漫天飞雪7808
幼苗
共回答了14个问题采纳率:100% 举报
f(x)=2sinxcosx-cos(2x+π/6)
f(x)=sin2x-cos2xcos(π/6)+sin2xsin(π/6)
f(x)=sin2x-(√3/2)cos2x+(1/2)sin2x
f(x)=(3/2)sin2x-(√3/2)cos2x
f'(x)=3cos2x+√3sin2x
令:f'(x)=0
即:3cos2x+√3sin2x=0
3cos2x=-√[3-3(cos2x)^2]
9(cos2x)^2=3-3(cos2x)^2
12(cos2x)^2=3
(cos2x)^2=1/4
cos2x=±1/2
x=[arccos(±1/2)]/2
x=±π/6
因为:x∈[0,2π/3]
所以:x=π/6
f(π/6)=2sin(π/6)cos(π/6)-cos(2×π/6+π/6)
f(π/6)=2×(1/2)×(√3/2)-cos(π/2)
f(π/6)=√3/2
当x=π/6时,f(x)有最大值,f(x)最大=√3/2
1年前
10