Đáp án:
`S={\frac{2kπ}{5}, π+2kπ-2 | k∈\mathbb{Z}}`
Giải:
`sin(2x-1)+sin(3x+1)=0`
⇔ `2sin\frac{[(2x-1)+(3x+1)]}{2}cos\frac{[(2x-1)-(3x+1)]}{2}=0`
⇔ `2sin\frac{5x}{2}cos\frac{-x-2}{2}=0`
⇔ `2sin\frac{5x}{2}cos\frac{x+2}{2}=0`
⇔ $\left [\begin{array}{l} sin\dfrac{5x}{2}=0 \\ cos\dfrac{x+2}{2}=0 \end{array} \right.$
⇔ $\left [\begin{array}{l} \dfrac{5x}{2}=kπ \\ \dfrac{x+2}{2}=\dfrac{π}{2}+kπ \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{2kπ}{5} \\ x=π+k2π-2 \end{array} \right. \ (k∈\mathbb{Z})$
Vậy `S={\frac{2kπ}{5}, π+2kπ-2 | k∈\mathbb{Z}}`
