Let $\tan \alpha = \sqrt 5$ $\therefore$ $\tan \theta = {{\tan \alpha - \tan 2\theta } \over {1 + \tan \alpha \tan 2\theta }}$ $\therefore$ $\tan \theta = \tan (\alpha - 2\theta )$ $\alpha - 2\theta = n\pi + \theta$ $\Rightarrow 3\theta = \alpha - n\pi$ $\Rightarrow \theta = {\alpha \over 3} - {{n\pi } \over 3}\,\,\,\,\,\,\,\,\,;\,n \in Z$ If $\theta \in [ - \pi ,\,\pi /2]$ then $n = 0,1,2,3,4$ are acceptable $\therefore$ 5 solutions.