`2sin(3x-[pi]/3)+sqrt3=0`

`<=> sin(3x-[pi]/3)=-[sqrt3]/2`

`<=> sin(3x-[pi]/3)=sin(-[pi]/3)`

`<=> 3x-[pi]/3=-[pi]/3+k2pi` hoặc `3x-[pi]/3=pi+[pi]/3+k2pi (k in ZZ)`

`<=> x=[k2pi]/3` hoặc `x=[5pi]/9+[k2pi]/3 (k in ZZ)`

Đề:

`-[pi]/2<x<[pi]/2`

Xét `x=[k2pi]/3`:

`-[pi]/2<[k2pi]/3<[pi]/2`

`<=> -[3pi]/2<k2pi<[3pi]/2`

`<=> -3/4<k<3/4`

`<=> k=0 -> x=0`

Xét `x=[5pi]/9+[k2pi]/3`:

`-[pi]/2<[5pi]/9+[k2pi]/3<[pi]/2`

`<=> -[19pi]/18<[k2pi]/3<-[pi]/18`

`<=> -[19pi]/6<k2pi<-[pi]/6`

`<=> -19/12<k<-1/12`

`<=> k=-1 -> x=-[pi]/9`

Vậy có 2 nghiệm thuộc khoảng `(-[pi]/2; [pi]/2)`: `x=0`, `x=-[pi]/9`