Solve the equation in the interval [0, 2pi]

Solve the equation in the interval [0, 2pi]
(cos(x))^2 = 1/9

I only got x = 1.230959
but if I can get help with the other answers.
I'd appreciate it
10 points.

cos x = ± 1/3

cos x = 1/3 => x = 1.230959 and 2π - 1.230959 radians

cos = -1/3 => x = 1.910633 and 2π - 1.910633