The minimum of cos²(120^o + x) + cos²(120^o - x) is

$\begin{array}{l}{\cos }^2\left({120}^{\circ }+\mathrm{x}\right)+{\cos }^2\left({120}^{\circ }-\mathrm{x}\right)=\frac{1+\cos \left({240}^{\circ }+2x\right)}{2}+\frac{1+\cos \left({240}^{\circ }-2\mathrm{x}\right)}{2}\\ =1+\frac{1}{2}\left[\cos \left({240}^{\circ }+2\mathrm{x}\right)+\cos \left({240}^{\circ }-2\mathrm{x}\right)\right]=1+\cos {240}^{\circ }\cos 2\mathrm{x}=1-\frac{1}{2}\cos 2\mathrm{x}\end{array}$
Minimum value $=1-\frac{1}{2}=\frac{1}{2}.$