2sin(2x+π/6)设X∈（0,π/4）,求函数f(x)的值域要过程x∈（0，π/4），所以2x∈（0,π/2)所以2x+π/6∈（π/6,2π/3)所以sin(2x+π/6)∈(1/2,1]所以2sin(2x+π/6)∈(1,2]所以f(x)的值域为(1,2]请问所以2x+π/6∈（π/6,2

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)=2sin(2x-π/6)在x∈【0,π/2】上两个零点,则m的取值范围f(x)=2sin(2x-π/6)-m 设x∈(0,π/2),则函数(sin²x+1/sin²)(cosx²+1/cos²x)的最小值是 设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为 设函数f(x)=sin(2wx+π/6)+1/2,（其中0 设X∈(0,3/π）,闭区间,求函数y=cos(2x-3/π)+2sin(x-6/π)的最值 f(x)=2sin(2x-π/3)+3,设0