## Precalculus (6th Edition)

$\displaystyle \{\frac{\pi}{12},\frac{13\pi}{12},\frac{11\pi}{12},\frac{23\pi}{12}\}$
... dividing with 2, $\displaystyle \cos 2x=\frac{\sqrt{3}}{2}$ If $0 \leq x < 2\pi$ then $0 \leq 2x < 4\pi$ So, $2x\in[0,4\pi)$, meaning that if 2x (a solution) is any number from $[0,2\pi)$, then $2x+2\pi$ is also a solution. Using the unit circle,$2x$ can be $\displaystyle \frac{\pi}{6}$ or $\displaystyle \frac{\pi}{6}+2\pi=\frac{13\pi}{6},$ (quadrant I), or $\displaystyle \frac{11\pi}{6}$ or $\displaystyle \frac{11\pi}{6}+2\pi=\frac{23\pi}{6}$ (quadrant IV) So, $2x\displaystyle \in\{ \frac{\pi}{6},\frac{13\pi}{6},\frac{11\pi}{6},\frac{23\pi}{6}\}$ x is half of 2x, so the corresponding values for x are half of the above values, $\displaystyle \frac{\pi}{6}\div 2=\frac{\pi}{12},$ $\displaystyle \frac{13\pi}{6}\div 2=\frac{13\pi}{12},$ $\displaystyle \frac{11\pi}{6}\div 2=\frac{11\pi}{12},$ $\displaystyle \frac{23\pi}{6}\div 2=\frac{23\pi}{12}.$ Solution set: $x\in\displaystyle \{\frac{\pi}{12},\frac{13\pi}{12},\frac{11\pi}{12},\frac{23\pi}{12}\}$