Đáp án:
$\begin{cases}\max y = \sqrt3 \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi\\ \min y =- \sqrt3 \Leftrightarrow x = \dfrac{7\pi}{6} + k2\pi\end{cases}\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\quad y = \cos x + \cos\left(x - \dfrac{\pi}{3}\right)$
$\to y = \cos x + \dfrac12\cos x + \dfrac{\sqrt3}{2}\sin x$
$\to y = \dfrac32\cos x + \dfrac{\sqrt3}{2}\sin x$
$\to y = \sqrt3\left(\dfrac{\sqrt3}{2}\cos x +\dfrac12\sin x\right)$
$\to y = \sqrt3\cos\left(x - \dfrac{\pi}{6}\right)$
Ta có:
$\quad -1 \leqslant \cos\left(x - \dfrac{\pi}{6}\right) \leqslant 1$
$\to - \sqrt3 \leqslant \sqrt3\cos\left(x - \dfrac{\pi}{6}\right) \leqslant \sqrt3$
$\to -\sqrt3 \leqslant y \leqslant \sqrt3$
Do đó:
$+)\quad \max y = \sqrt3 \Leftrightarrow \cos\left(x - \dfrac{\pi}{6}\right)= 1 \Leftrightarrow x = \dfrac{\pi}{6} + k2\pi\quad (k\in\Bbb Z)$
$+)\quad \min y =- \sqrt3 \Leftrightarrow \cos\left(x - \dfrac{\pi}{6}\right)= 1 \Leftrightarrow x = \dfrac{7\pi}{6} + k2\pi\quad (k\in\Bbb Z)$
