16cos⁴2x + 71 = 96sin²2x
16cos⁴2x + 71 = 96 - 96cos²2x
16cos⁴2x + 96cos²2x - 25 = 0
Пусть t = cos²2x, t ≥ 0
16t² + 96t - 25 = 0
D = 9216 + 1600 = 10816 = 104²
t1 = (-96 + 104)/32 = 8/32 = 1/4
t2 = (-96 - 104)/32 = -6,25 - посторонний корень.
Обратная замена:
cos²2x = 1/4
cos2x = -1/2 и cos2x = 1/2
2x = ±2π/3 + 2πn, n ∈ Z
x = ±π/3 + πn, n ∈ Z
и
2x = ±π/3 + 2πn, n ∈ Z
x = ±π/6 + πn, n ∈ Z
Ответ: x = ±π/3 + πn, n ∈ Z; ±π/6 + πn, n ∈ Z.