[m]y=arccos \frac{2}{x+2}[/m]
получаем двойное неравенство
[m] -1\leq\frac{2}{x+2}\leq 1[/m] , которое равносильно системе неравенств:
[m]\left\{\begin{matrix}\frac{2}{x+2}\leq 1\\\frac{2}{x+2}\geq -1 \end{matrix}\right.[/m][m]\left\{\begin{matrix}\frac{2}{x+2}-1\leq 0\\\frac{2}{x+2}+1\geq 0\end{matrix}\right.[/m][m]\left\{\begin{matrix}\frac{2-x-2}{x+2}\leq 0\\\frac{2+x+2}{x+2}+1\geq 0\end{matrix}\right.[/m][m]\left\{\begin{matrix}\frac{-x}{x+2}\leq 0\\\frac{x+4}{x+2}\geq 0\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}\frac{x}{x+2}\geq 0\\\frac{x+4}{x+2}\geq 0\end{matrix}\right.[/m]
О т в е т. (- ∞ ;-4] U [0;+ ∞ )