9. Find the smallest positive real numbers $x$ and $y$ such that $x^{2}-3 x+2.5=\sin y-0.75$.
Answer: $x=\frac{3}{2}, y=\frac{\pi}{2}$
Solution: The equation above can be rewritten as $\left(x-\frac{3}{2}\right)^{2}+1=\sin y$. Since the left side is always greater or equal to 1 and the right side is always less than or equal to 1 , then both sides must be equal to 1. This means $x=\frac{3}{2}$ and $y=(4 k+1) \frac{\pi}{2}$ for all $k \in \mathbb{Z}$. Choosing $k=0$ gives the smallest possible value of $y$.