#### Answer

$\displaystyle \frac{\sqrt{13}}{4}$

#### Work Step by Step

$\displaystyle \frac{\pi}{2} < x < \pi\qquad/\div 2$
$\displaystyle \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$,
so, $\displaystyle \frac{x}{2}$ is in quadrant I, where the cosine is positive.
Half-Angle Identity:
$\displaystyle \sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}=+\sqrt{\frac{1-(-\frac{5}{8})}{2}}$
$=\sqrt{\dfrac{\frac{13}{8}}{2} }=\sqrt{\dfrac{13}{16}} $
$=\displaystyle \frac{\sqrt{13}}{4}$