设y1=arccos(sinx),y2=arcsin(cosx)则cosy1=sinx (0≤y1≤π/2)siny2=cosx(-π/2≤y2≤π/2) ;∴y1=π/2-x+2kπ或x-π/2+2kπy2=π/2-x+2kπ或x+π/2+2kπ;当0≤y1,y2≤π/2时,即0≤π/2-x+2kπ≤π/2或0≤x-π/2+2k...