设函数f（x）=sin(2x-π/6)-m在区间【0,π/2】上有两个零点则m 【1/2,1）.设函数f（x）=sin(2x-π/6)-m在区间【0,π/2】上有两个,则m 【1/2,1）

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)=sinπ/6（x),则f(1)+f(2)+f(3)+…f(2008)=? 设函数f(x)=sinπ/6（x),则f(1)+f(2)+f(3)+…f(2008)=? 设函数f（x）=sin（2x+φ）（-π 设函数f(x)=sin(2x+φ）（-π