Solve the equation: $\;$ $4 \cos^3 x - 4 \cos^2 x - \cos \left(\pi + x\right) - 1 = 0$
Given equation: $\;\;\;$ $4 \cos^3 x - 4 \cos^2 x - \cos \left(\pi + x\right) - 1 = 0$
i.e. $\;$ $4 \cos^2 x \left(\cos x - 1\right) - \left(- \cos x\right) - 1 = 0$
i.e. $\;$ $4 \cos^2 x \left(\cos x - 1\right) + \cos x - 1 = 0$
i.e. $\;$ $\left(4 \cos^2 x + 1\right) \left(\cos x - 1\right) = 0$
$\implies$ $4 \cos^2 x + 1 = 0$ $\;$ or $\;$ $\cos x - 1 = 0$
When $\;$ $4 \cos^2 x + 1 = 0$
i.e. $\;$ $\cos^2 x = \dfrac{-1}{4}$
But square of any number cannot be negative.
$\therefore \;$ $\cos^2 x = \dfrac{-1}{4}$ $\;$ will not give a valid solution.
When $\;$ $\cos x - 1 = 0$
i.e. $\;$ $\cos x = 1 = \cos 0$
$\implies$ $x = 2 n \pi, \;\; n \in I$