Answer is -7/9

5 (tan²x - cos²x) = 2cos2x +9 

5 ( (sec²x - 1) - cos²x) = 2(2cos²x - 1) +9 

5 (1/cos²x - 1- cos²x) = 4cos²x - 2 + 9 

5 (1 - cos²x - cos⁴x ) = 4cos⁴x + 7cos²x 

You solve this we get = 9cos⁴x+12cos²x-5 

Therefore let cos⁴x =y² and cos²x = y. 

0=9y² + 12y - 5

You will get the answer by solving for y

5(tan² x - cos² x) = 2 cos 2x + 9

= 5(sec² x - 1 - cos² x) = 2(2cos² x - 1) + 9 (\(\because\) cos2x = 2cos² x - 1)

= 5 \((\cfrac{1}{cos^2x}- 1 -cos^2x)\) = 4 cos² x -2 + 9  (\(\because\) sec x = \(\cfrac{1}{cos x}\))

= 5(1 - cos² x - cos⁴x) = 4 cos⁴x +7 cos² x ( multiplying both sides by cos²x)

= - 5 cos⁴ x - 5 cos² x + 5 = 4 cos⁴x + 7 cos²x 

= 9 cos⁴x + 12 cos²x - 5 = 0 (by transposing )

which is a quadric equation in cos² x.

= 9 cos⁴x + 15 cos² x - 3 cos² x - 5 = 

= 3 cos² x(3 cos² x + 5) - 1(3 cos² x + 5) = 0

= (3 cos² x + 5) (3 cos² x - 1) = 0

= 3cos² x + 5 = 0 or 3 cos² x - 1 = 0

= cos² x = \(\frac{-5}3\) or cos² x = \(\frac{1}3\)

= cos²x = \(\frac{1}3\) = (\(\because\) cos² x ≥ 0)

= cos² x ≠ \(\frac{-5}3\) )

Now cos2 x = 2 cos² x - 1

= \(\frac{2}3\) - 1 = \(\frac{-1}3\)

\(\therefore\) cos4 x = 2cos²(2x) - 1 (\(\because\) cos 2\(\theta\) = 2cos2\(\theta\) - 1)

= 2 \((\frac{-1}3)^2\) - 1

= \(\frac{2}9\) - 1

= \(\frac{2-9}9\) = \(\frac{-7}9\)