Đáp án: $x\in (\dfrac23\pi+k\pi,\dfrac76\pi+k\pi)$
Giải thích các bước giải:
Ta có:
$y=\cos(2x-\dfrac{\pi}3)$
$\to y'=(\cos(2x-\dfrac{\pi}3))'$
$\to y'=-2\sin(2x-\dfrac{\pi}3)$
$\to y'>0$
$\to -2\sin(2x-\dfrac{\pi}3)>0$
$\to \sin(2x-\dfrac{\pi}3)<0$
$\to \pi+k2\pi<2x-\dfrac{\pi}3<2\pi+k2\pi$
$\to \dfrac43\pi+k2\pi<2x<\dfrac73\pi+k2\pi$
$\to \dfrac23\pi+k\pi<x<\dfrac76\pi+k\pi$
$\to$Hàm số đồng biến khi $x\in (\dfrac23\pi+k\pi,\dfrac76\pi+k\pi)$
