求函数y=sin^4(x)+2根号3sinxcosx-cos^4(x)的最小正周期和最小值；并写出该函数在

y=sin^4(x)+2根号3sinxcosx-cos^4(x)
=sin^4(x)-cos^4(x)+2根号3sinxcosx
=(sin²x+cos²x)(sin²x-cos²x)+2根号3sinxcosx
=2根号3sinxcosx-cos2x
=√3sin2x-cos2x
=2sin(2x-π/6)

最小正周期π和最小值-2
x∈[0,π]
sin(2x-π/6)的增区间（2x-π/6）∈[(2n-1/2)π,(2n+1/2)π]
x∈[(n-1/6)π,(n+1/3)π]

取交集得x∈[0,π/3]

y=sin^4(x)+2根号3sinxcosx-cos^4(x)
=（sin^2x+cos^2x)(sin^2x-cos^2x)+根号sin2x
=sin^2x-cos^2x+根号3sin2x
=根号3sin2x-cos2x
=2sin(2x-π/3)
最小整周期 π，最小值-2