Question

已知函数f(x)=-1+2根号3sinxcosx+2cos平方x 求f(x)最小正周期 单凋减区间

Answer 1

f(x)=-1+2√3sinxcosx+2cos²x
=√3(2sinxcosx)+(2cos²x-1)
=√3sin(2x)+cos(2x)
=2[(√3/2)sin(2x)+(1/2)cos(2x)]
=2sin(2x+π/6)
最小正周期：T=2π/2=π
2kπ+π/2≤2x+π/6≤2kπ+3π/2 (k∈Z)时，f(x)单调递减，此时kπ+π/6≤x≤kπ+2π/3 (k∈Z)
函数的单调递减区间为[kπ+π/6，kπ+2π/3] (k∈Z)

Answer 2

f(x)=-1+2根号3sinxcosx+2cos平方x
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
所以最小正周期是2π/2=π
单减区间
2kπ+π/2≤2x+π/6≤2kπ+3π/2
2kπ+π/3≤2x≤2kπ+4π/3
kπ+π/6≤x≤kπ+2π/3