Đáp án:
$\left[\begin{array}{l}x = \dfrac{5\pi}{4}\\x = \dfrac{9\pi}{4}\end{array}\right.$
Giải thích các bước giải:
$\sin2x = 1$
$\Leftrightarrow x = \dfrac{\pi}{4} + k\pi\quad (k\in\Bbb Z)$
Ta có:
$\pi \leq x \leq \dfrac{5\pi}{2}$
$\Leftrightarrow \pi \leq \dfrac{\pi}{4} + k\pi \leq \dfrac{5\pi}{2}$
$\Leftrightarrow \dfrac{3}{4}\leq k \leq \dfrac{9}{4}$
Do $k \in \Bbb Z$
nên $k = \left\{1;2\right\}$
$\Rightarrow \left[\begin{array}{l}x = \dfrac{5\pi}{4}\\x = \dfrac{9\pi}{4}\end{array}\right.$
