求函数y=sinx^4+2*根号3*sinxcosx-cosx^4最小正周期和最小值,并写出该函数在[0,π]上的单调区

y=sinx^4-cosx^4+2*√3*sinxcosx
=(sinx^2+cosx^2)(sinx^2-cosx^2)+√3sin2x
=√3sin2x+(sinx^2-cosx^2)
=√3sin2x-cos2x
=2sin(2x-π/6)
最小正周期 2π/2=π
最小值 -2
y=2sin(2x-π/6)的单调增区间[kπ-π/6,kπ+π/3]
单调减区间[kπ+π/3,kπ+5π/6]
故该函数在[0,π]上的单调增区间[0,π/3]和[5π/6,π]
单调减区间[π/3,5π/6]

y=(sinx^4 -cosx^4)+2√3*sinxcosx
=(sin^2 x -cos^2 x)(sin^2 x +cos^2 x)+√3sin2x
=(sin^2 x -cos^2 x)*1+√3sin2x
=cos2x +√3sin2x
=2[(1/2)*cos2x +(√3/2)*sin2x]
=2*[sin(π/6)*cos2x +cos(π...