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f(x)=sin(wx+g)
f(8/3)=1,8w/3+g=2kπ+π/2,k∈Z。。。。(1)
f(14/3)=0,14w/3+g=2kπ+π/2±π/2,k∈Z。。。。(2)
(1)-(2):2w=±π/2,w>0,所以w=π/4
代入(1):g=2kπ+π/2-2π/3,|g|<π/2,所以g=-π/6
f(x)=sin(πx/4-π/6)
g(x)=f(x)-2cos^2(πx/8)+1,x∈[2/3,2]
=sin(πx/4-π/6)-[2cos^2(πx/8)-1]
=sin(πx/4-π/6)-cos[2(πx/8)]
=sin(πx/4-π/6)-cos(πx/4)
=sin(πx/4)cos(π/6)-cos(πx/4)sin(π/6)-cos(πx/4)
=√3sin(πx/4)/2-3cos(πx/4)/2
=√3[sin(πx/4)/2-√3cos(πx/4)/2]
=√3[sin(πx/4)cos(π/3)-cos(πx/4)sin(π/3)]
=√3sin(πx/4-π/3)
x∈[2/3,2],则πx/4-π/3∈[-π/6,π/6]
所以g(x)=√3sin(πx/4-π/3)的值域为g(x)∈[-√3/2,√3/2]
f(8/3)=1,8w/3+g=2kπ+π/2,k∈Z。。。。(1)
f(14/3)=0,14w/3+g=2kπ+π/2±π/2,k∈Z。。。。(2)
(1)-(2):2w=±π/2,w>0,所以w=π/4
代入(1):g=2kπ+π/2-2π/3,|g|<π/2,所以g=-π/6
f(x)=sin(πx/4-π/6)
g(x)=f(x)-2cos^2(πx/8)+1,x∈[2/3,2]
=sin(πx/4-π/6)-[2cos^2(πx/8)-1]
=sin(πx/4-π/6)-cos[2(πx/8)]
=sin(πx/4-π/6)-cos(πx/4)
=sin(πx/4)cos(π/6)-cos(πx/4)sin(π/6)-cos(πx/4)
=√3sin(πx/4)/2-3cos(πx/4)/2
=√3[sin(πx/4)/2-√3cos(πx/4)/2]
=√3[sin(πx/4)cos(π/3)-cos(πx/4)sin(π/3)]
=√3sin(πx/4-π/3)
x∈[2/3,2],则πx/4-π/3∈[-π/6,π/6]
所以g(x)=√3sin(πx/4-π/3)的值域为g(x)∈[-√3/2,√3/2]
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