Problem 6. Calculate $2 \operatorname{arctg} 2+\arcsin \frac{4}{5}$.
Answer: $\pi$.

Solution. First solution. Let $\operatorname{arctg} 2$ be denoted by $\alpha$, and $\arcsin 4 / 5$ by $\beta$. Note that $\beta \in(0, \pi / 2)$, and $\operatorname{tg}^{2} \beta=\sin ^{2} \beta /\left(1-\sin ^{2} \beta\right)=4^{2} / 3^{2}$, from which $\operatorname{tg} \beta=4 / 3$; also $\operatorname{tg} \alpha=2, \alpha \in(0, \pi / 2)$.

We find:

$$
\begin{gathered}
\operatorname{tg}(2 \alpha)=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{2 \cdot 2}{1-2^{2}}=-\frac{4}{3} \\
\operatorname{tg}(2 \alpha+\beta)=\frac{\operatorname{tg}(2 \alpha)+\operatorname{tg} \beta}{1-\operatorname{tg}(2 \alpha) \operatorname{tg} \beta}=\frac{-4 / 3+4 / 3}{1-(-4 / 3) \cdot(4 / 3)}=0
\end{gathered}
$$

Finally, since $0<\alpha<\pi / 2,0<\beta<\pi / 2$, then $0<2 \alpha+\beta<3 \pi / 2$. Therefore, $2 \alpha+\beta=\pi$.

Second solution. Mark the points $O(0,0), A(3,0), B(3,4), C(-5,10)$, $D(-5,0)$ on the coordinate plane. Since the slope of the line $O B$ is $4 / 3$, and the slope of the line $B C$ is $-3 / 4$, we get that $\angle O B C=90^{\circ}$.

In triangle $O A B: \angle O A B=90^{\circ}, A B=4, B O=5$; thus, $\angle A O B=\arcsin 4 / 5$. In triangle $O B C: \angle O B C=90^{\circ}, B O=5, B C=10$; thus, $\angle B O C=\operatorname{arctg} 2$. In triangle $O C D: \angle O D C=90^{\circ}$, $D O=5, B C=10$; thus, $\angle C O D=\operatorname{arctg} 2$.

Therefore, $\arcsin 4 / 5+2 \operatorname{arctg} 2=\angle A O B+\angle B O C+\angle C O D=\angle A O D=\pi$.