D

- Hướng dẫn giải

\(\begin{array}{l}
\cos \left( {2x - \alpha } \right) = \cos x\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \alpha  = x + k2\pi \\
2x - \alpha  =  - x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\alpha }{3} + \frac{{k2\pi }}{3}\\
x = \alpha  + k2\pi 
\end{array} \right.\\
\cos 2x + \sqrt 3 \sin 2x - 2\cos x = 0\\
 \Leftrightarrow \frac{1}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x = \cos x\\
 \Leftrightarrow \cos \left( {2x - \frac{\pi }{3}} \right) = \cos x\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{3} + k2\pi \\
x = \frac{\pi }{9} + \frac{{k2\pi }}{3}
\end{array} \right.
\end{array}\)

Để hai phương trình tương đương cần có 

\(\left\{ \begin{array}{l}
\frac{\alpha }{3} = \frac{\pi }{9}\\
\alpha  = \frac{\pi }{3}
\end{array} \right. \Rightarrow \alpha  = \frac{\pi }{3}\)