`~rai~`
$y=cosx+cos(x-\dfrac{\pi}{3})$
$\Leftrightarrow y=cosx+cosxcos\dfrac{\pi}{3}+sinxsin\dfrac{\pi}{3}$
$\Leftrightarrow y=cosx+\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx$
$\Leftrightarrow y=\dfrac{3}{2}cosx+\dfrac{\sqrt{3}}{2}sinx$
$\Leftrightarrow y=\sqrt{3}(\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx)$
$\Leftrightarrow y=\sqrt{3}(cos\dfrac{\pi}{6}cosx+sin\dfrac{\pi}{6}sinx)$
$\Leftrightarrow y=\sqrt{3}cos(x-\dfrac{\pi}{6})$
$\text{Ta có:} -1\le cos(x-\dfrac{\pi}{6}) \le 1$
$\Leftrightarrow -\sqrt{3} \le \sqrt{3}cos(x-\dfrac{\pi}{6}) \le \sqrt{3}$
$\Leftrightarrow -\sqrt{3} \le y \le \sqrt{3}$
$\text{Min y=}-\sqrt{3} \Leftrightarrow cos(x-\dfrac{\pi}{6})=-1\Leftrightarrow x-\dfrac{\pi}{6}=\pi+k2\pi\Leftrightarrow x=\dfrac{7\pi}{6}+k2\pi$
$\text{Max y=}\sqrt{3} \Leftrightarrow cos(x-\dfrac{\pi}{6})=1\Leftrightarrow x-\dfrac{\pi}{6}=k2pi\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi$
$\text{Vậy Miny=} -\sqrt{3} \text{khi} x=\dfrac{7\pi}{6}+k2\pi;Maxy=\sqrt{3} \text{khi} x=\dfrac{\pi}{6}+k2\pi(k\in\mathbb{Z}).$
