正弦函数y=sinx的递减区间为[2kπ + π/2 ,2kπ + 3π/2] (k∈Z)
所以函数y=sin（x+π/3）的递减区间为
{x|2kπ + π/2≤x+π/3≤2kπ + 3π/2,k∈Z}
={x|2kπ + π/6≤x≤2kπ + 7π/6,k∈Z}
[π/6,π] 刚好落在递减区间内.