2sin^2x + căn3 sin 2x - 2(căn3 sinx + cosx) -7 =0

$\begin{array}{l} 2{\sin ^2}x + \sqrt 3 \sin 2x - 2\left( {\sqrt 3 \sin x + \cos x} \right) - 7 = 0\,\,\left( * \right)\\ Dat\,\sqrt 3 \sin x + \cos x = t\left( { - 2 \le t \le 2} \right)\,thi:\\ {t^2} = 3{\sin ^2}x + 2\sqrt 3 \sin x\cos x + {\cos ^2}x = 2{\sin ^2}x + \sqrt 3 \sin 2x + 1\\ \Rightarrow 2{\sin ^2}x + \sqrt 3 \sin 2x = {t^2} - 1\\ \Rightarrow \left( * \right) \Leftrightarrow {t^2} - 1 - 2t - 7 = 0 \Leftrightarrow {t^2} - 2t - 8 = 0 \Leftrightarrow \left[ \begin{array}{l} t = - 2\left( {TM} \right)\\ t = 4\left( {loai} \right) \end{array} \right.\\ \Rightarrow \sqrt 3 \sin x + \cos x = - 2 \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x = - 1\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{2}} \right) \Leftrightarrow x + \frac{\pi }{6} = - \frac{\pi }{2} + k2\pi \Leftrightarrow x = - \frac{{2\pi }}{3} + k2\pi \end{array}$

$\begin{array}{l} 2{\sin ^2}x + \sqrt 3 \sin 2x - 2\left( {\sqrt 3 \sin x + \cos x} \right) - 7 = 0\,\,\left( * \right)\\ Dat\,\sqrt 3 \sin x + \cos x = t\left( { - 2 \le t \le 2} \right)\,thi:\\ {t^2} = 3{\sin ^2}x + 2\sqrt 3 \sin x\cos x + {\cos ^2}x = 2{\sin ^2}x + \sqrt 3 \sin 2x + 1\\ \Rightarrow 2{\sin ^2}x + \sqrt 3 \sin 2x = {t^2} - 1\\ \Rightarrow \left( * \right) \Leftrightarrow {t^2} - 1 - 2t - 7 = 0 \Leftrightarrow {t^2} - 2t - 8 = 0 \Leftrightarrow \left[ \begin{array}{l} t = - 2\left( {TM} \right)\\ t = 4\left( {loai} \right) \end{array} \right.\\ \Rightarrow \sqrt 3 \sin x + \cos x = - 2 \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x = - 1\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{2}} \right) \Leftrightarrow x + \frac{\pi }{6} = - \frac{\pi }{2} + k2\pi \Leftrightarrow x = - \frac{{2\pi }}{3} + k2\pi \end{array}$