## Trigonometry (11th Edition) Clone

The solution set is: $\{\frac{5\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}\}$
$-2~cos~2x = \sqrt{3}$ $cos~2x = -\frac{\sqrt{3}}{2}$ $2x = arccos(-\frac{\sqrt{3}}{2})$ $2x = \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{17\pi}{6}, \frac{19\pi}{6}$ $x = \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}$ For the domain $[0, 2\pi]$, the solution set is: $\{\frac{5\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}\}$