答案没问题啊
∵1+sinacosa>0∴由所给不等式得到：sina-cosa≥0即：√2sin(a-π/4)≥0 sin(a-π/4)≥0 2kπ≤a-π/4≤2kπ+π解得：2kπ+π/4≤a≤2kπ+5π/4
