4sin^2x+cos4x=1 решите плз задачу

4(1 - cos^2x) + 8cos^4x - 8cos^2x + 1 = 1  4 - 4cos^2x + 8cos^4x - 8cos^2x = 0  8cos^4x - 12cos^2x + 4 = 0   /:4 2cos^4x - 3cos^2x + 1 = 0  Пусть cos^2x = t, тогда 2t^2  - 3t + 1 = 0  D = 9 - 8 =  1 t1 = ( 3 + 1)/4 =  1 t2 = ( 3 -  1)/4 = 1/2 cos^2x = 1 cosx = 1 x = 2pik, k ∈Z cosx = - 1 x = pi + 2pik, k∈Z cos^2x = 1/2 cosx = √2/2 x = ± arccos (√2/2) + 2pik, k ∈ Z cosx =  - √2/2 x = ± (pi - arccos √2/2) + 2pik, k ∈ Z