若函数y=cos(2x+&) (0
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若函数y=cos(2x+&)(0若函数y=cos(2x+&)(0若函数y=cos(2x+&)(0y=cos(2x+&)=sin(π/2-2x-&)=-sin(2x+&-π/2)奇函数所以&-π/2=k
若函数y=cos(2x+&) (0
若函数y=cos(2x+&) (0
若函数y=cos(2x+&) (0
y=cos(2x+&)
=sin(π/2-2x-&)
=-sin(2x+&-π/2)
奇函数所以
&-π/2=kπ
&=kπ+π/2
k=0
&=π/2
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