#### Answer

$\displaystyle \{\frac{7\pi}{6}, \ \ \frac{11\pi}{6}\}$

#### Work Step by Step

For a point on the unit circle, the coordinates are
($\cos x,\sin x)$, $x\in[0,2\pi)$
To solve the problem, find the points for which
the $y$-coordinate is $-\displaystyle \frac{1}{2}$, ($\sin x$=$-\displaystyle \frac{1}{2}$)
Using the given unit circle, we find such points
in quadrant III, for $x=\displaystyle \frac{7\pi}{6}$
in quadrant IV, for $x=\displaystyle \frac{11\pi}{6}$
Solution set: $\displaystyle \{\frac{7\pi}{6}, \ \ \frac{11\pi}{6}\}$