Question:

**Step 1:** Rewrite the given equation using the identity $\csc^2\theta = \frac{1}{\sin^2\theta}$. Let $x = \sin^2\theta$. Since $\theta \in (0, \pi)$, $\sin\theta \neq 0$, so $\sin^2\theta > 0$. Thus, $x > 0$. The equation becomes: $$2x + 2\sqrt{2} = \frac{3}{x}$$ **Step 2:** Multiply by $x$ to transform the equation into a quadratic form. $$2x^2 + 2\sqrt{2}x = 3$$ $$2x^2 + 2\sqrt{2}x - 3 = 0$$ **Step 3:** Solve the quadratic equation for $x$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=2$, $b=2\sqrt{2}$, $c=-3$. $$x = \frac{-2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4(2)(-3)}}{2(2)}$$ $$x = \frac{-2\sqrt{2} \pm \sqrt{8 + 24}}{4}$$ $$x = \frac{-2\sqrt{2} \pm \sqrt{32}}{4}$$ $$x = \frac{-2\sqrt{2} \pm 4\sqrt{2}}{4}$$ This yields two possible values for $x$: $$x_1 = \frac{-2\sqrt{2} + 4\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$$ $$x_2 = \frac{-2\sqrt{2} - 4\sqrt{2}}{4} = \frac{-6\sqrt{2}}{4} = -\frac{3\sqrt{2}}{2}$$ Since $x = \sin^2\theta$ must be positive, we reject $x_2$. Therefore, the only valid solution for $x$ is $x = \frac{\sqrt{2}}{2}$. **Step 4:** Substitute back $x = \sin^2\theta$ and solve for $\sin\theta$. $$\sin^2\theta = \frac{\sqrt{2}}{2}$$ Since $\theta \in (0, \pi)$, $\sin\theta$ must be positive. $$\sin\theta = \sqrt{\frac{\sqrt{2}}{2}} = \sqrt{\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt[4]{2}}$$ **Step 5:** Determine the number of solutions and their sum in the interval $(0, \pi)$. Let $k = \frac{1}{\sqrt[4]{2}}$. Since $\sqrt[4]{2} \approx 1.189$, we have $0 < k < 1$. For any value $k$ such that $0 < k < 1$, the equation $\sin\theta = k$ has exactly two solutions in the interval $(0, \pi)$. Let $\alpha = \arcsin\left(\frac{1}{\sqrt[4]{2}}\right)$. Then the two solutions are $\theta_1 = \alpha$ and $\theta_2 = \pi - \alpha$. The number of real solutions is 2. The sum of all solutions is $\theta_1 + \theta_2 = \alpha + (\pi - \alpha) = \pi$.