If $$2\cos^2 \theta + 3\sin \theta = 3$$, where $$0^\circ < \theta < 90^\circ$$, then what is the value of $$\sin^2 2\theta + \cos^2 \theta + \tan^2 2\theta + \cosec^2 2\theta$$?

$$2\cos^2 \theta + 3\sin \theta = 3$$
$$(2 - 2\sin^2 \theta) + 3\sin \theta - 3 = 0$$
$$2\sin^2 \theta - 3\sin \theta + 1 = 0$$
$$2\sin^2 \theta - 2\sin \theta - \sin \theta + 1 = 0$$
$$2\sin\theta(\sin\theta - 1) - 1(\sin\theta - 1) = 0$$
$$\sin\theta = 1 or \sin\theta = 1/2 $$ 
Here $$0^\circ < \theta < 90^\circ$$,
$$\sin\theta = 1/2$$
$$\theta = 30\degree$$
Now,
$$\sin^2 2\theta + \cos^2 \theta + \tan^2 2\theta + \cosec^2 2\theta$$
= $$\sin^2 2\times 30\degree + \cos^2 \times 30\degree + \tan^2 2\times 30\degree + \cosec^2 2\times 30\degree$$
= $$\sin^2 60\degree + \cos^2 30\degree + \tan^260\degree + \cosec^260\degree$$
= $$(\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 + (\sqrt{3})^2 + (\frac{2}{\sqrt{3}})^2$$
=$$\frac{3}{4} + \frac{3}{4} + 3 + \frac{4}{3} = \frac{35}{6}$$