$\cos (\frac{{17\pi }}{8} + x) = \cos (\frac{\pi }{8} + x)$
$\cos (\frac{{3\pi }}{8} - x) = \sin (\frac{\pi }{2} - \frac{{3\pi }}{8} + x) = \sin (\frac{\pi }{8} + x)$
$ \to \sin (\frac{\pi }{8} + x)\cos (\frac{\pi }{8} + x) = \frac{1}{4} \to \frac{1}{2}\sin (\frac{\pi }{4} + 2x) = \frac{1}{4} \to \sin (\frac{\pi }{4} + 2x) = \frac{1}{2}$
$\left. \begin{gathered}
\frac{\pi }{4} + 2x = 2k\pi + \frac{\pi }{6} \to x = k\pi - \frac{\pi }{{24}} \to x = - \frac{\pi }{{24}} \hfill \\
\frac{\pi }{4} + 2x = 2k\pi + \frac{{5\pi }}{6} \to x = k\pi + \frac{{7\pi }}{{24}} \to x = \frac{{7\pi }}{{24}} \hfill \\
\end{gathered} \right\}\frac{{6\pi }}{{24}} = \frac{\pi }{4}$