已知函数f(x)=根号3sinxcosx-cos^2x求f(x)的最小正周期和单调递增区间
问题描述：
已知函数f(x)=根号3sinxcosx-cos^2x
求f(x)的最小正周期和单调递增区间
答
f(x)=√3sinxcosx-cos²x
=√3/2(2sinxcosx)-1/2(2cos²x-1)-1/2
=√3/2sin2x-1/2cos2x-1/2
=cosπ/3sin2x-sinπ/3cos2x-1/2
=sin(2x-π/3)-1/2
可得最小正周期T=2π/|2|=π
因sinx在[2kπ-π/2,2kπ+π/2]上单调递增，可得：
2kπ-π/2≤2x-π/3≤2kπ+π/2
解得函数单调递增区间为：[kπ-π/12,kπ+5π/12]
答
∵f(x)=√3sinxcosx-cos^2x
=(√3/2)sin2x-(1/2)(cos2x+1)
=(√3/2)sin2x-(1/2)cos2x-(1/2)
=sin(2x-π/6)-1/2
∴f(x)的最小正周期T=2π/2=π,
∵sinx在[2kπ-π/2,2kπ+π/2]上单调递增,
∴2kπ-π/2≤2x-π/6≤2kπ+π/2
2kπ-π/3≤2x≤2kπ+2π/3
kπ-π/6≤x≤kπ+2π/6
即f(x)的单调递增区间是[ kπ-π/6,kπ+2π/6] （k∈z)
楼主注意：我是正确答案.