Đáp án:
\[x = - \frac{\pi }{{18}} + \frac{{2k\pi }}{3}\]
Giải thích các bước giải:
ĐKXĐ: \(2{\cos ^2}x + \sin x - 1 \ne 0\)
Ta có:
\(\begin{array}{l}
\frac{{\cos x - 2\sin x\cos x}}{{2{{\cos }^2}x + \sin x - 1}} = \sqrt 3 \\
\Leftrightarrow \frac{{\cos x - \sin 2x}}{{\left( {2{{\cos }^2}x - 1} \right) + \sin x}} = \sqrt 3 \\
\Leftrightarrow \frac{{\cos x - \sin 2x}}{{\cos 2x + \sin x}} = \sqrt 3 \\
\Leftrightarrow \cos x - \sin 2x = \sqrt 3 \cos 2x + \sqrt 3 \sin x\\
\Leftrightarrow \cos x - \sqrt 3 \sin x = \sqrt 3 \cos 2x + \sin 2x\\
\Leftrightarrow \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x = \frac{{\sqrt 3 }}{2}\cos 2x + \frac{1}{2}\sin 2x\\
\Leftrightarrow \cos x.\cos \frac{\pi }{3} - \sin x.\sin \frac{\pi }{3} = \cos 2x.\cos \frac{\pi }{6} + \sin \frac{\pi }{6}.\sin 2x\\
\Leftrightarrow \cos \left( {x + \frac{\pi }{3}} \right) = \cos \left( {2x - \frac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{3} = 2x - \frac{\pi }{6} + 2k\pi \\
x + \frac{\pi }{3} = - 2x + \frac{\pi }{6} + 2k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + 2k\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( L \right)\\
x = - \frac{\pi }{{18}} + \frac{{2k\pi }}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {t/m} \right)
\end{array} \right.
\end{array}\)
Vậy \(x = - \frac{\pi }{{18}} + \frac{{2k\pi }}{3}\)
