因为-π/4≤x≤π/3
-π/2≤2x≤2π/3
-π/6≤2x+π/3≤π
-1/2≤sin(2x+π/3)≤1
-3/2≤3sin(2x+π/3)≤3
所以-5/2≤3sin(2x+π/3)-1≤2
即值域为[-5/2,2]
若令t=2x+π/3,则t∈p[-π/6,π]
sint增区间为[-π/6,π/2)
减区间为[π/2,π]
取2x+π/3=π/2
x=π/12
即原函数增区间为[-π/4,π/12)
减区间为[π/12,π/3]