[tex]0 \leq \theta \leq 2\pi;\\ 2\cos(2\theta) + 4\cos(\theta) - 6 = 0|:2\\\cos(2\theta) + 2\cos(\theta) - 3 = 0\\2\cos^2{\theta} - 1 + 2\cos(\theta) - 3 = 0\\2\cos^2{\theta} + 2\cos(\theta) - 4 = 0\\[/tex]
[tex]Let \ t = \cos(\theta)'\\2t^2 + 2t - 4 = 0 => By \ the \ quadratic formula:\\t = \frac{-2 \pm \sqrt{4 - 4(2)(-4)}}{4} = \frac{-2 \pm \sqrt{36}}{4} = \frac{-2 \pm 6}{4} = \frac{-1 \pm 3}{2} => t \in \{1; -2\}\\\\\cos(\theta) = 1 => \theta \in \{0, \pi\} => the \ answer \ is \ C[/tex]
