Đáp án:
$y\in \left[-\sqrt 3;\sqrt 3\right]$
Giải thích các bước giải:
$y=\cos2x+\cos\left(2x-\dfrac{\pi}{3}\right)$
$=2\cos\left(\dfrac{2x+2x-\dfrac{\pi}{3}}{2}\right).\cos\left(\dfrac{2x-2x+\dfrac{\pi}{3}}{2}\right)$
$=2\cos\left(2x-\dfrac{\pi}{6}\right).\cos\left(\dfrac{\pi}{6}\right)$
$=2.\dfrac{\sqrt{3}}{2}.\cos\left(2x-\dfrac{\pi}{6}\right)$
$=\sqrt{3}\cos\left(2x-\dfrac{\pi}{6}\right)$
Ta có:
$-1\le \cos\left(2x-\dfrac{\pi}{6}\right)\le 1$
$⇔-\sqrt 3\le \sqrt{3}\cos\left(2x-\dfrac{\pi}{6}\right)\le \sqrt 3$
$⇔-\sqrt 3\le y\le \sqrt 3$
Vậy $y\in \left[-\sqrt 3;\sqrt 3\right]$.
