Đáp án:
$\begin{cases}miny = \dfrac{1}{6} \Leftrightarrow x = - \dfrac{\pi}{12} + k\dfrac{\pi}{3}\\maxy = \dfrac{7}{6} \Leftrightarrow x = \dfrac{\pi}{12} + k\dfrac{\pi}{3}\end{cases}\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$y = \dfrac{2}{3} + sin3xcos3x$
$=\dfrac{2}{3} + \dfrac{2sin3x.cos3x}{2}$
$= \dfrac{2}{3} + \dfrac{1}{2}sin6x$
Ta có: $-1 \leq sin6x \leq 1$
$\Leftrightarrow -\dfrac{1}{2} \leq sin6x \leq \dfrac{1}{2}$
$\Leftrightarrow \dfrac{2}{3} - \dfrac{1}{2} \leq \dfrac{2}{3} + \dfrac{1}{2}sin6x \leq \dfrac{2}{3} + \dfrac{1}{2}$
$\Leftrightarrow \dfrac{1}{6} \leq \dfrac{2}{3} + \dfrac{1}{2}sin6x \leq \dfrac{7}{6}$
Hay $\dfrac{1}{6} \leq y \leq \dfrac{7}{6}$
Vậy $miny = \dfrac{1}{6} \Leftrightarrow sin6x = -1 \Leftrightarrow x = - \dfrac{\pi}{12} + k\dfrac{\pi}{3} \,\,\,(k \in \Bbb Z)$
$maxy = \dfrac{7}{6} \Leftrightarrow sin6x - 1 \Leftrightarrow x = \dfrac{\pi}{12} + k\dfrac{\pi}{3} \,\,\,(k \in \Bbb Z)$
