Đáp án:
$\begin{array}{l}
y = \left( {m - 3} \right)x - \left( {2m + 1} \right)\cos x\\
\Rightarrow y' = m - 3 + \left( {2m + 1} \right).\sin x \le 0\forall x\\
\Rightarrow \left( {2m + 1} \right).\sin x \le 3 - m\forall x\\
+ Khi:m = - \dfrac{1}{2}\\
\Rightarrow 0 \le 3 + \dfrac{1}{2}\left( {tm} \right)\\
+ Khi:m \ge - \dfrac{1}{2}\\
\Rightarrow \sin x \le \dfrac{{3 - m}}{{2m + 1}}\forall x\\
Do:\sin x \le 1\forall x\\
\Rightarrow \dfrac{{3 - m}}{{2m + 1}} \ge 1\\
\Rightarrow \dfrac{{3 - m - 2m - 1}}{{2m + 1}} \ge 0\\
\Rightarrow 2 - 3m \ge 0\\
\Rightarrow m \le \dfrac{2}{3}\\
\Rightarrow - \dfrac{1}{2} < m \le \dfrac{2}{3}\\
+ Khi:m < - \dfrac{1}{2}\\
\Rightarrow \sin x \ge \dfrac{{3 - m}}{{2m + 1}}\forall x\\
Do:\sin x \ge - 1\forall x\\
\Rightarrow \dfrac{{3 - m}}{{2m + 1}} \le - 1\\
\Rightarrow \dfrac{{3 - m + 2m + 1}}{{2m + 1}} \le 0\\
\Rightarrow m + 4 \le 0\\
\Rightarrow m \le - 4\\
\text{Vậy}\,m \le - 4\,\text{hoặc}\, - \dfrac{1}{2} \le m \le \dfrac{2}{3}
\end{array}$
