Phương trình \(\cos x + \sqrt 3 \sin x = \sqrt 3 \) có \(\sqrt {{a^2} + {b^2}}  = \sqrt 4  = 2\)

\(\frac{1}{{\sqrt {{a^2} + {b^2}} }} = \frac{1}{2} = \sin \frac{\pi }{6}\)

Vậy:

 \(\begin{array}{l}\cos x + \sqrt 3 \sin x = \sqrt 3  \Leftrightarrow 2\sin \left( {x + \frac{\pi }{6}} \right) = \sqrt 3 \\ \Leftrightarrow \sin \left( {x + \frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2} = \sin \frac{\pi }{3} \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{6} = \frac{\pi }{3} + k2\pi \\x + \frac{\pi }{6} = \frac{{2\pi }}{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{2} + k2\pi \end{array} \right.,k \in \mathbb{Z}\end{array}\)