Решите с подробностями 2(sinx+cosx)+sin2x+1=0 2sinx*cosx+cos2x-1-sinx=0 sin^4x+cos^4x=1

1)2(sinx+cosx)+sin2x+1=0
2(sinx+cosx)+(sinx+cosx)²=0
(sinx+cosx)(2+sinx+cosx)=0
sinx+cosx=0
sinx+sin(π/2-x)=0
2sinπ/4cos(x-π/4)=0⇒cos(x-π/4=0⇒x-π/4=π/2+πn⇒x=3π/4+πn
2+sinx+cosx=0
2+2sinπ/4cos(x-π/4)=0
2+√2cos(x-π/4)=0⇒cos(x-π/4)=-√2∉[-1;1]-нет решения
2)2sinxcosx+cos2x-1=0
2sinxcosx-2sin²x=0 /2cos²x≠0
tg²x-tgx=0
tgx(tgx-1)=0
tgx=0⇒x=πn
tgx=1⇒x=π/4+πn
3)sin^4 x+cos^4 x=1
(1-cos2x)²/4+(1+cos2x)²/4=1
1-2cos2x+cos²2x+1+2cos2x+cos²2x=4
cos²2x=1
1+cos4x=2⇒cos4x=1⇒4x=2πn⇒x=πn/2