6cos2 x – 5 = cos 4x
Решите

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.