#### Answer

$ \displaystyle \frac{\sqrt{50-15\sqrt{10}}}{10}$

#### Work Step by Step

$\displaystyle \frac{\pi}{2} < x < \pi \qquad/\div 2$
$\displaystyle \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$
So, we see that
$x$ is in quadrant II, $\cos x$ is negative,
and $\displaystyle \frac{x}{2}$ is in quadrant I, $\displaystyle \cos\frac{x}{2}$ is positive.
$\displaystyle \cot x=-3\Rightarrow\tan x=-\frac{1}{3}$
$\displaystyle \sec^{2}x=\tan^{2}+1=\frac{1}{9}+1=\frac{10}{9}$
$\displaystyle \sec x=-\frac{\sqrt{10}}{3}$
$\displaystyle \cos x=-\frac{3}{\sqrt{10}}=-\frac{3\sqrt{10}}{10}$
Half-Angle Identity:
$\displaystyle \cos\frac{x}{2}=+\sqrt{\frac{1+\cos x}{2}}=\sqrt{\frac{1-\frac{3\sqrt{10}}{10}}{2}}$
$=\displaystyle \sqrt{\frac{10-3\sqrt{10}}{20}\cdot\frac{5}{5}}=\frac{\sqrt{50-15\sqrt{10}}}{10}$