Let S = [- π, (π)/(2)) - \ - (π)/(2), - (π)/(4), - (3π)/(4),...

Given,

$S = \left[- \pi, \dfrac{\pi}{2}\right) - \left\{- \dfrac{\pi}{2}, - \dfrac{\pi}{4}, - \dfrac{3\pi}{4}, \dfrac{\pi}{4}\right\}$ and the set $A = \left\{\theta \in S: tan\theta\left(1 + \sqrt{5}\tan \left(2\theta\right)\right) = \sqrt{5} -\text{ tan }\left(2\theta\right)\right\}$

Now let $tan\alpha = \sqrt{5}$

So, $tan\theta\left(1 + \sqrt{5}\tan \left(2\theta\right)\right) = \sqrt{5} - \tan \left(2\theta\right)$ becomes,

$⇒ tan\theta\left(1 + tan\alpha tan\left(2\theta\right)\right) = tan\alpha - \tan \left(2\theta\right)$

On rearranging we get,

$⇒ tan\theta = \dfrac{tan\alpha - \tan 2\theta}{1 + tan\alpha tan2\theta}$

On using formula $\tan \left(A - B\right) = \dfrac{tanA - tanB}{1 - tanAtanB}$ we get,

$⇒ tan\theta = \tan \left(\alpha - 2\theta\right)$

$⇒ \alpha - 2\theta = n\pi + \theta$

$⇒ 3\theta = \alpha - n\pi$

$⇒ \theta = \dfrac{\alpha}{3} - \dfrac{n\pi}{3}; n \in Z$

If $\theta \in [ - \pi, \pi / 2 )$ then $n = 0, 1, 2, 3, 4$ are acceptable solution

So, total $5$ solutions are possible.