设函数f(x)=2sinxcosx-cos(2x+π/6）当x属于[0,2/3π]时,求f(x)的最大值及此时x的值
3个回答
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
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