1. Solve the inequality $\operatorname{tg} \arccos x \leqslant \sin \operatorname{arctg} x$. Answer: $\left[-\frac{1}{\sqrt[4]{2}} ; 0\right) \cup\left[\frac{1}{\sqrt[4]{2}} ; 1\right]$.
Solution. Since

$$
\operatorname{tg} \arccos x=\frac{\sin \arccos x}{\cos \arccos x}=\frac{\sqrt{1-x^{2}}}{x}
$$

$\sin \operatorname{arctg} x=\operatorname{tg} \operatorname{arctg} x \cdot \cos \operatorname{arctg} x=\frac{x}{\sqrt{1+x^{2}}}$,

the inequality takes the form

$$
\begin{aligned}
& \frac{\sqrt{1-x^{2}}}{x} \leqslant \frac{x}{\sqrt{1+x^{2}}} \Leftrightarrow \frac{\sqrt{1-x^{4}}-x^{2}}{x \sqrt{1+x^{2}}} \leqslant 0 \quad \Leftrightarrow \\
& \Leftrightarrow \quad \frac{1-2 x^{4}}{x\left(\sqrt{1-x^{4}}+x^{2}\right)} \leqslant 0 \quad \Leftrightarrow \quad\left\{\begin{array}{l}
\frac{\left(x-\frac{1}{\sqrt[4]{2}}\right)\left(x+\frac{1}{\sqrt[4]{2}}\right)}{x} \geqslant 0 \\
-1 \leqslant x \leqslant 1
\end{array}\right.
\end{aligned}
$$

Applying the method of intervals, we obtain the answer from here.