设函数f(x)=sin^2x+根号3sinxcosx+3/2.若abc分别是△ABC的内角ABC所对的边a=2根号3,c=4,A为锐角,且f（A）是函数f(x)的在[0,π/2]最大值,则b=?f(x)怎么化简啊?我只化到f(x)=1-cos^2x+（根号3/2）sin2x+（3/2）就不

设函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,x属于[0,π/2],求f(x)的值域 已知函数f(x)=2根号3sin平方x-sin（2x-π/3) 设函数f(x)=cosωx(根号3×sinωx+cosωx)其中0 设函数f(x)=-x^2+x+7,若不等式f(3+2sinθ) 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+∮)(-兀 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)(0 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