Question

已知f(x)=根号3sinxcosx-cos^2x-1/2 求的函数f(x)最小正周期,单调增区间

Answer 1

f(x)=√3sinxcosx+cos^2x-1/2
=√亩带改3sin2x/2+(1+cos2x)/2-1/2
=sin(2x+π/6)
所以行判,最小正周期=2π/2=π
2x+π/迅判6∈[2kπ-π/2,2kπ+π/2](k∈Z)
单调增区间为[kπ-π/3,kπ+π/6](k∈Z)