f(x)=2√3sinxcosx+2cos²x-1
=√3sin(2x)+cos(2x)
=2[(√3/2)sin(2x)+(1/2)cos(2x)]
=2sin(2x+π/6)
x1∈[π/4，π/2]
2π/3≤2x1+π/6≤7π/6 cos(2x1+π/6)<0
f(x1)=2sin(2x1+π/6)=6/5
sin(2x1+π/6)=3/5
cos(2x1+π/6)=-√[1-sin²(2x1+π/6)]=-4/5
cos(2x1)=cos(2x1+π/6-π/6)
=cos(2x1+π/6)cos(π/6)+sin(2x1+π/6)sin(π/6)
=(-4/5)(√3/2)+(3/5)(1/2)
=(3-4√3)/10
f(x)=2根号3sinxcosx+2cos²x-1
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已知函数f(x)=2根号3sinxcosx+2cos^2x-1,求f(x)的单调递减区间
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