Khi đó: \ PT \Leftrightarrow \left( {\sin \,x + 1} \right)\left( {\sin \,x + \sqrt 3 \,cos\,x} \right) + 2co{s^2}x = 0.
\ \begin{array}{l} \;{\kern 1pt} \begin{array}{*{20}{l}} { \Leftrightarrow {{\sin }^2}x + \sqrt 3 \sin x{\mkern 1mu} {\kern 1pt} \cos x + \sin {\mkern 1mu} {\kern 1pt} x + \sqrt 3 {\mkern 1mu} {\kern 1pt} cos{\mkern 1mu} {\kern 1pt} x + 2co{s^2}x = 0}\\ { \Leftrightarrow 2cos\left( {2x - \frac{\pi }{3}} \right) + 4\cos \left( {x - \frac{\pi }{6}} \right) + 3 = 0}\\ { \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {t = \cos \left( {x - \frac{\pi }{6}} \right)\left( {t \in \left[ { - 1;1} \right]} \right)}\\ {{{\left( {2t + 1} \right)}^2} = 0} \end{array}} \right. \Leftrightarrow \cos \left( {x - \frac{\pi }{6}} \right) = - \frac{1}{2} = cos\left( { - \frac{{2\pi }}{3}} \right)} \end{array}.\\ \,\, \Leftrightarrow x - \frac{\pi }{6} = \left[ \begin{array}{l} - \frac{{2\pi }}{3} + k2\pi \\ \frac{{2\pi }}{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{6} + k2\pi \\ x = \frac{{5\pi }}{6} + k2\pi \end{array} \right. \end{array}.
Vậy PT có nghiệm \ S = \left\{ { - \frac{\pi }{6} + k2\pi ;\frac{{5\pi }}{6} + k2\pi ;k \in Z} \right\}.
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