Question:

Step 1: Rewrite the equation using $\csc^2\theta = \dfrac{1}{\sin^2\theta}$: $2\sin^2\theta + 2\sqrt{2} = \dfrac{3}{\sin^2\theta}$. Step 2: Let $u = \sin^2\theta$. Then $2u + 2\sqrt{2} = \dfrac{3}{u}$, giving $2u^2 + 2\sqrt{2}u - 3 = 0$. Step 3: Solve: $u = \dfrac{-2\sqrt{2} \pm \sqrt{8 + 24}}{4} = \dfrac{-2\sqrt{2} \pm \sqrt{32}}{4} = \dfrac{-2\sqrt{2} \pm 4\sqrt{2}}{4}$. Step 4: Two roots: $u = \dfrac{2\sqrt{2}}{4} = \dfrac{\sqrt{2}}{2}$ or $u = \dfrac{-6\sqrt{2}}{4} < 0$ (rejected since $u = \sin^2\theta \geq 0$). Step 5: So $\sin^2\theta = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$, giving $\sin\theta = \pm \left(\dfrac{1}{\sqrt{2}}\right)^{1/2}$. Step 6: Since $\theta \in (0, \pi)$, $\sin\theta > 0$, so $\sin\theta = \dfrac{1}{2^{1/4}}$. This gives exactly 2 solutions in $(0,\pi)$: $\theta_1$ and $\pi - \theta_1$. Step 7: Sum of solutions $= \theta_1 + (\pi - \theta_1) = \pi$. Options (a) and (c) are correct.