Given,
\( \tan x=\frac{3}{4}, \Pi < x < 3 \frac{\Pi}{2} \) \( \Rightarrow \frac{\Pi}{2} < \frac{x}{2} < 3 \frac{\Pi}{4} \) (belongs to quadrant II)
In the second quadrant, \( \cos \left(\frac{x}{2}\right) \) is negative.
\[
\begin{array}{l}
\sec 2 x=1+\tan 2 x=1+\left(\frac{3}{4}\right) 2=\frac{(16+9)}{16}=\frac{25}{16} \\
\sec x=\pm \frac{5}{4}
\end{array}
\]
Thus, \( \cos x=-\frac{4}{5}\left\{\right. \) since \( \left.\Pi < x < 3 \frac{\Pi}{2}\right\} \)
Using the formula, \( \cos 2 A=2 \cos 2 A-1 \),
\( 2 \cos 2\left(\frac{x}{2}\right)=1+\cos x=1+\left(-\frac{4}{5}\right) \)
\( 2 \cos 2\left(\frac{x}{2}\right)=\frac{1}{5} \)
\( \cos 2\left(\frac{x}{2}\right)=\frac{1}{10} \)
\( \cos \left(\frac{x}{2}\right)=\pm \frac{\sqrt{1}}{10}=-\frac{1}{\sqrt{10}}\left\{\right. \) since \( \frac{x}{2} \) lies in the quadrant II\}