设函数f(x)=sin(2x+φ)(-π

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设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+φ)(-π<φ<0),y=f(x)图像的一条对称轴是直线x=π/6
（1）求φ
（2）求函数y=f(x)的单调区间

设函数f(x)=sin(2x+φ)(-π
(1)f(x)的对称轴位于函数Y的最大值或最小值点的位置.而当x=π/6 ,f(π/6)=sin(π/3+φ),当π/3+φ=0时,f(x)有最大值1,此时φ=-π/3
(2)f(x)=sin(2x-π/3),当2x-π/3=2nπ时,x=nπ+π/6,f(x)有最大值,
当2x-π/3=(2n-1)π时,x=nπ-π/3,f(x)有最小值,
单调区间为：nπ-π/3≤x≤nπ+π/6,单调上升,
xnπ+π/6≤x≤nπ+2π/3,单调下降.

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