Dada a expressão: $$y = \sin{2x} - \sqrt{3}\cos{2x}$$Então,\begin{matrix} \dfrac{y}{2} = \dfrac{1}{2} \sin{2x} - \dfrac{\sqrt{3}}{2} \cos{2x} \end{matrix}O que equivale escrever:\begin{matrix} \dfrac{y}{2} = \cos{\left(\dfrac{\pi}{3}\right)}\sin{2x} - \sin{\left(\dfrac{\pi}{3}\right)} \cos{2x} \end{matrix}Confrome o seno da soma, segue:\begin{matrix} y = 2 \sin{\left(2x - \dfrac{\pi}{3}\right)} &,& y_{máx}\ \rightarrow \ \sin{\left(2x - \dfrac{\pi}{3}\right)} = 1 \end{matrix}Continuando,\begin{matrix} 2x - \dfrac{\pi}{3}= \dfrac{\pi}{2} + 2n\pi \end{matrix}Portanto,\begin{matrix} x = \dfrac{5\pi}{12} + n\pi &,& n \in \mathbb{Z} &\tiny{\blacksquare} \end{matrix}