函数y=sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)cos(x+π/3)的图象的一条对称轴的方

求详解过程，谢谢

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cos(x+π/3)
=sin[π/2-(x+π/历衡3)]
=sin(π/6-x)
=-sin(x-π/6)

所以y=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/物散6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/2)
=cos2x

cosx的对称轴用就是取最大或最罩烂氏小的地方，即x=kπ
所以此处是2x=kπ
x=kπ/2

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y=sin(3x+π/弯耐碧3)cos(x-π/6)+cos(3x+π/3)cos(x+π/3)
=sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)sin[π/2-(x+π/3)]
=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/2)
=cos2x
其中一条对埋举称轴的亩铅方程为x=π/2

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cos(x+π/3)=-sin(x-π/6)
∴sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)cos(x+π/3)
=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/橘陪2)
=cos(2x)
∴对称轴为2x=kπ，即x=kπ/2，k∈Z
若取圆颤蠢洞腊k=0，得x=0

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