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：最小正周期為T=2π/2=π
當(dāng)cos(2x-2π/3)=-1時(shí)有最小值為-2
2、 因余弦函數(shù)的遞增區(qū)間為：[2kπ-π,2kπ]
所以有：2kπ-π≤2x-2π/3≤2kπ
解得：kπ-π/6≤x≤kπ+π/3
因此可得該函數(shù)在〔0,π〕上的單調(diào)遞增區(qū)間為：(0,π/3]∪[5π/6,π)