Ta có: $\sin x \in [-1;1]$
$⇔ -1 \le \sin x \le 1$
$⇔ -1 \le \sin (2x+1) \le 1$
$⇔ -2 \le 2 \sin (2x+1) \le 2$
$⇔ 5 \ge 3-2 \sin (2x+1) \ge 1$
$\max = 5$ khi $2x+1 = -\dfrac{\pi}{2}+k2\pi ⇔ x=-\dfrac{3\pi}{4}+k\pi \ (k \in \mathbb{Z})$
$\min = 1$ khi $2x+1 = \dfrac{\pi}{2}+k2\pi ⇔ x = -\dfrac{\pi}{4}+k\pi \ (k \in \mathbb{Z})$
