## Precalculus (6th Edition)

$\displaystyle \{\frac{\pi}{12}, \ \ \frac{11\pi}{12}, \ \ \frac{13\pi}{12} , \ \ \frac{23\pi}{12}\}$.
As solved in exercise $2$, two angles (radian measures) with cosine equal to $\displaystyle \frac{1}{2}$ are $\displaystyle \{\frac{\pi}{6}, \ \ \frac{11\pi}{6}\}.$ Also, if $0 \leq x < 2\pi\qquad/\times 2$ then $0 \leq 2x < 4\pi.$ So either $2x= \displaystyle \frac{\pi}{6}\quad$or$\quad 2x= \displaystyle \frac{\pi}{6}+2\pi$ 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 $x= \displaystyle \frac{\pi}{12}\quad$or$\quad x= \displaystyle \frac{\pi}{12}$+$\displaystyle \pi=\frac{13\pi}{12}$ or $x=\displaystyle \frac{11\pi}{12}\quad$or$\quad x= \displaystyle \frac{11\pi}{12}$+$\displaystyle \pi=\frac{23\pi}{12}$ Solution set: $\displaystyle \{\frac{\pi}{12}, \ \ \frac{11\pi}{12}, \ \ \frac{13\pi}{12} , \ \ \frac{23\pi}{12}\}$.