(b) $2{\sin ^2}x + {\sin ^2}2x = 2$ ……$(i)$

and $\sin 2x + \cos 2x = \tan x$…..$(ii)$

Solving $(i)$, ${\sin ^2}2x = 2{\cos ^2}x$

==>$2{\cos ^2}x\cos 2x = 0$

==>$x = (2n + 1)\frac{\pi }{2}{\rm{ or }}x = (2n + 1)\frac{\pi }{4}$

$\therefore $ Common roots are $(2n \pm 1)\frac{\pi }{4}$

Solving $(ii)$, $\frac{{2\tan x + 1 – {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \tan x$

$ \Rightarrow $ ${\tan ^3}x + {\tan ^2}x – \tan x – 1 = 0$

$ \Rightarrow $ $({\tan ^2}x – 1)\,(\tan x + 1) = 0$ 

$ \Rightarrow $ $x = m\pi \pm \frac{\pi }{4}$

Trick : For $n = 0$, option $(a)$ gives $\theta = – \frac{\pi }{2}$ which satisfies the equation $(i)$ but does not satisfy the $(ii)$. 

Now option $(b) $ gives $\theta = \frac{\pi }{4}$ which satisfies both the equations.