Đáp án:
\[\left[ \begin{array}{l}
x = k4\pi \\
x = \dfrac{{10\pi }}{3} + k4\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{3}} \right) + \dfrac{{\sqrt 3 }}{2} = 0\\
\Leftrightarrow \sin \left( {\dfrac{x}{2} - \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \sin \left( {\dfrac{x}{2} - \dfrac{\pi }{3}} \right) = \sin \left( { - \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} - \dfrac{\pi }{3} = - \dfrac{\pi }{3} + k2\pi \\
\dfrac{x}{2} - \dfrac{\pi }{3} = \pi - \left( { - \dfrac{\pi }{3}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} - \dfrac{\pi }{3} = - \dfrac{\pi }{3} + k2\pi \\
\dfrac{x}{2} - \dfrac{\pi }{3} = \dfrac{{4\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} = k2\pi \\
\dfrac{x}{2} = \dfrac{{5\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k4\pi \\
x = \dfrac{{10\pi }}{3} + k4\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
