Đáp án:
\(\left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2{\cos ^2}x - \left( {2 + \sqrt 3 } \right)\cos x + \sqrt 3 = 0\\
\Leftrightarrow 2{\cos ^2}x - 2\cos x - \sqrt 3 \cos x + \sqrt 3 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x - 2\cos x} \right) + \left( { - \sqrt 3 \cos x + \sqrt 3 } \right) = 0\\
\Leftrightarrow 2\cos x\left( {\cos x - 1} \right) - \sqrt 3 \left( {\cos x - 1} \right) = 0\\
\Leftrightarrow \left( {\cos x - 1} \right)\left( {2\cos x - \sqrt 3 } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x - 1 = 0\\
2\cos x - \sqrt 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \cos 0\\
\cos x = \cos \dfrac{\pi }{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
