设函数f(x)=2sin(πx/2+π/5),若对任意x∈R都有f(x1)≤f(x)≤f(x2)成立,则|x1-x2|的最小值为

设函数 f(x)=sin(2x+y),(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f x=SIN(2X+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+ φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+ φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(wx+t)(-π/2 设函数f(x)=sin(wx+t)(-π/2 设函数f (x)=cos(2x-π/3)-2sin平方x (1)求函数f(x 设函数f(x)=sin(3x)+|sin(3x)|,函数的最小正周期为什么是2π? 设函数f(x)=sin(πx/2+π/4)x 设关于X的函数f(x)=sin(2x+a)(-π