2. In the set of real numbers, solve the system of equations:

$$
\begin{aligned}
x+y+\cos z & =3 \\
2 x-y+\sin \{z\} & =1 \\
x-3 y-\operatorname{arctg} z^{2} & =-2
\end{aligned}
$$

(With $\{z\}$ we denote the fractional part of the number $z$, i.e., $\{z\}=z-\lfloor z\rfloor$.)
2. From the first equation, we have $x+y=3-\cos z \geqslant 2$.

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-49.jpg?height=266&width=449&top_left_y=910&top_left_x=1059)

Dr $20172 \mathrm{~A} 1$ Since $0 \leqslant\{z\}<1<\frac{\pi}{2}$, we conclude $\sin \{z\} \geqslant 0$, so from the second equation we get $2 x-y=1-\sin \{z\} \leqslant 1$. From the third equation, we have $x-3 y=-2+\operatorname{arctg} z^{2} \geqslant-2+\operatorname{arctg} 0=-2$ (since $\operatorname{arctg}$ is an increasing function and $z^{2} \geqslant 0$). Multiplying the inequality $x+y \geqslant 2$ by 3 and adding $x-3 y \geqslant-2$ gives $4 x \geqslant 4$, i.e., $x \geqslant 1$, and multiplying $2 x-y \leqslant 1$ by -3 and adding $x-3 y \geqslant-2$ gives $-5 x \geqslant-5$, i.e., $x \leqslant 1$, so it must be $x=1$. Now the obtained inequalities reduce to $y \geqslant 1$ and $y \leqslant 1$, so $y=1$. Finally, from the last equality in the problem statement, we get $\operatorname{arctg} z^{2}=0$, i.e., $z=0$, so the only solution to the given system is $(x, y, z)=(1,1,0)$.