Задание №89456 ЕГЭ Профильной математике

Уравнение вида $\left|\mathrm{f}\right(\mathrm{x}\left)\right|=\mathrm{g}\left(\mathrm{x}\right)$, равносильно совокупности:

$\left\{\begin{array}{l}\mathrm{g}\left(\mathrm{x}\right)\ge 0\\ \left[\begin{array}{c}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right); \\ \mathrm{f}\left(\mathrm{x}\right)=-\mathrm{g}\left(\mathrm{x}\right).\end{array}\right.\end{array}\right.$

$$\begin{gathered} \left\{\begin{array}{l}\mathrm{x}+2\ge 0\\ \left[\begin{array}{c}{\mathrm{x}}^2+4\mathrm{x}+2=\mathrm{x}+2\\ {\mathrm{x}}^2+4\mathrm{x}+2=-\mathrm{x}-2\end{array}\right.\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{x}\ge -2\\ \left[\begin{array}{c}{\mathrm{x}}^2+3\mathrm{x}=0\\ {\mathrm{x}}^2+5\mathrm{x}+4=0\end{array}\right.\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{x}\ge -2\\ \left[\begin{array}{c}\mathrm{x}(\mathrm{x}+3)=0\\ (\mathrm{x}+4)(\mathrm{x}+1)=0\end{array}\right.\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{x}\ge -2\\ \left[\begin{array}{c}\mathrm{x}=0\\ \mathrm{x}=-3\\ \mathrm{x}=-4\\ \mathrm{x}=-1\end{array}\right.\end{array}\right. \\ \left[\begin{array}{c}\mathrm{x}=-1\\ \mathrm{x}=0.\end{array}\right. \end{gathered}$$

б) Отберём корни, принадлежащие отрезку  $[-\frac{\mathrm{\pi }}{3};-\frac{\mathrm{\pi }}{4}].$

Применим значение $\mathrm{\pi }=3,14.$

Тогда промежуток принимает вид $[-\frac{3,14}{3};-\frac{3,14}{4}].$

$$\begin{gathered} -\frac{3,14}{3}\le -1\le -\frac{3,14}{4} \Rightarrow -1\in \left[\begin{array}{cc}-\frac{\mathrm{\pi }}{3};& -\frac{\mathrm{\pi }}{4}\end{array}\right]. \\ 0\notin \left[\begin{array}{cc}-\frac{\mathrm{\pi }}{3};& -\frac{\mathrm{\pi }}{4}\end{array}\right]. \end{gathered}$$

Ответ: а) $-1; 0;$ б) $-1.$