Помогите, пожалуйста, решить уравнение! 1+sin2x=cosx+sinx

   sin²x+cos²x+2sin x*cos x=cjs x+sin x    (sin x+cos x)²-(cos x+sin x)=0     sin x+cos x) (sin x+cos x-1)=0      1)sin x+cos x=0  /cos x     tg x=-1   x=-π/4=πn  n∈z      2)sin x+cos x=1    sin x+sin(π/2-x)=1   2sin(x+π/2-x)/2cos (x-π/2+x)/2=1         2sinπ/4*cos(x-π/4)=1    2*√2/2*cos(x-π/4)=1   √2cos(x-π/4)=1    cos (x-π/4)=√2/2    x-π/4= +-π/4+2πn    x=+-π/4+π/4+2πn      x1=π/4+π/4+2πn=π/2+2πn. n∈z   x2=-π/4+π/4+2πk=2πk.  k∈z