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   /:42cos^4x - 3cos^2x + 1 = 0 Пусть cos^2x = t, тогда2t^2  - 3t + 1 = 0 D = 9 - 8 =  1t1 = ( 3 + 1)/4 =  1t2 = ( 3 -  1)/4 = 1/2cos^2x = 1cosx = 1x = 2pik, k ∈Zcosx = - 1x = pi + 2pik, k∈Zcos^2x = 1/2cosx = √2/2x = ± arccos (√2/2) + 2pik, k ∈ Zcosx =  - √2/2x = ± (pi - arccos √2/2) + 2pik, k ∈ Z