设f(x)=√3sin2wx+1,其中w>0.

f(x)=根号3sin2wx+cos2wx = 2(cosπ/6 * sin2wx + sinπ/6 * cos2wx) = 2sin(2wx + π/6)
(1)
f(x)的图像中相邻的两条对称轴间的距离不小于π/2 那么 T/2 >= π/2
而 T = 2π/2w = π/w
所以有
π/w >= π/2
=>
w

1.
向量m=(1,2cosωx),n=(√3sin2ωx,-cosωx)
f(x)=mn=√3sin2ωx-2(cosωx)^2
=√3sin2ωx-cos2ωx-1
=2sin(2ωx-π/6)-1
图像相邻2对称轴距离π/2
说明T/2=π/2
故T=π
所以T=2π/2ω=π
那么ω=1
2.
由1知道f(x)=2sin(2x-π/6)-1
x∈[π/4,π/2]
2x-π/6∈[π/3,5π/6]
sin(2x-π/6)∈[1/2,1]
所以f(x)∈[0,1]
所以f(x)的最大值是1,最小值是0.
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f(x+π/2+π/2)=-f(x+π/2)=-(-f(x))=f(x),所以f(x)周期为π,所以2π/2w=π,所以w=1

f(x)=√3/2(1-coswx)+1/2sinwx=1/2sinwx-√3/2coswx+√3/2=sin(wx-π/3)+√3/2
因为周期为π,所以w=2
所以,f(x)=sin(2x-π/3)+√3/2
单调增区间：2kπ-π/2

1.f(x)=mn=√3sin2wx-2(coswx)^2=√3sin2wx-(1+cos2wx)=2(√3/2sin2wx-1/2cos2wx)-1
=2sin(2wx-π/6)-1
对称轴 2wx--π/6=kπ+π/2 x=kπ/2w+1/3π
相邻两条对称轴距 π/2w=π/2 w=1
2.f(x)=2sin(2x-π/6)-1
π/4

[1+cos(2ωx)]/2+[(√3)sin(2ωx)]/2,
原式=(1/2)+(1/2)cos(2ωx)+(√3/2)sin(2ωx)
=(1/2)+cos(2ωx)cos(π/3)+sin(2ωx)sin(π/3)
=cos(2ωx-π/3)+1/2.

(x)=√3sin2wx+2cos²wx =√3sin2wx+(2cos²wx-1)+1 =√3sin2wx+cos2wx+1 =2(√3/2sin2wx+1/2cos2wx)+1 =2(sin2wxcosπ/6+cos2wxsinπ/6)+1 =2sin(2wx+π/6)+1∵T=2π/2w=π∴w=1
x∈[0.π/2]∴2x+π/6∈[π/6,7π/6]∴sin(2x+π/6)∈[-1/2,1]∴f(x)∈[0,3]
f(X)=2sin2x向左移π/6

2倍根号3coswxsinwx
=2√3coswxsinwx
=√3(2coswxsinwx)
=√3(sin2wx)