Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $3 \cos \left(2x\right) = \sin x + 2$
$3 \cos \left(2x\right) = \sin x + 2$
i.e. $\;$ $3 \left(1 - 2 \sin^2 x\right) = \sin x + 2$
i.e. $\;$ $3 - 6 \sin^2 x = \sin x + 2$
i.e. $\;$ $6 \sin^2 x + \sin x - 1 = 0$
i.e. $\;$ $6 \sin^2 x + 3 \sin x - 2 \sin x - 1 = 0$
i.e. $\;$ $3 \sin x \left(2 \sin x + 1\right) - 1 \left(2 \sin x + 1\right) = 0$
i.e. $\;$ $\left(3 \sin x - 1\right) \left(2 \sin x + 1\right) = 0$
i.e. $\;$ $3 \sin x - 1 = 0$ $\;$ OR $\;$ $2 \sin x + 1 = 0$
$\implies$ $\sin x = \dfrac{1}{3}$ $\;$ OR $\;$ $\sin x = - \dfrac{1}{2}$
i.e. $\;$ $\sin x = \dfrac{1}{3}$ $\;$ OR $\;$ $\sin x = \sin \left(- \dfrac{\pi}{6}\right)$
i.e. $\;$ $x = n \pi + \left(-1\right)^n \times \arcsin \left(\dfrac{1}{3}\right), \;\;\; n \in Z$
OR $\;$ $x = m \pi + \left(- 1\right)^m \times \left(- \dfrac{\pi}{6}\right), \;\;\; m \in Z$
In the range $\left[0, 2 \pi \right)$,
when $n = 0$, $\;$ $x = \arcsin \left(\dfrac{1}{3}\right)$
when $n = 1$, $\;$ $x = \pi - \arcsin \left(\dfrac{1}{3}\right)$
when $m = 0$, $\;$ $x = - \dfrac{\pi}{6}$
$- \dfrac{\pi}{6}$ $\;$ is the same as $\;$ $2 \pi - \dfrac{\pi}{6} = \dfrac{11 \pi}{6}$
when $m = 1$, $\;$ $x = \pi + \dfrac{\pi}{6} = \dfrac{7 \pi}{6}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = \arcsin \left(\dfrac{1}{3}\right), \; \pi - \arcsin \left(\dfrac{1}{3}\right), \; \dfrac{7 \pi}{6}, \; \dfrac{11 \pi}{6}$