Đáp án: $\left[ \begin{array}{l}
x = \frac{\pi }{6} + \frac{{k\pi }}{2}\\
x = \frac{{ - \pi }}{6} + \frac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
2\cos 4x + 1 = 0\\
\Rightarrow \cos 4x = \frac{{ - 1}}{2}\\
\Rightarrow \left[ \begin{array}{l}
4x = \frac{{2\pi }}{3} + k2\pi \\
4x = - \frac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + \frac{{k\pi }}{2}\\
x = \frac{{ - \pi }}{6} + \frac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
