If 5(tan^{2} x – cos^{2}x) = 2 cos 2x + 9, then the value of cos 4x is

- -3/5
- 1/3
- 2/9
- -7/9

**Solution**

5(sec^{2} x – 1 – cos^{2} x) = 2(2cos^{2} x – 1) + 9

Let cos^{2} x = t

5(1/t - 1 - t) = 2(2t - 1) + 9

5(1 – t – t^{2}) = 4t^{2} + 7t

9t^{2} + 12t – 5 = 0

t = 1/3, -5/3

cos^{2} x = 1/3

cos 2x = 2/3 - 1 = -1/3

cos 4x = -7/9

**The correct option is D.**