Ta có: $y = \sqrt{3}.cosx - sinx$
$\frac{y}{2} = \frac{\sqrt{3}}{2}.cosx - \frac{1}{2}sinx$
$\frac{y}{2} = sin\frac{\pi}{3}.cosx - cos\frac{\pi}{3}.sinx$
$\frac{y}{2} = sin(\frac{\pi}{3} - x)$
$y = 2sin(\frac{\pi}{3} - x)$
Với $x ∈ [0; \frac{\pi}{2}]$ , ta có:
$-\frac{1}{2} ≤ sin(\frac{\pi}{3} - x) ≤ \frac{\sqrt{3}}{2}$
$⇔ -1 ≤ 2sin(\frac{\pi}{3} - x) ≤ \sqrt{3}$
$⇔ -1 ≤ y ≤ \sqrt{3}$
Với $y = -1, x = \frac{\pi}{2} $
Với $y = \sqrt{3}, x = 0 $
