f (x)=根号3⼀2sinxcosx- 3⼀2sin^2x+3⼀4,求单调递增区间

2025-12-14 16:09:15
推荐回答(1个)
回答1:

f (x)=√3/2sinxcosx- 3/2sin^2x+3/4
=√3/2sinxcosx- 3/2(1-cos2x)/2+3/4
=√3/2sinxcosx- 3/4(1-cos2x)+3/4
=√3/2sinxcosx- 3/4+3/4cos2x+3/4
=√3/4sin2x+3/4cos2x
=√3/2(1/2sin2x+√3/2cos2x)
=√3/2(sin2xcosπ/3+cos2xsinπ/3)
=√3/2sin(2x+π/3)
2kπ-π/2<2x+π/3<2kπ+π/2,单调递增区间
2kπ-5π/6<2x<2kπ+π/6
kπ-5π/12