f(x)=1+2√3sinxcosx+2cos^2x1.f(x)的最小正周期2.f(x)的单调递减区间

问题描述:

f(x)=1+2√3sinxcosx+2cos^2x
1.f(x)的最小正周期
2.f(x)的单调递减区间

f(x)=1+2√3sinxcosx+2cos^2x
=1+√3sin2x+2cos^2x
=1+2sin(2x+π/6)
所以f(x)的最小正周期T=2π/2=π
2kπ+π/2<=2x+π/6<=2kπ+3π/2
解得kπ+π/6<=x<=kπ+2π/3
所以f(x)的单调递减区间[kπ+π/6 ,kπ+2π/3] k属于整数