Đáp án: $x = k2\pi \left( {k \in Z} \right)$
Giải thích các bước giải:
ĐKXĐ: $x \ne \left\{ {\pi + k2\pi ;\dfrac{\pi }{3} + k2\pi ;\dfrac{{ - \pi }}{3} + k2\pi \left( {k \in Z} \right)} \right\}$
Ta có:
$\begin{array}{l}
\dfrac{{1 + \cos x + \cos 2x + \cos 3x}}{{2{{\cos }^2}x + \cos x - 1}} = \dfrac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow \dfrac{{\cos x + 2{{\cos }^2}x + 4{{\cos }^3}x - 3\cos x}}{{2{{\cos }^2}x + \cos x - 1}} = \dfrac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow \dfrac{{4{{\cos }^3}x + 2{{\cos }^2}x - 2\cos x}}{{2{{\cos }^2}x + \cos x - 1}} = \dfrac{2}{3}\left( {3 - \sqrt 3 \sin x} \right)\\
\Leftrightarrow 3\cos x - \left( {3 - \sqrt 3 \sin x} \right) = 0\\
\Leftrightarrow \sqrt 3 \cos x + \sin x = \sqrt 3 \\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos x + \dfrac{1}{2}\sin x = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{6} = \dfrac{{ - \pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right) \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \left( l \right)\\
x = k2\pi
\end{array} \right.\left( {k \in Z} \right) \Leftrightarrow x = k2\pi \left( {k \in Z} \right)
\end{array}$
Vậy phương trình có họ nghiệm: $x = k2\pi \left( {k \in Z} \right)$
