已知f(x)=2sin(wx+派/6),y=f(x)的图像与直线y=1的两个相邻交点的距离最小值等于派/3,则f(x)的

f(x)=2sin(wx+π/6)跟y=1的交点,
先求出其横坐标,
2sin(wx+π/6)=1 即 sin(wx+π/6)=1/2=sinπ/6=sin5π/6
wx+π/6=π/6+2kπ 或者 wx+π/6=5π/6+2kπ 化简得
x1=2kπ/w 或 x2=2π/3w+2kπ/w
依题意,当k取同样的值的时候,x1和x2相差最小,为π/3
所以2π/3w=π/3,解得w=2,
所以f(x)=2sin(2x+π/6)
单调增区间为 2x+π/6∈[-π/2+2mπ,π/2+2mπ],即x∈[-π/3+mπ,π/6+mπ]
单调减区间为 2x+π/6∈[π/2+2mπ,3π/2+2mπ] ,即x∈[π/6+mπ,2π/3+mπ]