若函数f(x)=2sin(2x+φ)(|φ|0)的图像有相同的对称中心,则φ=?应该是g(x)=co

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设对称中心为(a,0),则
g(0)= -1/2 → g(2a)=sin(2wa-π/6)= 1/2
2wa-π/6=π/6+2kπ
wa=π/6+kπ
而f(0)=2sin(φ),则f(a)=0,f(2a)=2sin(4a+φ)= -2sin(φ)
所以 sin(4a+φ) + sin(φ) =0
2sin(2a+φ)·cos(2a)=0
则cos(2a)=0,
则2a=π/2 +kπ,a=π/4 +kπ/2
f(a)=2sin(2a+φ)=2sin(π/2 +kπ +φ)
=0→sin(π/2 +φ)=0
→π/2 +φ =k1π,φ =k1π -π/2
由于|φ|