi}{5}\right]
$$

from which, taking into account the decreasing nature of the arccosine,

$$
x \in\left[\cos \frac{2 \pi}{5} ; \cos \frac{8 \pi}{25}\right) \cup\left(\cos \frac{8 \pi}{25} ; \cos \frac{\pi}{5}\right]
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_84c54589ca5e85eb41a6g-2.jpg?height=466&width=417&top_left_y=875&top_left_x=1522)

Answer to variant 2: $\left[\sin \frac{\pi}{5} ; \sin \frac{8 \pi}{25}\right) \cup\left(\sin \frac{8 \pi}{25} ; \sin \frac{2 \pi}{5}\right]$.

Answer to variant 3: $\left[\cos \frac{2 \pi}{5} ; \cos \frac{8 \pi}{25}\right) \cup\left(\cos \frac{8 \pi}{25} ; \cos \frac{\pi}{5}\right]$.

Answer to variant 4: $\left[\sin \frac{\pi}{5} ; \sin \frac{8 \pi}{25}\right) \cup\left(\sin \frac{8 \pi}{25} ; \sin \frac{2 \pi}{5}\right]$.