Đáp án:
\[\left[ \begin{array}{l}
x = - \frac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x - 6\sin x - 4\cos x - 5 = 0\\
\Leftrightarrow \left( {{{\cos }^2}x - {{\sin }^2}x} \right) - 6\sin x - 4\cos x - 5 = 0\\
\Leftrightarrow \left( {{{\cos }^2}x - 4\cos x + 4} \right) - \left( {{{\sin }^2}x + 6\sin x + 9} \right) = 0\\
\Leftrightarrow {\left( {\cos x - 2} \right)^2} - {\left( {\sin x + 3} \right)^2} = 0\\
\Leftrightarrow \left( {\cos x - 2 - \sin x - 3} \right)\left( {\cos x - 2 + \sin x + 3} \right) = 0\\
\Leftrightarrow \left( {\cos x - \sin x - 5} \right)\left( {\cos x + \sin x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x - \sin x = 5\\
\cos x + \sin x = - 1
\end{array} \right. \Leftrightarrow \cos x + \sin x = - 1\\
\Leftrightarrow \frac{{\sqrt 2 }}{2}\cos x + \frac{{\sqrt 2 }}{2}\sin x = - \frac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin \left( { - \frac{\pi }{4}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{4} = - \frac{\pi }{4} + k2\pi \\
x + \frac{\pi }{4} = \frac{{5\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.
\end{array}\)
