最小正周期=π；
当：2kπ-π/2≤2x-π/6≤2kπ+π/2
kπ-π/6≤x≤kπ+π/3
所以增区间为：[kπ-π/6,kπ+π/3]
3.x∈{0,π/2}时,-π/6≤2x-π/6≤5π/6
f(x)的最小值为-2=1+a;a=-3看明白了太谢谢你了。