Đáp án:
$\begin{array}{l}
Do: - 1 \le \cos x \le 1\\
\Rightarrow 0 \le {\cos ^2}x \le 1\\
Do:2\left( {{m^2} + 1} \right).{\cos ^2}x = {m^2} + m + 4\\
\Rightarrow {\cos ^2}x = \dfrac{{{m^2} + m + 4}}{{2{m^2} + 2}}\\
\Rightarrow 0 \le \dfrac{{{m^2} + m + 4}}{{2{m^2} + 2}} \le 1\\
\Rightarrow \dfrac{{{m^2} + m + 4}}{{2{m^2} + 2}} - 1 \le 0\\
\Rightarrow \dfrac{{{m^2} + m + 4 - 2{m^2} - 2}}{{2{m^2} + 2}} \le 0\\
\Rightarrow \dfrac{{ - {m^2} + m + 2}}{{2{m^2} + 2}} \le 0\\
\Rightarrow - {m^2} + m + 2 \le 0\left( {do:2{m^2} + 2 > 0} \right)\\
\Rightarrow {m^2} - m - 2 \ge 0\\
\Rightarrow \left( {m - 2} \right)\left( {m + 1} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
m \ge 2\\
m \le - 1
\end{array} \right.\\
\text{Vậy}\,m \ge 2\,\text{hoặc}\,m \le - 1
\end{array}$
