Question

已知函数f(X)=根号3cos²x+sinxcosx-2分之根号3,x∈R求最小正周期和f(x)单调区间

Answer 1

解 f(x)=√3cos²x+sinxcosx-√3/2

=√3*(1+cos2x)/2+(1/2)sin2x-√3/2

=(1/2)sin2x+(√3/2)cos2x

=sin(2x+π/3)

∴T=π

单增区间：-π/2+2kπ≤2x+π/3≤π/2+2kπ,k∈Z

-5π/6+2kπ≤2x≤π/6+2kπ,k∈Z

-5π/12+kπ≤x≤π/12+2kπ,k∈Z

即为：[-π/12+kπ,5π/12+kπ],k∈Z