4cos(4x-1)+12sin^2(4x-1) =11

Решение
4cos(4x-1)+12sin^2(4x-1) =11
4cos(4x-1)+12*(1 - сos²(4x-1)) = 11
4cos(4x-1) + 12  - 12сos²(4x-1)  = 11
12cos²(4x - 1) - 4cos(4x - 1) - 1 = 0
cos(4x - 1) = t
12t² - 4t - 1 = 0
D = 16 + 4*12*1 = 64
t₁ = (4 - 8)/24 = - 4/24 = - 1/6
t₂ = (4 + 8)/24 = 12/24 = 1/2
1)  cos(4x - 1) = - 1/6
4x - 1 = (+ -) arccos(- 1/6) + 2πk, k ∈ Z
4x  = (+ -) arccos(- 1/6) + 1 + 2πk, k ∈ Z
x₁  = (+ -) *(1/4)*arccos(- 1/6) + 1/4 + πk/2, k ∈ Z
2)   cos(4x - 1) = 1/2
4x - 1 = (+ -) arccos(1/2) + 2πn, n ∈ Z
4x  = (+ -) (π/3) + 1 + 2πn, n ∈ Z
x₂  = (+ -) (π/12) + 1/4 + πn/2, n ∈ Z