Solve the equation: $\;$ $1 + \cos x + \cos 2 x = 0$
Given equation: $\;\;\;$ $1 + \cos x + \cos 2 x = 0$
i.e. $\;$ $1 + \cos x + 2 \cos^2 x - 1 = 0$
i.e. $\;$ $2 \cos^2 x + \cos x = 0$
i.e. $\;$ $\cos x \left(2 \cos x + 1\right) = 0$
i.e. $\;$ $\cos x = 0$ $\;$ or $\;$ $2 \cos x + 1 = 0$
Case 1: $\;$ $\cos x = 0 = \cos \left(\dfrac{\pi}{2}\right)$
i.e. $\;$ $x = \left(2n + 1\right) \times \dfrac{\pi}{2}, \;\;\; n \in Z$
Case 2: $\;$ $2 \cos x + 1 = 0$
i.e. $\;$ $\cos x = \dfrac{-1}{2} = \cos \left(\pi - \dfrac{\pi}{3}\right)$
i.e. $\;$ $\cos x = \cos \left(\dfrac{2 \pi}{3}\right)$
i.e. $\;$ $x = 2 m \pi \pm \dfrac{2 \pi}{3}, \;\;\; m \in Z$