Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\sin \left(2x\right) = \tan x$
$\sin \left(2x\right) = \tan x$
i.e. $\;$ $2 \sin x \cos x = \dfrac{\sin x}{\cos x}$
i.e. $\;$ $2 \sin x \cos^2 x = \sin x$
i.e. $\;$ $2 \sin x \left(1 - \sin^2 x\right) = \sin x$
i.e. $\;$ $\sin x - 2 \sin^3 x = 0$
i.e. $\;$ $\sin x \left(1 - 2 \sin^2 x \right) = 0$
i.e. $\;$ $\sin x = 0$ $\;$ OR $\;$ $1 - 2 \sin^2 x = 0$
i.e. $\;$ $\sin x = 0$ $\;$ OR $\;$ $\sin^2 x = \dfrac{1}{2}$
i.e. $\;$ $\sin x = 0$ $\;$ OR $\;$ $\sin x = \pm \dfrac{1}{\sqrt{2}}$
Now, $\sin x = 0$ $\implies$ $x = n \pi, \;\;\; n \in Z$
OR $\;$ $\sin x = \dfrac{1}{\sqrt{2}}$ $\implies$ $x = m \pi + \left(-1\right)^m \times \left(\dfrac{\pi}{4}\right), \;\;\; m \in Z$
OR $\;$ $\sin x = - \dfrac{1}{\sqrt{2}}$ $\implies$ $x = q \pi + \left(-1\right)^q \times \left(-\dfrac{\pi}{4}\right), \;\;\; q \in Z$
In the range $\left[0, 2 \pi \right)$,
when $n = 0$, $\;$ $x = 0$
when $n = 1$, $\;$ $x = \pi$
when $m = 0$, $\;$ $x = \dfrac{\pi}{4}$
when $m = 1$, $\;$ $x = \pi - \dfrac{\pi}{4} = \dfrac{3 \pi}{4}$
when $q = 0$, $\;$ $x = - \dfrac{\pi}{4}$
$\left[- \dfrac{\pi}{4} \text{ is the same as } 2 \pi - \dfrac{\pi}{4} = \dfrac{7 \pi}{4} \right]$
when $q = 1$, $\;$ $x = \pi + \dfrac{\pi}{4} = \dfrac{5 \pi}{4}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = 0, \; \dfrac{\pi}{4}, \; \dfrac{3 \pi}{4}, \; \pi, \; \dfrac{5 \pi}{4}, \; \dfrac{7 \pi}{4}$