All possible values of θ ∈ [0, 2π] for which 2θ + 2θ > 0 lie in :

$⇒ \sin 2\theta + \dfrac{\sin 2\theta}{\cos 2\theta} > 0$

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

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

$⇒ 1 - 2\sin^{2}\theta \ne 0 ⇒ sin\theta \ne \pm \dfrac{1}{\sqrt{2}}$

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

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

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

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

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