from your intimate knowlege of the unit circle, you should already know that sine has a value of $-\dfrac{1}{2}$ at the angles $\dfrac{7\pi}{6}$ in quad III and $\dfrac{11\pi}{6}$ in quad IV.

$0 \le x < 2\pi \implies 0 \le 3x < 6\pi$

$3x = \bigg[\dfrac{7\pi}{6} \, , \, \dfrac{11\pi}{6} \, , \, \dfrac{19\pi}{6} \, , \, \dfrac{23\pi}{6} \, , \, \dfrac{31\pi}{6}\, , \, \dfrac{35\pi}{6} \bigg]$

finish solving for $x$.

For the second problem, note the double angle identity $\cos(2x)=2\cos^2{x}-1$ ...

$2\cos^2{x}-1=\cos{x}$

$2\cos^2{x}-\cos{x}-1=0$

$(2\cos{x}+1)(\cos{x}-1)=0$

use the zero product property ...