Ta có $\sin 2x=1 \Leftrightarrow 2x= \frac{\pi}{2}+k2\pi\Leftrightarrow x=\frac{\pi}{4}+k\pi,(k\in Z)$Theo giả thiết nghiệm $x\in [\frac{\pi}{2};5\pi]$ nên:$\frac{\pi}{2}\leq x\leq 5\pi$$\Leftrightarrow\frac{\pi}{2}\leq \frac{\pi}{4}+ k\pi\leq 5\pi $$\Leftrightarrow \frac{\pi}{4}\leq k\pi\leq \frac{19\pi}{4}$$\Leftrightarrow \frac{1}{4}\leq k\leq \frac{19}{4}$$\Leftrightarrow k=1;2;3;4 (k\in Z)$$\Leftrightarrow x\in \left \{ \frac{5\pi}{4};\frac{9\pi}{4};\frac{13\pi}{4};\frac{17\pi}{4}\left. \right \} \right.$
Ta có $\sin 2x=1 \Leftrightarrow 2x= \frac{\pi}{2}+k2\pi\Leftrightarrow x=\frac{\pi}{4}+k\pi,(k\in Z)$Theo giả thiết nghiệm $x\in [\frac{\pi}{2};5\pi]$ nên:$\frac{\pi}{2}\leq x\leq 5\pi$$\Leftrightarrow\frac{\pi}{2}\leq \frac{\pi}{4}+ k\pi\leq 5\pi $$\Leftrightarrow \frac{\pi}{4}\leq k\pi\leq \frac{19\pi}{4}$$\Leftrightarrow \frac{1}{4}\leq k\leq \frac{19}{4}$$\Leftrightarrow k=1;2;3;4 (k\in Z)$$\Leftrightarrow x\in \left \{ \frac{5\pi}{4};\frac{9\pi}{4};\frac{13\pi}{4};\frac{17\pi}{4}\left. x \right \} \right.$
Ta có $\sin 2x=1 \Leftrightarrow 2x= \frac{\pi}{2}+k2\pi\Leftrightarrow x=\frac{\pi}{4}+k\pi,(k\in Z)$Theo giả thiết nghiệm $x\in [\frac{\pi}{2};5\pi]$ nên:$\frac{\pi}{2}\leq x\leq 5\pi$$\Leftrightarrow\frac{\pi}{2}\leq \frac{\pi}{4}+ k\pi\leq 5\pi $$\Leftrightarrow \frac{\pi}{4}\leq k\pi\leq \frac{19\pi}{4}$$\Leftrightarrow \frac{1}{4}\leq k\leq \frac{19}{4}$$\Leftrightarrow k=1;2;3;4 (k\in Z)$$\Leftrightarrow x\in \left \{ \frac{5\pi}{4};\frac{9\pi}{4};\frac{13\pi}{4};\frac{17\pi}{4}\left. \right \} \right.$