Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\sqrt{2} \; \cos \left(x\right) - \sqrt{2} \; \sin \left(x\right) = 1$
$\sqrt{2} \; \cos \left(x\right) - \sqrt{2} \; \sin \left(x\right) = 1$
i.e. $\;$ $\cos x - \sin x = \dfrac{1}{\sqrt{2}}$
i.e. $\;$ $\dfrac{1}{\sqrt{2}} \; \cos x - \dfrac{1}{\sqrt{2}} \; \sin x = \dfrac{1}{2}$
i.e. $\;$ $\cos \left(\dfrac{\pi}{4}\right) \; \cos x - \sin \left(\dfrac{\pi}{4}\right) \; \sin x = \dfrac{1}{2}$
i.e. $\;$ $\cos \left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{2} = \cos \left(\dfrac{\pi}{3}\right)$
i.e. $\;$ $x + \dfrac{\pi}{4} = 2 n \pi \pm \dfrac{\pi}{3}$
i.e. $\;$ $x = 2 n \pi + \dfrac{\pi}{3} - \dfrac{\pi}{4}$ $\;$ OR $\;$ $x = 2 n \pi - \dfrac{\pi}{3} - \dfrac{\pi}{4}$
i.e. $\;$ $x = 2 n \pi + \dfrac{\pi}{12}, \;\;\; n \in Z$ $\;$ OR $\;$ $x = 2 n \pi - \dfrac{7 \pi}{12}, \;\;\; n \in Z$
In the range $\left[0, 2 \pi \right)$,
when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{12}$ $\;$ OR $\;$ $x = - \dfrac{7 \pi}{12}$ $\;\;$ $\left[- \dfrac{7\pi}{12} \text{ is the same as } 2 \pi - \dfrac{7\pi}{12} = \dfrac{17 \pi}{12} \right]$
when $\;$ $n = 1$, $\;$ $x = 2 \pi + \dfrac{\pi}{12} = \dfrac{25 \pi}{12} \notin \left[0, 2 \pi \right)$ $\;$ OR $\;$ $x = 2 \pi - \dfrac{7 \pi}{12} = \dfrac{17 \pi}{12}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = \dfrac{\pi}{12}, \; \dfrac{17 \pi}{12}$