2) sin^4x+cos^4x=sin2x-1/2
[b]sin^2x-sqrt(2)cos(2x-π/4)=1[/b]
sin^2x-sqrt(2)cos(2x-π/4)-1=0
-cos^2x-sqrt(2)cos(2x-π/4)=0
(1+cos2x)/2+sqrt(2)cos(2x-π/4)=0
1+cos2x+2sqrt(2)cos(2x)sin(π/4)+2sqrt(2)sin(2x)cos(π/4)=0
1+cos2x+2sqrt(2)cos2x*sqrt(2)/2+2sqrt(2)sin2x*sqrt(2)/2=0
1+cos2x+2cos2x+2sin2x=0
sin^2x+cos^2x+3cos^2x-3sin^2x+4sinxcosx=0
2sin^2x-4sinxcosx-4cos^2x=0 / 2cos^2x≠0
tg²x-2tgx-2=0
Замена tgx=t
t²-2t-2=0
D=4+8=12
t1=(2-2sqrt(3))/2=1-sqrt(3) ⇒ tgx=1-sqrt(3) ⇒ x=arctg(1-sqrt(3))+πn
t2=(2+2sqrt(3))/2=1+sqrt(3) ⇒ tgx=1+sqrt(3) ⇒ x=arctg(1+sqrt(3))+πn
2)
[b]sin^4(x)+cos^4(x)=sin2x-1/2[/b]
(sin^2(x) + cos^2(x))^2 - 2sin^2(x)cos^2(x) = sin2x - 1/2
1 - (1/2)sin^2(2x) = sin2x - 1/2
sin^2(2x) + 2sin2x - 3 = 0;
D/4 = 1 + 3 = 4
sinx = -1±2
Подходит только sin x = 1
x = Pi/2 + 2k*Pi.