$\begin{array}{*{20}{l}}
{\rm{\;}}&{\frac{{\left( {1 + \tan x} \right)\left( {1 - \tan x} \right)\left( {cos{\rm{ }}x - \sin x} \right)}}{{2\tan x}} = \frac{{1 - 2{{\sin }^2}x}}{{\left( {cos{\rm{ }}x + \sin x} \right)}}}\\
{{\rm{\;\;}}}&{\frac{{\left( {1 - {{\tan }^2}x} \right)}}{{2\tan x}} = \frac{{1 - 2{{\sin }^2}x}}{{\left( {cos{\rm{ }}x + \sin x} \right)\left( {cos{\rm{ }}x - \sin x} \right)}}}\\
{}&{\frac{{\left( {1 - {{\tan }^2}x} \right)}}{{2\tan x}} = \frac{{1 - 2{{\sin }^2}x}}{{\left( {cos{{\rm{ }}^2}x - {{\sin }^2}x} \right)}}}\\
{}&{\frac{1}{{\tan 2x}} = \frac{{cos{\rm{ 2}}x}}{{cos{\rm{ 2}}x}}}\\
{\rm{\;}}&\begin{array}{l}
\tan 2x = 1 \to 2x = k\pi + \frac{\pi }{4} \to x = \frac{{k\pi }}{2} + \frac{\pi }{8}\\
x = \frac{\pi }{8},\frac{{5\pi }}{8} \to \frac{\pi }{8} + \frac{{5\pi }}{8} = \frac{{6\pi }}{8} = \frac{{3\pi }}{4}
\end{array}
\end{array}$