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Đặt :$\begin{array}{l}
t = \cos X, - 1 \le t \le 1\\
\Rightarrow c{\rm{os}}2X + m.\cos X + 4 = 2{\cos ^2}X - 1 + m.\cos X + 4 = 2{t^2} + mt + 3
\end{array}$
Hàm số $(1)$ xác định
$\begin{array}{l}
\forall X \in R\\
\Leftrightarrow 2{t^2} + mt + 3 > 0\\
\forall t \in \left[ { - 1,1} \right]
\end{array}$
Đặt
$f(t) > 0\forall t \in \left[ { - 1,1} \right]$
$ \Leftrightarrow \left[ \begin{array}{l}
\Delta < 0(2)\\
\left\{ \begin{array}{l}
\Delta \ge 0\\
\left[ \begin{array}{l}
- 1 < 1 < {t_1} < {t_2}\\
{t_1} \le {t_2} < - 1 < 1
\end{array} \right.
\end{array} \right.(3)
\end{array} \right.$
Ta có:
$\Delta = {m^2} - 24;f(1) = m + 5;f( - 1) = - m + 5$
(HV-18)
$\begin{array}{l}
(2) \Leftrightarrow - 2\sqrt 6 < m < 2\sqrt 6 \\
(3) \Leftrightarrow \left\{ \begin{array}{l}
\left\{ {m \le - 2\sqrt 6 } \right. \vee m \ge 2\sqrt 6 \vee m \ge 2\sqrt 6 \\
\left\{ \begin{array}{l}
f(1) > 0\\
\frac{s}{2} - 1 > 0
\end{array} \right. \vee \left\{ \begin{array}{l}
f( - 1) > 0\\
\frac{s}{2} + 1 < 0
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
f(1) > 0\\
\frac{s}{2} - 1 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m + 5 > 0\\
- \frac{m}{4} - 1 > 0
\end{array} \right. \Leftrightarrow - 5 < m < - 4\\
\left\{ \begin{array}{l}
f( - 1) > 0\\
\frac{s}{2} + 1 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- m + 5 > 0\\
- \frac{m}{4} + 1 < 0
\end{array} \right. \Leftrightarrow 4 < m < 5\\
suy\,\,ra\,\,\,(3) \Leftrightarrow \left[ \begin{array}{l}
- 5 < m \le - 2\sqrt 6 \\
2\sqrt 6 \le m \le 5
\end{array} \right.(5)
\end{array}$
Hợp các tập nghiệm ở $(4)$và $(5)$ ta có
$ - 5 < m < 5$. Vậy ${\rm{D = ( - 5, 5)}}$
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