2cos²x +2√3 sinx. cosx +1=3(sinx+√3cosx)

$\begin{array}{l} 2{\cos ^2}x + 2\sqrt 3 \sin x\cos x + 1 = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\  \Leftrightarrow 1 + \cos 2x + \sqrt 3 \sin 2x + 1 = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\  \Leftrightarrow 2 = 3\left( {\sin x + \sqrt 3 \cos x} \right) - \left( {\cos 2x + \sqrt 3 \sin 2x} \right)\\  \Leftrightarrow 1 = 3\cos \left( {x - \dfrac{\pi }{6}} \right) - \cos \left( {2x - \dfrac{\pi }{3}} \right)\\  \Leftrightarrow 1 = 3\cos \left( {x - \dfrac{\pi }{6}} \right) - \left( {2{{\cos }^2}\left( {x - \dfrac{\pi }{6}} \right) - 1} \right)\\  \Leftrightarrow 3\cos \left( {x - \dfrac{\pi }{6}} \right) - 2{\cos ^2}\left( {x - \dfrac{\pi }{6}} \right) = 0\\  \Leftrightarrow \cos \left( {x - \dfrac{\pi }{6}} \right)\left( {3 - 2\cos \left( {x - \dfrac{\pi }{6}} \right)} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} \cos \left( {x - \dfrac{\pi }{6}} \right) = 0\\ 3 - 2\cos \left( {x - \dfrac{\pi }{6}} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{\pi }{6} = \dfrac{\pi }{2} + k\pi \\ \cos \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{3}{2}(PTVN) \end{array} \right.\\  \Rightarrow x = \dfrac{{2\pi }}{3} + k\pi \left( {k \in \mathbb{Z}} \right)\\  \Rightarrow S = \left\{ {\dfrac{{2\pi }}{3} + k\pi |k \in \mathbb{Z}} \right\} \end{array}$