y=sin^4x+2√3sinxcosx-cos^4x1,求最小正周期和最小值2,写出该函数在〔0,π〕上的单调递增区间

y=sin^4x+2√3sinxcosx-cos^4x
=sin^4x-cos^4x+√3sin2x
=(sin²x+cos²x)(sin²x-cos²x)+√3sin2x
=sin²x-cos²x+√3sin2x
=-cos2x+√3sin2x
=2(√3/2sin2x-1/2cos2x)
=2sin(2x-π/6)
最小正周期是2π/2=π
最小值是 -2

0≤2x-π/6≤π/2
故π/12≤x≤π/3
故在[0,π]上单调递增区间为：[π/12,π/3]

y=sin^4x+2√3sinxcosx-cos^4x
=(sin^4x-cos^4x)+2√3sinxcosx
=(sin^2x-cos^2x)(sin^2x+cos^2x)+2√3sinxcosx
=(sin^2x-cos^2x)+2√3sinxcosx
=√3sin2x-cos2x
=2sin(2x-π/6)
1.
最小正周期T=2π/2=π 最小值-2
2.
y=2sin(2x-π/6) 的单调递增区间
2kπ-π/2kπ-π/6在〔0，π〕上的单调递增区间 【0，π/3】 和 【5π/6,π】

1、
y=sin^4x+2√3sinxcosx-cos^4x
=(sin^2x+cos^2x)(sin^2x-cos^2x)+2√3sinxcosx
=-cos2x+√3sin2x
=2(cos2π/3cos2x+sin2π/3sin2x)
=2cos(2x-2π/3)
可得：最小正周期为T=2π/2=π
当cos(2x-2π/3)=-1时有最小值为-2
2、 因余弦函数的递增区间为：[2kπ-π,2kπ]
所以有：2kπ-π≤2x-2π/3≤2kπ
解得：kπ-π/6≤x≤kπ+π/3
因此可得该函数在〔0,π〕上的单调递增区间为：(0,π/3]∪[5π/6,π)