The sum of solutions of the equation

${{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right) - \left\{ {{\pi \over 4}, - {\pi \over 4}} \right\}$$ is :

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