Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\tan \left(2x\right) - 2 \cos x = 0$
$\tan \left(2x\right) - 2 \cos x = 0$
i.e. $\;$ $\dfrac{\sin \left(2x\right)}{\cos \left(2x\right)} - 2 \cos x = 0$
i.e. $\;$ $\dfrac{2 \sin x \; \cos x}{1 - 2 \sin^2 x} - 2 \cos x = 0$
i.e. $\;$ $\cos x \left(\sin x - 1 + 2 \sin^2 x\right) = 0$
i.e. $\;$ $\cos x \left(2 \sin^2 x + 2 \sin x - \sin x - 1\right) = 0$
i.e. $\;$ $\cos x \left[2 \sin x \left(\sin x + 1\right) - 1 \left(\sin x + 1\right)\right] = 0$
i.e. $\;$ $\cos x \left(2 \sin x - 1\right) \left(\sin x + 1\right) = 0$
i.e. $\;$ $\cos x = 0$ $\;$ OR $\;$ $2 \sin x - 1 = 0$ $\;$ OR $\;$ $\sin x + 1 = 0$
i.e. $\;$ $\cos x = 0$ $\;$ OR $\;$ $\sin x = \dfrac{1}{2}$ $\;$ OR $\;$ $\sin x = - 1$
Now, $\;$ $\cos x = 0$ $\implies$ $x = 2 n \pi \pm \dfrac{\pi}{2}, \;\;\; n \in Z$
OR, $\;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$ $\implies$ $x = m \pi + \left(-1\right)^m \times \left(\dfrac{\pi}{6}\right), \;\;\; m \in Z$
OR, $\;$ $\sin x = - 1 = \sin \left(- \dfrac{\pi}{2}\right)$ $\implies$ $x = p \pi + \left(- 1\right)^p \times \left(- \dfrac{\pi}{2}\right), \;\;\; p \in Z$
In the range $\left[0, 2 \pi \right)$,
when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{2}$
when $\;$ $m = 0$, $\;$ $x = \dfrac{\pi}{6}$
when $\;$ $m = 1$, $\;$ $x = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$
when $\;$ $p = 0$, $x = - \dfrac{\pi}{2}$
$\left[- \dfrac{\pi}{2} \text{ is the same as } 2 \pi - \dfrac{\pi}{2} = \dfrac{3 \pi}{2} \right]$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = \dfrac{\pi}{6}, \; \dfrac{\pi}{2}, \; \dfrac{5 \pi}{6}, \; \dfrac{3 \pi}{2}$