#### Answer

$\displaystyle \{\frac{7\pi}{12}, \ \ \frac{11\pi}{12}, \ \ \frac{19\pi}{12} , \ \ \frac{23\pi}{12}\}$

#### Work Step by Step

As solved in exercise $3$,
two angles (radian measures) with cosine equal to $\displaystyle \frac{1}{2}$ are
$\displaystyle \{\frac{7\pi}{6}, \ \ \frac{11\pi}{6}\}.$
Also, if $0 \leq x < 2\pi\qquad/\times 2$
then
$0 \leq 2x < 4\pi.$
So either
$2x= \displaystyle \frac{7\pi}{6}\quad$or$\quad 2x= \displaystyle \frac{7\pi}{6}+2\pi$
or
$2x= \displaystyle \frac{11\pi}{6}\quad$or$\quad 2x= \displaystyle \frac{11\pi}{6}+2\pi$
$... $divide both sides by 2 in each equation
$x= \displaystyle \frac{7\pi}{12}\quad$or$\quad x= \displaystyle \frac{7\pi}{12}$+$\displaystyle \pi=\frac{19\pi}{12}$
or
$ x=\displaystyle \frac{11\pi}{12}\quad$or$\quad x= \displaystyle \frac{11\pi}{12}$+$\displaystyle \pi=\frac{23\pi}{12}$
Solution set: $\displaystyle \{\frac{7\pi}{12}, \ \ \frac{11\pi}{12}, \ \ \frac{19\pi}{12} , \ \ \frac{23\pi}{12}\}$.