The solution of the equation ( ^ 2 x - 1) ^ - 1 = 1 + 2x is

\dfrac{1}{\tan^{2} x - 1} = 1 + \cos 2x$<br />$⇒ \dfrac{\cos^{2} x}{\sin^{2} x - \cos^{2} x} = 1 + \cos 2x$<br />$⇒ \dfrac{\left(1 + \cos 2x\right)}{2\left(- \cos 2x\right)} - \left(1 + \cos 2x\right) = 0$<br />\left(1 + \cos 2x\right)\left(- \dfrac{1}{2\cos 2x} - 1\right) = 0$
⇒ 1 + \cos 2x = 0 or $1 + \dfrac{1}{2\cos 2x} = 0$
⇒ \cos 2x = - 1 or $- \dfrac{1}{2}$
$⇒ 2x = 2n\pi \pm \pi$ or $2n\pi \pm \dfrac{2\pi}{3}$
$⇒ x = n\pi \pm \dfrac{\pi}{2}$ or $n\pi \pm \dfrac{\pi}{3}$
But for $x = n\pi \pm \dfrac{\pi}{2}, cos x = 0$ and hence $tan x$ is not defined
$⇒ x = n\pi \pm \dfrac{\pi}{3}$