Đáp án:
$\begin{array}{l}
- \dfrac{\pi }{4} \le x \le \dfrac{\pi }{4}\\
\Rightarrow - \dfrac{1}{{\sqrt 2 }} \le \sin \,x \le \dfrac{1}{{\sqrt 2 }}\\
\Rightarrow - \dfrac{1}{{\sqrt 2 }} \le t \le \dfrac{1}{{\sqrt 2 }}\left( {t = \sin x} \right)\\
{\sin ^2}x + \left( {m - 1} \right)\sin x - m = 0\\
\Rightarrow {t^2} + \left( {m - 1} \right).t - m = 0\\
\Rightarrow {t^2} - t + m\left( {t - 1} \right) = 0\\
\Rightarrow m = - t\left( {do:t \ne 1} \right)\\
Do: - \dfrac{1}{{\sqrt 2 }} \le t \le \dfrac{1}{{\sqrt 2 }}\\
\Rightarrow - \dfrac{1}{{\sqrt 2 }} \le - t \le \dfrac{1}{{\sqrt 2 }}\\
\Rightarrow - \dfrac{1}{{\sqrt 2 }} \le m \le \dfrac{1}{{\sqrt 2 }}\\
hay\, - \dfrac{{\sqrt 2 }}{2} \le m \le \dfrac{{\sqrt 2 }}{2}
\end{array}$
