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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