Giải giúp em với mn ạ

Đáp án:

 `A. 0`

Giải thích các bước giải:

`sin2x=-1/2`

`<=>`\(\left[ \begin{array}{l}2x=-\dfrac{\pi}{6}+k2\pi\\2x=\pi+\dfrac{\pi}{6}+k2\pi\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{array} \right.\)`(kinZZ)`

Do `x in(0;pi/2)=>`\(\left[ \begin{array}{l}0<-\dfrac{\pi}{12}+k\pi<\dfrac{\pi}{2}\\0<\dfrac{7\pi}{12}+k\pi<\dfrac{\pi}{2}\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}\dfrac{1}{12}<k<\dfrac{7}{12}\\-\dfrac{7}{12}<k<-\dfrac{1}{12}\end{array} \right.\)`=>k in∅ `

`toA`