已知函数f(x)=2sin(2wx+π/3)(w>

2个回答

f(x)=2sin(2wx+π/3)(w>0),T=2π/2w=π/w=π,w=1,f(x)=2sin(2x+π/3),
h（x）=g(x)-f（x）=cos^2（2x+π/3）-2sin(2x+π/3)=1-sin^2（2x+π/3）-2sin(2x+π/3)
=-sin^2（2x+π/3）-2sin(2x+π/3)+1=-[sin(2x+π/3)+1]^2+2,
,x属于[0,π/2],0=

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