The solution set of the equation $\left(2\cos x-1\right)\left(3+2\cos x\right)=0$  in $0\le x\le 2\pi$  is

$\cos x=\frac{1}{2}$    or $\cos x=-\frac{3}{2}$

Case-1:

$\cos x=\frac{1}{2}$

$x=\frac{\pi }{3},\frac{5\pi }{3}$

Case-2:

$\cos x=-\frac{3}{2}$  is not possible. $\left(\because -1\le \cos x\le 1\right)$