Question

已知函数f(x)=2根号3sinxcosx+2cos^2x-1,求f(x)的单调递减区间

Answer 1

f(x)=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(√3/2*sin2x+1/2*cos2x)
=2sin(2x+π/6)
由图像得：π/2+2kπ≤2x+π/6≤3π/2+2kπ，k为整数
即π/6+kπ≤x≤4π/3+kπ，k为整数
则单调减区间为[π/6+kπ，4π/3+kπ]，k为整数

Answer 2

f(x)=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
令2x+π/6=z
sinz的单调递减区间为[2kπ+π/2,2kπ+3π/2](k∈Z)
∴2kπ+π/2≤2x+π/6≤2kπ+3π/2
∴kπ+π/6≤x≤kπ+2π/3 (k∈Z)
∴f(x)在区间[kπ+π/6，kπ+2π/3](k∈Z)是单调递减