note that $\displaystyle \displaystyle \sin^2{u} = \frac{1 - \cos(2u)}{2}$
$\displaystyle \displaystyle \frac{1 - \cos{x}}{1 - \cos{2x}} = 1 - \cos{x}
$
since $\displaystyle \sin{x} \ne 0$ , $\displaystyle \cos{x} \ne 1$ ...
$\displaystyle \displaystyle \frac{1}{1 - \cos(2x)} = 1$
$\displaystyle \cos(2x) = 0$
$\displaystyle \displaystyle 2x = \frac{\pi}{2} + k\pi \, ; \, k \in \mathbb{Z}$
$\displaystyle \displaystyle x = \frac{\pi}{4} + \frac{k\pi}{2}$