$\begin{array}{l}
\cos 2x - \sin 2x - 1 = 0\\
\Leftrightarrow \cos 2x - \sin 2x = 1\\
\Leftrightarrow \sqrt 2 \cos \left( {2x + \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow \cos \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
2x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
k = 1 \Rightarrow x = \dfrac{\pi }{4} + \pi = \dfrac{{5\pi }}{4} \in S \to D\\
17)\sin 2x + \sqrt 3 \cos 2x = \sqrt 3 \\
\Leftrightarrow 2\sin \left( {2x + \dfrac{\pi }{3}} \right) = \sqrt 3 \\
\Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k\pi
\end{array} \right.\\
k = 0 \Rightarrow x = \dfrac{\pi }{6} \in \left( {0;\dfrac{\pi }{2}} \right)\\
k = 1,x = \pi \notin \left( {0;\dfrac{\pi }{2}} \right),x = \dfrac{{7\pi }}{6} \notin \left( {0;\dfrac{\pi }{2}} \right)\\
{\kern 1pt} \to 1\,gt \to A
\end{array}$
