设f(x)=sin(2x+ π/6 )+2msinxcosx,x∈R．当m=0时,求f（x）在[0,π/3]内的最小值及相应的x的值；

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)=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f(x)={sinπx(x 设f(x)={sinπx(x 设 f(x)= {sinπx(x 设f(x)={sinπx(x