Solve the equation: $\;$ $\cos x + 2 \cos 2x = 1$
Given equation: $\;\;\;$ $\cos x + 2 \cos 2x = 1$
i.e. $\;$ $\cos x + 2 \left(2 \cos^2 x -1\right) = 1$
i.e. $\;$ $\cos x + 4 \cos^2 x - 2 = 1$
i.e. $\;$ $4 \cos^2 x + \cos x - 3 = 0$
i.e. $\;$ $4 \cos^2 x + 4 \cos x - 3 \cos x - 3 = 0$
i.e. $\;$ $4 \cos x \left(\cos x + 1\right) - 3 \left(\cos x + 1\right) = 0$
i.e. $\;$ $\left(4 \cos x - 3\right) \left(\cos x + 1\right) = 0$
i.e. $\;$ $4 \cos x - 3 = 0$ $\;$ or $\;$ $\cos x + 1 = 0$
Case 1: $\;$ $4 \cos x - 3 = 0$
i.e. $\;$ $\cos x = \dfrac{3}{4}$
i.e. $\;$ $x = 2 n \pi \pm \cos^{-1} \left(\dfrac{3}{4}\right), \;\;\; n \in Z$
Case 2: $\;$ $\cos x + 1 = 0 = 0$
i.e. $\;$ $\cos x = -1 = \cos \left(\pi\right)$
i.e. $\;$ $x = 2 m \pi + \pi, \;\;\; m \in Z$