cos4x =6cos²x -5 ;
cos2*2x =6*(1+cos2x)/2 -5 ;
2cos²2x -1 =3+3cos2x -5;
2cos²2x -3cos2x +1 =0 ; * * * замена t = cos2x ; |t|≤1 * * *
2t² -3t+1 =0 ;
D =3² -4*2*1 =1.
t₁=(3-1)/4 =1/2 ;
t₂=(3+1)/4 =1.
а) cos2x = 1/2 ⇒2x = ±π/3 +2πk , k∈Z ⇔x = ±π/6 +πk , k∈Z.
б) cos2x =1⇒2x =2πk⇔x =πk , k∈Z.
