The domain of the function ${\cos }^{-1}{\log }_2\left(x^2+5x+8\right)$ is

${\cos }^{-1}{\log }_2\left(x^2+5x+8\right)$ is real if 

$\begin{array}{l}-1\le {\log }_2\left(x^2+5x+8\right)\le 1\\ \Rightarrow \frac{1}{2}\le x^2+5x+8\le 2\\ x^2+5x+8={\left(x+\frac{5}{2}\right)}^2+\frac{7}{4}>\frac{1}{2}\text{ for all }x.\\ x^2+5x+8\le 2\\ \Rightarrow (x+3)(x+2)\le 0\Rightarrow -3\le x\le -2\end{array}$