Solutions:

\( \dfrac{\cos x}{1+\sin x} = |\tan 2x|; \ x \in \left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right) - \left\{ -\dfrac{\pi}{4}, \dfrac{\pi}{4} \right\} \)

\( 2x \in \left( -\pi,\pi \right) - \left\{ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right\} \)

Case I: \( 2x \in \left( -\pi,-\dfrac{\pi}{2} \right) \cup \left( 0,\dfrac{\pi}{2} \right) \):

\( \dfrac{\cos x}{1+\sin x} = \tan 2x = \dfrac{2\sin x \cos x}{1-2\sin^2 x} \ \{ \because \ \tan 2x = \dfrac{\sin 2x}{\cos 2x} \} \)

\( \Rightarrow \ \cos x \,[1-2\sin^2 x] = \cos x \,[2\sin x(1+\sin x)] \)

\( \Rightarrow \ \cos x = 0 \ \text{(OR)} \ 1-2\sin^2 x = 2\sin x + 2\sin^2 x \)

\( \Rightarrow \ x \in \phi \ \text{(OR)} \ 4\sin^2 x + 2\sin x - 1 = 0; \ x \in \left( -\dfrac{\pi}{2}, -\dfrac{\pi}{4} \right) \cup \left( 0, \dfrac{\pi}{4} \right) \)

i.e., \( \sin x = \dfrac{\sqrt{5}-1}{4}, \ -\dfrac{\sqrt{5}+1}{4} \Rightarrow x = 18^\circ, -54^\circ \)

Case II: \( 2x \in \left( -\pi,0 \right) \cup \left( \dfrac{\pi}{2},\pi \right) \):

\( \Rightarrow \ \cos x \,[1-2\sin^2 x] = -\cos x \,[2\sin x(1+\sin x)] \)

\( \Rightarrow \ \cos x = 0 \ \text{(OR)} \ 1-2\sin^2 x = -2\sin x - 2\sin^2 x \)

\( \Rightarrow \ x \in \phi \ \text{(OR)} \ \sin x = -\dfrac{1}{2}; \ x \in \left( -\dfrac{\pi}{4},0 \right) \cup \left( \dfrac{\pi}{4}, \dfrac{\pi}{2} \right) \)

\( \Rightarrow \ x = -30^\circ \)

Hence, (Case I) \( \cup \) (Case II) \( \Rightarrow x = 18^\circ, -54^\circ, -30^\circ \)

Sum \( = -66^\circ = -\dfrac{11\pi}{30} \).

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