Solve the equation $\;$ $\sin \left(2x\right) = \cos x$ $\;$ giving the exact solutions which lie in $\left[0, 2\pi \right)$.
$\sin \left(2x\right) = \cos x$
i.e. $\;$ $2 \sin x \cdot \cos x - \cos x = 0$
i.e. $\;$ $\cos x \left(2 \sin x - 1\right) = 0$
$\implies$ $\cos x = 0$ $\;$ OR $\;$ $2 \sin x - 1 = 0$
Now, $\cos x = 0 = \cos \left(\dfrac{\pi}{2}\right)$
$\left\{\text{Principal value of } \cos x \text{ lies in } \left[0, \pi\right] \text{ i.e. in first or second quadrants} \right\}$
$\therefore \;$ $x = 2 n \pi \pm \dfrac{\pi}{2}, \;\;\; n \in Z$
$\therefore \;$ In the range $\left[0, 2\pi \right)$,
when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{2}$
when $\;$ $n = 1$, $\;$ $x = 2 \pi - \dfrac{\pi}{2} = \dfrac{3 \pi}{2}$
Now, $2 \sin x - 1 = 0$ $\implies$ $\sin x = \dfrac{1}{2}$ $\;\;\;$ [i.e. $\sin x$ is positive]
$\left\{\text{Principal value of } \sin x \text{ lies in } \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \text{ i.e. in first or fourth quadrants} \right\}$
$\therefore \;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$
$\implies$ $x = m \pi + \left(- 1\right)^m \times \dfrac{\pi}{6}, \;\;\; m \in Z$
$\therefore \;$ In the range $\left[0, 2\pi \right)$,
when $\;$ $m = 0$, $\;$ $x = \dfrac{\pi}{6}$
when $\;$ $m = 1$, $\;$ $x = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = \dfrac{\pi}{6}, \; \dfrac{\pi}{2}, \; \dfrac{5 \pi}{6}, \; \dfrac{3 \pi}{2}$