Official solution
\[{{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|\]
\[ \Rightarrow {{{{\cos }^2}x/2 - {{\sin }^2}x/2} \over {(\cos x/2 + \sin x/2)^2}} = \left| {\tan 2x} \right|\]
\[ \Rightarrow \] \[{{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}} = \left| {\tan 2x} \right|\]
\[ \Rightarrow \] \[{{1 - \tan {x \over 2}} \over {1 + \tan {x \over 2}}} = \left| {\tan 2x} \right|\]
\[ \Rightarrow \] \[{{\tan {\pi \over 4} - \tan {x \over 2}} \over {\tan {\pi \over 4} + \tan {x \over 2}}} = \left| {\tan 2x} \right|\]
\[ \Rightarrow {\tan ^2}\left( {{\pi \over 4} - {\pi \over 2}} \right) = {\tan ^2}2x\]
\[ \Rightarrow 2x = n\pi \pm \left( {{\pi \over 4} - {\pi \over 2}} \right)\]
\[ \Rightarrow x = {{ - 3\pi } \over {10}},{{ - \pi } \over 6},{\pi \over {10}}\]
or sum \[ = {{ - 11\pi } \over 6}\].