If 5(tan2x - cos2x) = 2cos2x + 9, then the value of cos4x is

We know that tan2x = sec2x and cos2x = 2cos²x - 1
∴ 5(sec2x - cos2x) = 2(2cos²x - 1) + 9
Take cos2x = t ⇒ 5(1/t - t) = 2(2t - 1) + 9 ⇒ 5/t - 5t = 4t + 7 ⇒ 5 - 5t² = 4t² + 7t ⇒ 9t² + 7t - 5 = 0 ⇒ 9t² + 12t - 5t - 5 = 0 ⇒ 3t(3t + 5) - 1(3t + 5) = 0 ⇒ (3t + 5)(3t - 1) = 0 ⇒ t = -5/3 or t = 1/3
t ≠ -5/3 as the range of cos2x is [0, 1]
Hence, cos2x = 1/3
Now, cos4x = 2cos²2x - 1 = 2[(1/3)²] - 1 = 2(1/9) - 1 = 2/9 - 1 = -7/9
Option D is the correct answer.