Đáp án:

$ \left[ \begin{array}{l} - 2\sqrt 2 < m < 2\sqrt 2 \\ - 3 \le m \le - 2\sqrt 2 \\ 2\sqrt 2 \le m \le 3 \end{array} \right.$

Lời giải:

\(\begin{array}{l} 1)\,\,\,f\left( x \right) = \dfrac{{3x}}{{\sqrt {2{{\sin }^2}x - m\sin x + 1} }}\\ \text{Hàm số xác định khi }\Leftrightarrow 2{\sin ^2}x - m\sin x + 1 > 0\,\,\,\left( * \right)\\ \text{Đặt }t = \sin x,\,\,\,\, - 1 \le t \le 1.\\ \Rightarrow \left( * \right) \Leftrightarrow 2{t^2} - mt + 1 > 0\,\,\,\forall t \in \left[ { - 1;\,\,1} \right]\\ \Leftrightarrow \left[ \begin{array}{l} \Delta < 0\\ \left\{ \begin{array}{l} \Delta \ge 0\\ - 1 \le {t_1} < {t_2} \le 1 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {m^2} - 8 < 0\\ \left\{ \begin{array}{l} {m^2} - 8 > 0\\ \left( {{t_1} - 1} \right)\left( {{t_2} - 1} \right) \ge 0\\ \left( {{t_1} + 1} \right)\left( {{t_2} + 1} \right) \ge 0 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 2\sqrt 2 < m < 2\sqrt 2 \\ \left\{ \begin{array}{l} \left[ \begin{array}{l} m \ge 2\sqrt 2 \\ m \le - 2\sqrt 2 \end{array} \right.\\ {t_1}{t_2} - \left( {{t_1} + {t_2}} \right) + 1 \ge 0\\ {t_1}{t_2} + \left( {{t_1} + {t_2}} \right) + 1 \ge 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2\sqrt 2 < m < 2\sqrt 2 \\ \left\{ \begin{array}{l} \left[ \begin{array}{l} m \ge 2\sqrt 2 \\ m \le - 2\sqrt 2 \end{array} \right.\\ \dfrac{1}{2} - \dfrac{m}{2} + 1 \ge 0\\ \dfrac{1}{2} + \dfrac{m}{2} + 1 \ge 0 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 2\sqrt 2 < m < 2\sqrt 2 \\ \left\{ \begin{array}{l} \left[ \begin{array}{l} m \ge 2\sqrt 2 \\ m \le - 2\sqrt 2 \end{array} \right.\\ m \le 3\\ m \ge - 3 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2\sqrt 2 < m < 2\sqrt 2 \\ - 3 \le m \le - 2\sqrt 2 \\ 2\sqrt 2 \le m \le 3 \end{array} \right..\,\, \end{array}\)