## Algebra 2 (1st Edition)

$x=\pm \dfrac{4 \pi}{3}+4k \pi; k \in z$
$2 \cos \dfrac{x}{2}+1=0$ This gives: $2 \cos \dfrac{x}{2}=-1$ and $\cos \dfrac{x}{2}=-\dfrac{1}{2}$ or, $\dfrac{x}{2}=\pm \dfrac{2 \pi}{3}+2 k \pi; k \in z$ Hence, we have $x=\pm \dfrac{4 \pi}{3}+4k \pi; k \in z$