The sum of solutions of the equation ( x)/(1 + x) = | 2x|, x ∈ (-...

We have,

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

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

Case-$1:$ When $0 \le x < \dfrac{\pi}{4} \text{ and } - \dfrac{\pi}{2} < x < - \dfrac{\pi}{4}$.

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

$⇒ \dfrac{\cos x}{1 + \sin x} = \dfrac{\sin 2x}{\cos 2x}$

$⇒ \dfrac{\cos x}{1 + \sin x} = \dfrac{2\sin x \cos x}{\cos^{2}x - \sin^{2}x}$

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

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

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

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

Then,

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

And,

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

$⇒ x = \dfrac{\pi}{10}, \dfrac{- 3\pi}{10}$

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

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

$⇒ \dfrac{\cos x}{1 + \sin x} = \dfrac{- \sin 2x}{\cos 2x}$

$⇒ \dfrac{\cos x}{1 + \sin x} = \dfrac{- 2\sin x \cos x}{\cos^{2}x - \sin^{2}x}$

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

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

$⇒ \cos x ( 1 + 2\sin x ) = 0$

$⇒ \cos x = 0 ⇒ x \in ϕ$

And, $\sin x = \dfrac{- 1}{2} ⇒ x = \dfrac{- \pi}{6}$

Sum of solutions

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