Đáp án:
\(m \in \left( { - \infty ; - 1} \right] \cup \left( {0; + \infty } \right)\).
Giải thích các bước giải:
\(m\sin x + \left( {m + 1} \right)\cos x = \dfrac{m}{{\cos x}}\)
ĐK: \(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \,\,\left( {k \in Z} \right)\).
\(\begin{array}{l}pt \Leftrightarrow m\sin x\cos x + \left( {m + 1} \right){\cos ^2}x = m\\ \Leftrightarrow \dfrac{m}{2}\sin 2x + \left( {m + 1} \right)\dfrac{{1 + \cos 2x}}{2} = m\\ \Leftrightarrow m\sin 2x + \left( {m + 1} \right)\cos 2x + m + 1 = m\\ \Leftrightarrow m\sin 2x + \left( {m + 1} \right)\cos 2x = - 1\end{array}\)
Phương trình đã cho có nghiệm
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}{m^2} + {\left( {m + 1} \right)^2} \ge {\left( { - 1} \right)^2}\\m\sin \left( {\pi + k2\pi } \right) + \left( {m + 1} \right)\cos \left( {\pi + k2\pi } \right) \ne - 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{m^2} + {m^2} + 2m + 1 \ge 1\\ - m - 1 \ne - 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2{m^2} + 2m \ge 0\\m \ne 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}m \ge 0\\m \le - 1\end{array} \right.\\m \ne 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}m > 0\\m \le - 1\end{array} \right.\end{array}\)
Vậy \(m \in \left( { - \infty ; - 1} \right] \cup \left( {0; + \infty } \right)\).
