$\sin 2 \theta+\tan 2 \theta>0$

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

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

Note : $\cos 2 \theta \neq 0$

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

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

$\Rightarrow \tan 2 \theta>0 \quad($ 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 $\sin \theta \neq \pm \frac{1}{\sqrt{2}} ;$ which has been already considered