1. **Problem statement:** Find the exact y-coordinates of points A and B where the curve $$x=\frac{2}{3}\sin(3y+\frac{\pi}{4})$$ intersects the y-axis, given the domain $$\frac{\pi}{12} < y < \frac{3\pi}{4}$$. 2. **Key fact:** At the y-axis, $$x=0$$. 3. **Set up the equation:** $$0=\frac{2}{3}\sin\left(3y+\frac{\pi}{4}\right) \implies \sin\left(3y+\frac{\pi}{4}\right)=0$$ 4. **Solve for the argument:** $$3y+\frac{\pi}{4} = n\pi, \quad n \in \mathbb{Z}$$ 5. **Express y:** $$y = \frac{n\pi - \frac{\pi}{4}}{3} = \frac{n\pi}{3} - \frac{\pi}{12}$$ 6. **Find values of n such that $$y$$ lies in $$\left(\frac{\pi}{12}, \frac{3\pi}{4}\right)$$:** - For $$n=1$$: $$y = \frac{\pi}{3} - \frac{\pi}{12} = \frac{4\pi}{12} - \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}$$ which is in the interval. - For $$n=2$$: $$y = \frac{2\pi}{3} - \frac{\pi}{12} = \frac{8\pi}{12} - \frac{\pi}{12} = \frac{7\pi}{12}$$ which is also in the interval. - For $$n=0$$: $$y = 0 - \frac{\pi}{12} = -\frac{\pi}{12}$$ which is outside the interval. 7. **Therefore, the points are:** - Point A: $$y=\frac{\pi}{4}$$ - Point B: $$y=\frac{7\pi}{12}$$ **Final answer:** $$\boxed{A: y=\frac{\pi}{4}, \quad B: y=\frac{7\pi}{12}}$$ Note: Adding $$2\pi$$ to $$y$$ is not necessary because the domain restricts $$y$$ to $$\left(\frac{\pi}{12}, \frac{3\pi}{4}\right)$$, and the solutions above satisfy this.