函数y=(根号3)Sin2x+Cos2x的最大值与最小值,并指出相应的x取值?

y=√3sin2x+cos2x
=2(√3/2*sin2x+1/2*cos2x)
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6)
最大值=2
此时2x+π/6=π/2+2kπ,k∈Z
x=π/6+kπ,k∈Z
最小值=-2
此时2x+π/6=3π/2+2kπ,k∈Z
x=2π/3+kπ,k∈Z