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
= \(\frac{2-9}9\) = \(\frac{-7}9\)