f(a)=3,sin(2a+π/6)=1/2,2a+π/6=5π/6,a=π/3
f(x)=2sin(2x+π/6)+2,2kπ-π/2<2x+π/6<2kπ+π/2,kπ-π/3<x<kπ+π/6
k=0时,-π/3<x<π/6递增,k=1时2π/3<x<7π/6递增,x属于[0,π ]时,求函数f(x)的单调递增区间
(0,π/6)和(2π/3,π)