Question

已知函数f（x）=sinxcosx-√3cos^2x2-√3/2 (X∈R)

1.求f(x)的周期
2.求f（x）最小值及f（x）取最小值时自变量x的集合
3.求f（x）的单调增区间

Answer 1

f(x)=sinxcosx-√3cos²x-√3/2
=(1/2)*sin2x-(√3/2)cos2x-√3
=sin(2x-π/3)-√3
(1)
f(x)的最小正周期是T=2π/2=π
(2)
f(x)的最小值是-1-√3
当2x-π/3=2kπ-π/2,k∈Z
即x=kπ-π/12,k∈Z
所以取最小值时自变量x的集合是{x|x=kπ-π/12,k∈Z}
(3)
f（x）的单调增区间:
令2kπ-π/2＜2x-π/3＜2kπ+π/2,k∈Z
kπ-π/12＜x＜kπ+5π/12,k∈Z
所以f(x)的单调增区间是(kπ-π/12,kπ+5π/12),k∈Z