已知f(x)=cos²x+2√3*sinxcosx-sin²x+a(x∈R).(1)求函数(x)的周

y=cos²x+2√3*sinxcosx-sin²x+a
=cos²x-sin²x+2√3*sinxcosx+a
=cos2x+√3sin2x+a
=2(1/2cos2x+√3/2sin2x)+a
=2[sin(π/6)cos2x+cos(π/6)sin2x]+a
=2sin(2x+π/6)+a
得周期T=(2π/2)=π
当x在[ 0,(π/2)]时,2x+π/6在 [ (π/6),(2π/3)]包含 (π/2)
所以当 2x+(π/6)=π/2时
f(x)有最大值为2sin(2x+π/6)+a=2sin(π/2)+a=2*1+a=4
a=2