We are being asked to show that the global maximum of ${\displaystyle f(x)=\sin(x)+{\sqrt {3}}\cos(x)}$ is at most equal to ${\displaystyle 2}$ and its global minimum is at least equal to ${\displaystyle -2}$. For this, note that the function is periodic with period ${\displaystyle 2\pi }$ and therefore suffices to find the global maximum and the global minimum on the closed interval ${\displaystyle [0,2\pi ]}$

First determine the critical points in the interval by computing the derivative and setting it equal to 0:

${\displaystyle f'(x)=\cos(x)-{\sqrt {3}}\cdot \sin(x),}$

so ${\displaystyle f'(x)=0}$ implies that ${\displaystyle \tan(x)={\frac {1}{\sqrt {3}}}}$. The only solutions of thid equation in the interval ${\displaystyle (0,2\pi )}$ are ${\displaystyle x={\frac {\pi }{6}}}$ and ${\displaystyle x=\pi +{\frac {\pi }{6}}}$. It remains to check the value of ${\displaystyle f(x)}$ at the endpoints of the interval and at the two critical points:

${\displaystyle f(0)={\sqrt {3}}{\text{, }}f\left({\frac {\pi }{6}}\right)={\frac {1}{2}}+{\sqrt {3}}\cdot {\frac {\sqrt {3}}{2}}=2.}$

${\displaystyle f\left(\pi +{\frac {\pi }{6}}\right)=-{\frac {1}{2}}+{\sqrt {3}}\cdot {\frac {-{\sqrt {3}}}{2}}=-2{\text{ and }}f(2\pi )={\sqrt {3}}.}$

Hence, ${\displaystyle \color {blue}{\text{the global maximum is }}2{\text{ and the global minimum is }}-2}$, as desired.