Question

求f(x)=-cos²x+根号3sinxcosx的最小正周期和单调增区间

Answer 1

f(x)=-cos²x+√3sinxcosx
=-1/2(1+cos2x)+√3/2sin2x
=√3/2sin2x-1/2cos2x-1/2
=sin(2x-π/6)-1/2
f(x)
最小正周期
T=2π/2=π
由2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z
得kπ-π/6≤x≤kπ+π/3,k∈Z
f(x)递增区间为[kπ-π/6,kπ+π/3],k∈Z
由2kπ+π/2≤2x-π/6≤2kπ+3π/2,k∈Z
得kπ+π/3≤x≤kπ+5π/6,k∈Z
f(x)递减区间为[kπ+π/3,kπ+5π/6],k∈Z

Answer 2

f(x)=1/2-1/2cos2x+√3/2(2sinxcosx)=1/2-1/2cos2x+√3/2sin2x=1/2+(√3/2sin2x-1/2cos2x)=1/2+(sin2xcosπ/6-cos2xsinπ/6)=1/2+sin(2x-π/6),所以最小正周期为2π/2=π
f(x)的单调增区间为-π/2+2kπ<2x-π/6<π/2+2kπ,即-π/6+kπ
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