已知：2sin2x=sin2x+sin2x,且1+sin2x=(sinx+cosx)²
那么：原式=-(cosx-sinx-2sin2x+2-2)
=-[(-√2*sin(x-45)+2(1-sin2x)]+2
=-[-√2*sin(x-45)+2(sinx-cosx)²]+2
=√2*sin(x-45)-4sin²(x-45)+2
根据令t=sin(x-45).讨论即可