$2\,{\sin ^2}\theta \, – \,\cos \,2\theta \, = \,0$

$ \Rightarrow \,2\,{\sin ^2}\theta \, – \,(1 – 2\,{\sin ^2}\theta )\, = \,0$

$ \Rightarrow \,2\,{\sin ^2}\theta \, – \,1 + 2\,{\sin ^2}\theta \, = \,0$

$ \Rightarrow \,4\,{\sin ^2}\theta \, = 1 \Rightarrow \,\sin \theta \, = \, \pm \,\frac{1}{2}$

$\therefore \,\,\theta \, = \,\frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4},\,\theta  \in \,\,[0\,,\,2\,\pi ]$

$\therefore \,\,\theta \, = \,\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}$

Now $2\,{\cos ^2}\,\theta \, – \,3\,\sin \,\theta  = 0$

$ \Rightarrow \,\,2(1 – \,{\sin ^2}\,\theta )\, – \,3\,\sin \,\theta  = 0$

$ \Rightarrow \, – \,2\,{\sin ^2}\,\theta \, – \,3\,\sin \,\theta \, + \,2 = 0$

$ \Rightarrow \, – \,2\,{\sin ^2}\,\theta \, – \,4\,\sin \,\theta  + \sin \,\theta \, + \,2 = 0$

$ \Rightarrow \,2\,{\sin ^2}\,\theta \, – \,\,\sin \,\theta  + 4\sin \,\theta \, – \,2 = 0$

$ \Rightarrow \sin \,\theta (\,2\,\sin \,\theta \, – 1) + 2(\,2\,\sin \,\theta \, – 1) = 0\,$

$ \Rightarrow \sin \,\theta \, = \frac{1}{2}\,,\, – \,2$

But $\sin \,\theta \, = \, – \,2\,$ , is not possible

$\therefore \,\,\sin \,\theta \, = \,\frac{1}{2},\,$   $ \Rightarrow \,\,\theta \, = \,\frac{\pi }{6},\frac{{5\pi }}{6}$

Hence, there are two common solution, there each of the statement $-1$ and $2$ are true but statement $-\,2$ is not a correct explanation for statment $-\,1$.