1. **State the problem:** Find the y-coordinates of points A and B where the curve $x = \frac{2}{3} \sin(3y + \frac{\pi}{4})$ intersects the y-axis, i.e., where $x=0$. 2. **Set up the equation for intersection:** Since $x=0$ on the y-axis, set $$0 = \frac{2}{3} \sin\left(3y + \frac{\pi}{4}\right)$$ which implies $$\sin\left(3y + \frac{\pi}{4}\right) = 0$$ 3. **Solve the trigonometric equation:** The general solutions for $\sin \theta = 0$ are $$\theta = n\pi, \quad n \in \mathbb{Z}$$ So, $$3y + \frac{\pi}{4} = n\pi$$ 4. **Isolate $y$:** $$y = \frac{n\pi - \frac{\pi}{4}}{3} = \frac{n\pi}{3} - \frac{\pi}{12}$$ 5. **Apply domain restriction:** Given $\frac{\pi}{12} < y < \frac{3\pi}{4}$, find integer $n$ such that $$\frac{\pi}{12} < \frac{n\pi}{3} - \frac{\pi}{12} < \frac{3\pi}{4}$$ Add $\frac{\pi}{12}$ to all parts: $$\frac{\pi}{6} < \frac{n\pi}{3} < \frac{10\pi}{12} = \frac{5\pi}{6}$$ Multiply all parts by $\frac{3}{\pi}$: $$\frac{3}{6} < n < \frac{15}{6}$$ Simplify: $$0.5 < n < 2.5$$ 6. **Find integer values of $n$:** Possible integers are $n=1$ and $n=2$. 7. **Calculate $y$ for each $n$:** - For $n=1$: $$y = \frac{\pi}{3} - \frac{\pi}{12} = \frac{4\pi}{12} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}$$ - For $n=2$: $$y = \frac{2\pi}{3} - \frac{\pi}{12} = \frac{8\pi}{12} - \frac{\pi}{12} = \frac{7\pi}{12}$$ 8. **Identify points:** - Point A has $y = \frac{\pi}{4}$ - Point B has $y = \frac{7\pi}{12}$ **Final answers:** $$\boxed{y_A = \frac{\pi}{4}, \quad y_B = \frac{7\pi}{12}}$$