Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
\frac{{1 + \cos x + \cos 2x + \cos 3x}}{{2{{\cos }^2}x + \cos x - 1}} = \frac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
DK:2{\cos ^2}x + \cos x - 1 \ne 0 \Leftrightarrow \left( {\cos x + 1} \right)\left( {2\cos x - 1} \right) \ne 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\cos x \ne 0\\
\cos x \ne \frac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \frac{\pi }{2} + k\pi \\
x \ne \pm \frac{\pi }{3} + k2\pi
\end{array} \right.\\
PT \Leftrightarrow \frac{{\left( {1 + \cos 2x} \right) + \left( {\cos x + \cos 3x} \right)}}{{2{{\cos }^2}x + \cos x - 1}} = \frac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow \frac{{2{{\cos }^2}x + 2\cos 2x\cos x}}{{2{{\cos }^2}x + \cos x - 1}} = \frac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow \frac{{2\cos x\left( {\cos x + \cos 2x} \right)}}{{2{{\cos }^2}x + \cos x - 1}} = \frac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow \frac{{2\cos x\left( {\cos x + 2{{\cos }^2}x - 1} \right)}}{{2{{\cos }^2}x + \cos x - 1}} = \frac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow 2\cos x = \frac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow 3\cos x = 3 - \sqrt 3 \sin x \Leftrightarrow \sqrt 3 \sin x + 3\cos x = 3\\
\Leftrightarrow \frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = \frac{{\sqrt 3 }}{2} \Leftrightarrow \sin \left( {x + \frac{\pi }{3}} \right) = \sin \frac{\pi }{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{3} = \frac{\pi }{3} + k2\pi \\
x + \frac{\pi }{3} = \pi - \frac{\pi }{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \frac{\pi }{3} + k2\pi \left( {loai} \right)
\end{array} \right.\\
Vay\,x = k2\pi
\end{array}\]
