已知函数f（x）=sin^2x+2根号3sinxcosx+3cos^2x求下详细答案,

答：f(x)=sin²x+2√3sinxcosx+3cos²x
=1+√3sin2x+cos2x+1
=2*[(√3/2)sin2x+(1/2)cos2x]+2
=2sin(2x+π/6)+2
（1）f(a)=2sin(2a+π/6)+2=3
sin(2a+π/6)=1/2
因为：0