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 - 12 сos² (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