Đáp án:$\left[ \begin{array}{l}
x = \frac{{2\pi }}{3} + k2\pi \\
x = k2\pi
\end{array} \right.\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
\sqrt 3 {\mathop{\rm s}\nolimits} {\rm{inx}} + c{\rm{osx = 1}}\\
\Rightarrow \frac{{\sqrt 3 }}{2}{\mathop{\rm s}\nolimits} {\rm{inx}} + \frac{1}{2}\cos x = \frac{1}{2}\\
\Rightarrow \sin \left( {\frac{\pi }{3}} \right).{\mathop{\rm s}\nolimits} {\rm{inx}} + c{\rm{os}}\frac{\pi }{3}.c{\rm{os}}x = \frac{1}{2}\\
\Rightarrow c{\rm{os}}\left( {x - \frac{\pi }{3}} \right) = cos\frac{\pi }{3}\\
\Rightarrow \left[ \begin{array}{l}
x - \frac{\pi }{3} = \frac{\pi }{3} + k2\pi \\
x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{2\pi }}{3} + k2\pi \\
x = k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
