All possible values of $\theta \in \left[0,2\pi \right]$ for which $\sin 2\theta +\tan 2\theta >0$ lie in :

$\Rightarrow \sin 2\theta +\frac{\sin 2\theta }{\cos 2\theta }>0$

$\Rightarrow \sin 2\theta \frac{\left(\cos 2\theta +1\right)}{\cos 2\theta }>0\Rightarrow \tan 2\theta \left(2{\cos }^2\theta \right)>0$

Note: $\cos 2\theta \ne 0$

$\Rightarrow 1-2{\sin }^2\theta \ne 0\Rightarrow \sin \theta \ne \pm \frac{1}{\sqrt{2}}$

Now, $\tan 2\theta \left(1+\cos 2\theta \right)>0$

$\Rightarrow \tan 2\theta >0$ (as $\cos 2\theta +1>0$)

$\Rightarrow 2\theta \in \left(0,\frac{\pi }{2}\right)\cup \left(\pi ,\frac{3\pi }{2}\right)\cup \left(2\pi ,\frac{5\pi }{2}\right)\cup \left(3\pi ,\frac{7\pi }{2}\right)$

$\Rightarrow \theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$

As $\sin \theta \ne \pm \frac{1}{\sqrt{2}};$ which has been already considered