If 5( ^ 2 x - ^ 2 x) = 2 2x + 9, then the value of 4x is

$5\left(\dfrac{1 - \cos 2x}{1 + \cos 2x} - \dfrac{1 + \cos 2x}{2}\right) = 2\cos 2x + 9$

Let, $\cos 2x = y$

$∴ 5( \dfrac{1 - y}{1 + y} - \dfrac{1 + y}{2} ) = 2y + 9$

$⇒ 5\left[- {y}^{2} - 4y + 1\right] = 4{y}^{2} + 22y + 18$

$⇒ 9{y}^{2} + 42y + 13 = 0$

$⇒ y = - \dfrac{1}{3}$ or $y = - \dfrac{13}{3}$ (Not possible)

Now, $\cos 4x = 2\cos^{2} 2x - 1$

$⇒ \cos 4x = 2{\left(- \dfrac{1}{3}\right)}^{2} - 1$

$⇒ \cos 4x = - \dfrac{7}{9}$