Đáp án:
$\begin{array}{l}
\text{Đặt}:\cos x + \sqrt 3 \sin x + 1 = a\\
\Rightarrow \cos x + \sqrt 3 \sin x = a - 1\\
Dk:1 + {\left( {\sqrt 3 } \right)^2} \ge {\left( {a - 1} \right)^2}\\
\Leftrightarrow {\left( {a - 1} \right)^2} \le 4\\
\Leftrightarrow - 2 \le a - 1 \le 2\\
\Leftrightarrow - 1 \le a \le 3\\
Pt:\cos x + \sqrt 3 \sin x = \dfrac{3}{{\cos x + \sqrt 3 \sin x + 1}}\\
\Leftrightarrow a - 1 = \dfrac{3}{a}\\
\Leftrightarrow a\left( {a - 1} \right) = 3\\
\Leftrightarrow {a^2} - a - 3 = 0\\
\Leftrightarrow {\left( {a - \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} + 3 = \dfrac{{13}}{4}\\
\Leftrightarrow a = \dfrac{{\sqrt {13} + 1}}{2}\left( {do: - 1 \le a \le 3} \right)\\
\Leftrightarrow \cos x + \sqrt 3 \sin x = \dfrac{{\sqrt {13} + 1}}{2} - 1\\
\Leftrightarrow \cos x + \sqrt 3 \sin x = \dfrac{{\sqrt {13} - 1}}{2}\\
\Leftrightarrow \dfrac{1}{2}.\cos x + \dfrac{{\sqrt 3 }}{2}\sin x = \dfrac{{\sqrt {13} - 1}}{4}\\
\Leftrightarrow \sin \left( {{{30}^0} + x} \right) = \dfrac{{\sqrt {13} - 1}}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x + {30^0} = {41^0} + k{.360^0}\\
x + {30^0} = {180^0} - {41^0} + k{.360^0}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = {11^0} + k{.360^0}\\
x = {109^0} + k{.360^0}
\end{array} \right.
\end{array}$
