Đáp án: `S={\frac{11π}{4};\frac{19π}{4};\frac{27π}{4}}`
Giải:
Ta có:
`sin(x-\frac{π}{4})=1`
⇔ `x-\frac{π}{4}=\frac{π}{2}+k2π`
⇔ `x=\frac{3π}{4}+k2π \ (k∈\mathbb{Z})`
Mặt khác:
`π≤x≤7π`
⇔ `π≤\frac{3π}{4}+k2π≤7π`
⇔ `1≤\frac{3}{4}+2k≤7`
⇔ `\frac{1}{4}≤2k≤\frac{25}{4}`
⇔ `\frac{1}{8}≤k≤\frac{25}{8}`
Mà `k∈\mathbb{Z}` nên `k∈{1;2;3}`
*) `k=1 → x=\frac{3π}{4}+2π=\frac{11π}{4}`
*) `k=2 → x=\frac{3π}{4}+4π=\frac{19π}{4}`
*) `k=3 → x=\frac{3π}{4}+6π=\frac{27π}{4}`
Vậy `S={\frac{11π}{4};\frac{19π}{4};\frac{27π}{4}}`
