\[\begin{array}{l}
\;\left( {1 - m} \right)sinx - m - 1 = 0\\
+ )voi:1 - m = 0 < = > m = 1\\
= > 0.{\mathop{\rm s}\nolimits} {\rm{inx}} - 1 - 1 = 0\\
< = > 0 = 2\left( {ktm} \right)\\
+ )voi:m \ne 1\\
= > \left( {1 - m} \right){\mathop{\rm s}\nolimits} {\rm{inx}} = m + 1\\
= > {\mathop{\rm s}\nolimits} {\rm{inx}} = \frac{{m + 1}}{{1 - m}}\\
Do: - 1 \le {\mathop{\rm s}\nolimits} {\rm{inx}} \le {\rm{1}}\\
{\rm{ = > - 1}} \le \frac{{m + 1}}{{1 - m}} \le 1\\
= > \left\{ \begin{array}{l}
\frac{{m + 1}}{{1 - m}} + 1 \ge 0\\
\frac{{m + 1}}{{1 - m}} - 1 \le 0
\end{array} \right.\\
= > \left\{ \begin{array}{l}
\frac{{m + 1 + 1 - m}}{{1 - m}} \ge 0\\
\frac{{m + 1 - 1 + m}}{{1 - m}} \le 0
\end{array} \right.\\
= > \left\{ \begin{array}{l}
\frac{2}{{1 - m}} \ge 0\\
\frac{{2m}}{{1 - m}} \le 0
\end{array} \right.\\
= > \left\{ \begin{array}{l}
1 - m > 0\\
2m \le 0
\end{array} \right.\\
= > \left\{ \begin{array}{l}
m < 1\\
m \le 0
\end{array} \right.\\
= > m \le 0
\end{array}\]

vậy m$\le$0 thì phương trình (1-m)sinx-m-1=0  có nghiệm