f(x)=sin2x+2sin(π/4-x)cos(π/4-x) =sin2x+sin[2(π/4-x)] =sin2x+sin(π/2-2x)] =sin2x+cos2x =√2πsin(2x＋π/4)∴f(x)的最小正周期T＝2π/2＝π-π/12≤x≤π/2-π/6≤2x≤ππ/12≤2x+π/4≤...-π/2≤x≤π/2-π≤2x≤π-3π/4≤2x+π/4≤5π/4-1≤sin(2x+π/4)≤1值域[-√2,√2]這后面的我沒看懂！我修改了下哦 你題目里面-π/12≤x≤π/2這樣的 不好意思-π/6≤2x≤ππ/12≤2x+π/4≤5π/4這里都是好理解的-√2/2≤sin(2x+π/4)≤1主要是通過圖像來看的所以值域[-1,√2]可是為什么-π/6≤2x≤π就到了π/12≤2x+π/4≤5π/4呢？這是怎么算的?。?π/6≤2x≤π-π/6+π/4≤2x+π/4≤π+π/4π/12≤2x+π/4≤5π/4