We have,

$\frac{\cos x}{1+\sin x}=\left|\tan 2x\right|$

where, $x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{\frac{\pi }{4},\frac{-\pi }{4}\right\}$

Case-$1:$ When $0\le x<\frac{\pi }{4} \& -\frac{\pi }{2}<x<-\frac{\pi }{4}$.

$\frac{\cos x}{1+\sin x}=\tan 2x$

$\Rightarrow \frac{\cos x}{1+\sin x}=\frac{\sin 2x}{\cos 2x}$

$\Rightarrow \frac{\cos x}{1+\sin x}=\frac{2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}$

$\Rightarrow \cos x\left({\cos }^{2}x-{\sin }^{2}x\right)=2\sin x\cos x\left(1+\sin x\right)$

$\Rightarrow \cos x\left(1-2{\sin }^{2}x\right)=\cos x\left(2\sin x+2{\sin }^{2}x\right)$

$\Rightarrow \cos x\left(-4{\sin }^{2}x-2\sin x+1\right)=0$

$\Rightarrow \cos x\left(4{\sin }^{2}x+2\sin x-1\right)=0$

Then,

$\cos x=0$ (Not possible, since $0\le x<\frac{\pi }{4} \& -\frac{\pi }{2}<x<-\frac{\pi }{4}$)

And,

$\sin x=\frac{-2\pm 2\sqrt{5}}{8}=\frac{-1\pm \sqrt{5}}{4}$

$\Rightarrow x=\frac{\pi }{10},\frac{-3\pi }{10}$

Case-$2:$ When $\frac{\pi }{4}<x<\frac{\mathrm{\pi }}{2}$ and $\frac{-\pi }{4}<x<0$, then

$\frac{\cos x}{1+\sin x}=-\tan 2x$

$\Rightarrow \frac{\cos x}{1+\sin x}=\frac{-\sin 2x}{\cos 2x}$

$\Rightarrow \frac{\cos x}{1+\sin x}=\frac{-2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}$

$\Rightarrow \cos x\left({\cos }^{2}x-{\sin }^{2}x\right)=-2\sin x\cos x\left(1+\sin x\right)$

$\Rightarrow \cos x\left(1-2{\sin }^{2}x\right)=\cos x\left(-2\sin x-2{\sin }^{2}x\right)$

$\Rightarrow \cos x(1+2\sin x)=0$

$\Rightarrow \cos x=0\Rightarrow x\in \varphi$

And, $\sin x=\frac{-1}{2}\Rightarrow x=\frac{-\pi }{6}$

Sum of solutions

$=\frac{\pi }{10}-\frac{3\pi }{10}-\frac{\pi }{6}=\frac{-11\pi }{30}$