Đáp án: ${x_{\min }} = x = \dfrac{\pi }{8}$
Giải thích các bước giải:
$\begin{array}{l}
2\sin \left( {4x - \dfrac{\pi }{3}} \right) - 1 = 0\\
\Leftrightarrow \sin \left( {4x - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
4x - \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\
4x - \dfrac{\pi }{3} = \pi - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{7\pi }}{6} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)\\
x > 0;{x_{\min }} = x = \dfrac{\pi }{8}
\end{array}$
