Cho phương trình cos 2x + ( {4m + 1} sin x - ( {2{m^2} + m + 1} = 0

\[\cos 2x + \left( {4m + 1} \right)\sin x - \left( {2{m^2} + m + 1} \right) = 0\]

\[ \Leftrightarrow 1 - 2{\sin ^2}x + \left( {4m + 1} \right)\sin x - 2{m^2} - m - 1 = 0\]

\[ \Leftrightarrow \left( {\sin x - m} \right)\left( {2\sin x - 2m - 1} \right) = 0\]

\[ \Leftrightarrow \left[ \begin{array}{l}\sin x = m\\\sin x = m + \frac{1}{2}\end{array} \right.\]

TH1:\[\left\{ \begin{array}{l} - 1 < m \le \frac{1}{2}\\\frac{{\sqrt 3 }}{2} < m + \frac{1}{2} < 1\end{array} \right. \Leftrightarrow \frac{{\sqrt 3 - 1}}{2} < m < \frac{1}{2}\].

TH2: \[\left\{ \begin{array}{l}m + \frac{1}{2} = 1\\\frac{{\sqrt 3 }}{2} < m < 1\end{array} \right.{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \]vô nghiệm.

TH3: \[\left\{ \begin{array}{l}\frac{1}{2} < m + \frac{1}{2} \le \frac{{\sqrt 3 }}{2}\\\frac{1}{2} < m \le \frac{{\sqrt 3 }}{2}\end{array} \right.{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \]vô nghiệm.

Đáp số: \[\frac{{\sqrt 3 - 1}}{2} < m < \frac{1}{2}\].