1. **State the problem:** Given acute angles $A$ and $B$ such that $\sin(A-B)=0$ and $2\cos(A+B)-1=0$, find the values of $A$ and $B$. 2. **Use the given equations:** - From $\sin(A-B)=0$, we know that $A-B = n\pi$ for some integer $n$. Since $A$ and $B$ are acute (between 0 and $\frac{\pi}{2}$), the only possibility is $A-B=0$, so $A = B$. 3. **Use the second equation:** $$2\cos(A+B)-1=0 \implies 2\cos(2A)-1=0$$ 4. **Solve for $A$:** $$2\cos(2A) = 1 \implies \cos(2A) = \frac{1}{2}$$ 5. **Recall cosine values:** $$\cos(2A) = \frac{1}{2} \implies 2A = \frac{\pi}{3} \text{ or } 2A = \frac{5\pi}{3}$$ Since $A$ is acute, $2A = \frac{\pi}{3}$ is valid. 6. **Find $A$ and $B$:** $$A = \frac{\pi}{6} = 30^\circ$$ Since $A = B$, we have $$B = 30^\circ$$ **Final answer:** $$A = 30^\circ, \quad B = 30^\circ$$