5 (tan²x – cos²x) = 2cos 2x + 9

$\Rightarrow \text{5 }(\frac{1-\cos 2\mathrm{x}}{1+\cos 2\mathrm{x}}-\frac{1+\cos 2\mathrm{x}}{2})=\text{ 2cos 2x }+\text{ 9}$

$\Rightarrow \text{5}\left(\frac{2-2\cos 2\mathrm{x}-(1+2\cos 2\mathrm{x}+{\cos }^{2}2\mathrm{x})}{2(1+\cos 2\mathrm{x})}\right)=\text{ 2cos 2x }+\text{ 9}$

⇒ 5 (1 – 4cos 2x – cos²2x) = 2 (2cos²2x + 11cos 2x + 9)

⇒ 9cos²2x + 42cos 2x + 13 = 0

⇒ 9cos²2x + 3cos 2x + 39cos 2x + 13 = 0

⇒ (3cos 2x + 1) (3cos 2x + 13) = 0

$\Rightarrow \text{\hspace{0.17em}cos 2x }=\frac{-1}{3}$            $(\because –1\le \cos 2\mathrm{x}\le 1)$

$\therefore \text{cos 4x }={\text{ 2cos}}^{\text{2}}\text{2x }–\text{1 }= \frac{2}{9}–\text{ 1 }=\frac{–7}{9}.$