Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\cos \left(2x\right) = 2 - 5 \cos x$
$\cos \left(2x\right) = 2 - 5 \cos x$
i.e. $\;$ $2 \cos^2 x - 1 = 2 - 5 \cos x$
i.e. $\;$ $2 \cos^2 x + 5 \cos x - 3 = 0$
i.e. $\;$ $2 \cos^2 x + 6 \cos x - \cos x - 3 = 0$
i.e. $\;$ $\cos x \left(2 \cos x - 1\right) + 3 \left(2 \cos x - 1\right) = 0$
i.e. $\;$ $\left(\cos x + 3\right) \left(2 \cos x - 1\right) = 0$
i.e. $\;$ $\cos x + 3 = 0$ $\;$ OR $\;$ $2 \cos x - 1 = 0$
Now, $\cos x + 3 = 0$ $\implies$ $\cos x = -3$, $\;$ is not possible $\;$ $\because \; -1 \leq \cos x \leq 1$
$2 \cos x - 1 = 0$ $\implies$ $\cos x = \dfrac{1}{2}$
$\left\{\text{Principal value of } \cos x \text{ lies in } \left[0, \pi\right] \text{ i.e. in first or second quadrants} \right\}$
$\therefore \;$ $\cos x = \dfrac{1}{2} = \cos \left(\dfrac{\pi}{3}\right)$
$\implies$ $x = 2 n \pi \pm \dfrac{\pi}{3}, \;\;\; n \in Z$
$\therefore \;$ In the range $\left[0, 2\pi \right)$,
when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{3}$
when $\;$ $n = 1$ $\;$ $x = 2 \pi - \dfrac{\pi}{3} = \dfrac{5 \pi}{3}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = \dfrac{\pi}{3}, \; \dfrac{5 \pi}{3}$