解
|sinx+cosx|-1>=0
|sinx+cosx|>=1
|√2sin(x+π/4)|>=1
2sin^2(x+π/4)>=1
sin^2(x+π/4)>=1/2
所以有
sin(x+π/4)>=√2/2或 sin(x+π/4)<=-√2/2
x+π/4>=π/4+2kπ或x+π/4<=-π/4+2kπ,k∈Z
所以x>=2kπ或x<=-π/2+2kπ,k∈Z
解
|sinx+cosx|-1>=0
|sinx+cosx|>=1
|√2sin(x+π/4)|>=1
2sin^2(x+π/4)>=1
sin^2(x+π/4)>=1/2
所以有
sin(x+π/4)>=√2/2或 sin(x+π/4)<=-√2/2
x+π/4>=π/4+2kπ或x+π/4<=-π/4+2kπ,k∈Z
所以x>=2kπ或x<=-π/2+2kπ,k∈Z