The sum of solutions of the equation $$\frac{\cos x}{1+\sin x} = |\tan 2x|$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) - \left\{-\frac{\pi}{4}, \frac{\pi}{4}\right\}$$ is:

We have to solve the equation

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

for all $$x$$ lying in the open interval $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$ except the two points $$x=-\dfrac{\pi}{4},\,\dfrac{\pi}{4}$$ and then add all admissible solutions.

Since $$x$$ is between $$-\dfrac{\pi}{2}$$ and $$\dfrac{\pi}{2},$$ we know that $$\cos x>0.$$ Also $$\sin x\in(-1,1),$$ so $$1+\sin x>0.$$ Hence the left-hand side is always positive. Therefore the absolute value on the right-hand side can be removed by writing

$$\frac{\cos x}{1+\sin x}=|\tan 2x|\quad\Longleftrightarrow\quad \frac{\cos x}{1+\sin x}=\pm\tan 2x.$$

Rationalising the denominator of the left side is convenient. Multiplying numerator and denominator by $$1-\sin x$$ we use $$1-\sin^2 x=\cos^2 x$$ to get

$$\frac{\cos x}{1+\sin x}=\frac{\cos x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\frac{\cos x(1-\sin x)}{\cos^2 x}=\frac{1-\sin x}{\cos x}.$$

So the equation becomes

$$\frac{1-\sin x}{\cos x}=\pm\tan 2x. \quad -(1)$$

Putting $$s=\sin x$$ and $$c=\cos x=\sqrt{1-s^2}$$ (positive), the left hand side is $$\dfrac{1-s}{c}.$$ Using the double-angle identity $$\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}$$ together with $$\tan x=\dfrac{s}{c},$$ we obtain

$$\tan 2x=\frac{2\,\dfrac{s}{c}}{1-\dfrac{s^2}{c^2}} =\frac{2s c}{1-2s^2}.$$

Substituting these expressions in (1) and squaring both sides (because of the $$\pm$$) gives

$$\left(\frac{1-s}{c}\right)^2=\left(\frac{2s c}{1-2s^2}\right)^2.$$

Multiplying by $$c^2$$ (which is positive) yields

$$ (1-s)^2=\frac{4s^2c^4}{(1-2s^2)^2}.$$

But $$c^2=1-s^2,$$ so $$c^4=(1-s^2)^2.$$ Hence

$$ (1-s)^2=\frac{4s^2(1-s^2)^2}{(1-2s^2)^2}.$$

Taking square roots on both sides we get two cases:

$$ (1-s)(1-2s^2)=\; 2s(1-s^2),\qquad (A) $$

$$ (1-s)(1-2s^2)=-2s(1-s^2).\qquad (B) $$

Expanding case (A):

$$ (1-s)(1-2s^2)=1-s-2s^2+2s^3, $$

and

$$ 2s(1-s^2)=2s-2s^3. $$

Setting them equal,

$$1-s-2s^2+2s^3=2s-2s^3,$$

which simplifies to

$$4s^3-2s^2-3s+1=0.$$

Dividing by $$(s-1)$$ (because the polynomial vanishes at $$s=1$$) we get

$$4s^2+2s-1=0.$$

Using the quadratic formula,

$$s=\frac{-2\pm\sqrt{4+16}}{8}=\frac{-1\pm\sqrt5}{4}.$$

The root $$s=1$$ is discarded because $$x=\dfrac{\pi}{2}$$ is not in the open interval. Thus from (A) we obtain

$$s_1=\frac{-1+\sqrt5}{4},\qquad s_2=\frac{-1-\sqrt5}{4}.$$

Now expand case (B):

$$ 1-s-2s^2+2s^3=-2s+2s^3,$$

which reduces to

$$-2s^2+s+1=0\;\Longleftrightarrow\;2s^2-s-1=0.$$

Again applying the quadratic formula,

$$s=\frac{1\pm3}{4}\;\Longrightarrow\;s=1,\;s=-\frac12.$$

Once more $$s=1$$ is not allowed, so the third admissible value is

$$s_3=-\frac12.$$

Since $$|s|\lt1,$$ each value of $$s=\sin x$$ corresponds to a unique $$x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$ Converting back to angles:

$$\sin x=\frac{-1+\sqrt5}{4}\; \Longrightarrow\; x=\arcsin\!\left(\frac{-1+\sqrt5}{4}\right)=\frac{\pi}{10},$$

because $$\sin\dfrac{\pi}{10}=0.3090\dots=\dfrac{-1+\sqrt5}{4}.$$

$$\sin x=\frac{-1-\sqrt5}{4}\; \Longrightarrow\; x=\arcsin\!\left(\frac{-1-\sqrt5}{4}\right)=-\frac{3\pi}{10},$$

since $$\sin\!\left(-\dfrac{3\pi}{10}\right)=-0.8090\dots=\dfrac{-1-\sqrt5}{4}.$$

$$\sin x=-\frac12\; \Longrightarrow\; x=\arcsin\!\left(-\frac12\right)=-\frac{\pi}{6}.$$

All three angles lie inside $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$ and are different from $$\pm\dfrac{\pi}{4},$$ so they are genuine solutions. Finally we add them:

$$x_1+x_2+x_3=\frac{\pi}{10}+\left(-\frac{3\pi}{10}\right)+\left(-\frac{\pi}{6}\right) =-\frac{2\pi}{10}-\frac{\pi}{6} =-\frac{\pi}{5}-\frac{\pi}{6} =-\frac{6\pi}{30}-\frac{5\pi}{30} =-\frac{11\pi}{30}.$$

Hence, the correct answer is Option 4.