I want to minimize a real valued function $f(x) = \frac{2}{3} \cos(x)+1+ \frac{1}{9}\cos^2(x),$ but it turns out that the minimum occurs at $x = \pi - \arccos 3$.
I am not sure what to conclude from this fact. Does the minimum now occur at $x = \pi$, or how should I interpret the fact that $\arccos$ is only defined in the interval $[-1,1]$?