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