4cos^4x -3cos2x-1=0 решить уравнение,заранее !

4cos⁴x-3cos2x-1=0
cos2x=2cos²x-1
4(cos²x)²-3*(2cos²x-1)-1=0
cos²x=t, t∈[ 0 ; 1 ]
4t²-6t+2=0
2t²-3t+1=0
D=1
t₁=1/2, t₂=1

t₁=1/2
cos²x=1/2, cosx=+-√(1/2),  cosx=+-√2/2
1. cosx=√2/2, x₁=+-π/4+2πn, n∈Z
2. cosx=-√2/2, x=+-(π-arccos√2/2)+2πn, n∈Z, x₂=+-3π/4+2πn, n∈z
t₂=1
cos²x=1, cosx=+-1
cosx=-1, x₃=π+2πn, n∈Z
cosx=1, x₄=2πn, n∈Z