求函数y=√3cos(3π/2+2x)+cos^2x-sin^2x的周期,当x取何值时,y取最大值,最小值

y=√3cos(3π/2+2x)+cos²x-sin²x
=√3sin(2x)+cos(2x)
=2[√3sin(2x)/2+cos(2x)/2]
=2[sin(2x)cos(π/6)+cos(2x)sin(π/6)]
=2sin(2x+π/6)
T=2π/2=π
y的最小正周期是π
当2x+π/6=2kπ+π/2,即x=kπ+π/6,k∈Z时,y有最大值2
当2x+π/6=2kπ-π/2,即x=kπ-π/3,k∈Z时,y有最大值-2