已知函数fx=sin(x-兀/3)+根号3cos(x-兀/3)

已知函数fx=sin(x-兀/3)+根号3cos(x-兀/3)

fx=sin(x-π/3)+√3cos(x-π/3)
=sinx*1/2-cosx*√3/2+√3*cosx*1/2+√3*sinx*√3/2
=2sinx
f(x)的增区间是[0,π/2]和,3π/2，2π]
(2)
gx=(1+sinx)fx
=2sinx+2sin²x
=2sin²x+2sinx
=2(sinx+1/2)²-1/2
∵-1<=sinx<=1
∴g(x)最大值=2*(3/2)²-1/2=4
最小值=0-1/2=-1/2
∴值域是[-1/2,4]
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f(x)=sin(x-π/3)+√3cos(x-π/3)=2[1/2*sin(x-π/3)+√3/2*cos(x-π/3)]=2[cosπ/3*sin(x-π/3)+sinπ/3*cos(x-π/3)]
f(x)=2sin(x-π/3+π/3)=2sinx
所以在x∈[0,π/2]∪[3π/2，2π]单调递增区间
（2）g(x)= (1+sinx)f(x)=(1+sinx)2sinx
=2sin²x+sinx=2(sinx+1/2)²-1/4 因为sinx∈[-1,1]
∴ g(x)∈[-1/4,17/4]