Lời giải chi tiết:

ĐKXĐ: $\left\{ \begin{array}{l}\sin 2x - 1 \ne 0\\\cos 4x - 1 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sin 2x - 1 \ne 0\\\cos 4x \ge 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sin 2x - 1 \ne 0\,\,\left( 1 \right)\\\cos 4x = 1\,\,\,\,\,\,\,\,\left( 2 \right)\end{array} \right.$

Giải (1):

$\Leftrightarrow \sin 2x \ne 1 \Leftrightarrow 2x \ne \frac{\pi }{2} + k2\pi  \Leftrightarrow x \ne \frac{\pi }{4} + k\pi$

Giải (2): $\cos 4x = 1 \Leftrightarrow 4x = k2\pi  \Leftrightarrow x = \frac{{k\pi }}{2}$.

Vậy $D = \left\{ {\frac{{k\pi }}{2}|k \in \mathbb{Z}} \right\}$.

Chọn B.