Giải thích các bước giải:
\(\begin{array}{l}
y = \frac{{m\sin x + 1}}{{\cos x + 2}}\\
\Rightarrow y\cos x + 2y = m\sin x + 1\\
\Leftrightarrow m\sin x - y\cos x = 2y - 1\\
De\;ton\;tai\;x:\\
{m^2} + {y^2} \ge {(2y - 1)^2} \Leftrightarrow 4{y^2} - 4y + 1 \le {y^2} + {m^2}\\
\Leftrightarrow 3{y^2} - 4y + 1 - {m^2} \le 0\\
De\;y\;dat\;gia\;tri\;lon\;nhat\;thi\;pt\;3{y^2} - 4y + 1 - {m^2} = 0\;co\;2\;nghiem\;pb\\
\Rightarrow \Delta ' > 0 \Leftrightarrow 4 - 3.(1 - {m^2}) > 0 \Leftrightarrow m \in R\\
Theo\;vi - et\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{4}{3}\\
{x_1}.{x_2} = \frac{{ - {m^2}}}{3}
\end{array} \right.\\
De\;y\max > 2 \Rightarrow {x_2} > 2\\
{x_1} + {x_2} < 2 + 2 \Rightarrow {x_1} < 2 \Rightarrow ({x_1} - 2)({x_2} - 2) < 0 \Leftrightarrow {x_1}.{x_2} - 2({x_1} + {x_2}) + 4 < 0\\
\Leftrightarrow \frac{{ - {m^2}}}{3} - \frac{8}{3} + 4 < 0 \Rightarrow \left[ \begin{array}{l}
m > 2\\
m < - 2
\end{array} \right.\\
m \in ( - 2018;2018) \Rightarrow co\;4030\;gia\;tri
\end{array}\)
