Question

已知函数f(x)=2根号3sinxcosx-2cos²x+1

Answer 1

f(x)=2√3sinxcosx-2cos^2x+1
=√3*2sinxcosx-(2cos^2x-1)
=√3sin(2x)-cos(2x)
=2sin(2x-π/6)
∴f(x)的最小正周期=2π/2=π
f(α)=2sin(2α-π/6)=
-1
sin(2α-π/6)=
-1/2
2α-π/6=
-π/6+kπ,k∈Z
∵α∈(0,π/2)
∴2α∈(0,π)
2α-π/6∈(-π/6,5π/6)
∴sin(2α-π/6)∈(-1/2,1/2)
而1/2不在区间内
∴不存在α使得f(α)=
-1
本题若有疑问请追问，若理解请采纳，谢谢~~

Answer 2

f(x)=1/2cos2x+根号3/2sin2x-1-cos2x
=√3/2
×sin2x－1/2
×cos2x
－1
＝sin(2x-π/6)
－1
当sin(2x-π/6)＝1时
f(x)有最大值＝0
f(a)=-1/5
则，sin(2a-π/6)-1＝-1/5
sin(2a-π/6)=4/5
所以，cos(2a+π/3)
＝cos(2a-π/6＋π/2)
＝-sin(2a-π/6)
=-4/5
f(a)=-1/5
则，sin(2a-π/6)-1＝-1/5
cos(2a+π/3)=-4/5
π/3
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