Đáp án: `x=\frac{π}{7}+k2π` hoặc `x=\frac{8π}{21}+\frac{k2π}{3} \ (k∈\mathbb{Z})`
Giải:
Ta có:
`sin(2x-\frac{π}{7})-sinx=0`
⇔ `sin(2x-\frac{π}{7})=sinx`
⇔ $\left [\begin{array}{l} 2x-\dfrac{π}{7}=x+k2π \\ 2x-\dfrac{π}{7}=π-x+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{π}{7}+k2π \\ 3x=π+\dfrac{π}{7}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{π}{7}+k2π \\ 3x=\dfrac{8π}{7}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{π}{7}+k2π \\ x=\dfrac{8π}{21}+\dfrac{k2π}{3} \end{array} \right. \ (k∈\mathbb{Z})$
Vậy `x=\frac{π}{7}+k2π` hoặc `x=\frac{8π}{21}+\frac{k2π}{3} \ (k∈\mathbb{Z})`
