Đáp án:
$\begin{cases}\min y = -1 \Leftrightarrow x = -\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\\max y = \Leftrightarrow x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\end{cases}\quad(k\in\Bbb Z)$
Giải thích các bước giải:
$y = 3\sin3x + 2$
Ta có:
$\quad -1 \leq \sin3x \leq 1$
$\to -3\leq 3\sin3x \leq 3$
$\to -1 \leq 3\sin3x +2 \leq 5$
$\to -1\leq y \leq 5$
Vậy $\begin{cases}\min y = -1 \Leftrightarrow \sin3x = -1\Leftrightarrow x = -\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\\max y = 5 \Leftrightarrow \sin3x = 1 \Leftrightarrow x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\end{cases}\,\,(k\in\Bbb Z)$
