Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\cos \left(4x\right) = \cos \left(2x\right)$
$\cos \left(4x\right) = \cos \left(2x\right)$
i.e. $\;$ $\cos \left(4x\right) - \cos \left(2x\right) = 0$
i.e. $\;$ $- 2 \sin \left(\dfrac{4x + 2x}{2}\right) \; \sin \left(\dfrac{4x - 2x}{2}\right) = 0$
i.e. $\;$ $\sin \left(3 x\right) \; \sin \left(x\right) = 0$
i.e. $\;$ $\sin \left(3 x\right) = 0$ $\;$ OR $\;$ $\sin \left(x\right) = 0$
i.e. $\;$ $3x = n \pi, \;\;\; n \in Z$ $\;$ OR $\;$ $x = m \pi, \;\;\; m \in Z$
i.e. $\;$ $x = \dfrac{n \pi}{3}, \;\;\; n \in Z$ $\;$ OR $\;$ $x = m \pi, \;\;\; m \in Z$
In the range $\left[0, 2 \pi \right)$,
when $\;$ $n = 0$, $\;$ $x = 0$
when $\;$ $n = 1$, $\;$ $x = \dfrac{\pi}{3}$
when $\;$ $n = 2$, $\;$ $x = \dfrac{2 \pi}{3}$
when $\;$ $n = 3$, $\;$ $x = \pi$
when $\;$ $n = 4$, $\;$ $x = \dfrac{4 \pi}{3}$
when $\;$ $n = 5$, $\;$ $x = \dfrac{5 \pi}{3}$
when $\;$ $m = 0$, $\;$ $x = 0$
when $\;$ $m = 1$, $\;$ $x = \pi$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = 0, \; \dfrac{\pi}{3}, \; \dfrac{2 \pi}{3}, \; \pi, \; \dfrac{4 \pi}{3}, \; \dfrac{5 \pi}{3}$