4sin^2x+cos4x=1 решите плз задачу

cos4x=8cos^4x-8cos^2x+1

Пж по прытче тест

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, kZ

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