已知函数f(x)=2sin(11π/6 -2x)+3+m (1)求函数f(x)的单调递增区间 (2)若f(x)的定义域为

f(x)=2sin(11π/6 -2x)+3+m=2sin(-π/6 -2x)+3+m=-2sin(2x +π/6 )+3+m
(1)因为-sinx的递增区间为2kπ+π/2≤x≤2kπ+3π/2,k∈Z,
所以2kπ+π/2≤2x +π/6≤2kπ+3π/2,k∈Z,
解出x得单增区间（亲,请写成集合或者区间形式）.
(2) π/2≤x≤π,
7π/6≤2x +π/6≤13π/6,
-1≤sin(2x +π/6)≤1/2
2+m≤f(x)≤5+m,
2+m=2, and 5+m=5,
m=0.