In $ \triangle ABC, \ \cos(2A \minus{} B) \plus{} \sin(A\plus{}B) \equal{} 2$ and $ AB\equal{}4.$ What is $ BC?$ 

$ \textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}$
1. Given the equation \(\cos(2A - B) + \sin(A + B) = 2\), we know that the maximum values of \(\cos x\) and \(\sin y\) are both 1. Therefore, for the sum to be 2, both \(\cos(2A - B)\) and \(\sin(A + B)\) must be equal to 1.
   
2. Since \(\cos(2A - B) = 1\), it implies that \(2A - B = 360^\circ k\) for some integer \(k\). Given that \(2A - B < 360^\circ\), we have \(2A - B = 0\), which simplifies to:
   \[
   2A = B
   \]

3. Similarly, since \(\sin(A + B) = 1\), it implies that \(A + B = 90^\circ + 360^\circ m\) for some integer \(m\). Given that \(A + B < 360^\circ\), we have \(A + B = 90^\circ\).

4. Substituting \(B = 2A\) into \(A + B = 90^\circ\), we get:
   \[
   A + 2A = 90^\circ \implies 3A = 90^\circ \implies A = 30^\circ
   \]
   Therefore, \(B = 2A = 60^\circ\).

5. Now, we know that \(\triangle ABC\) is a 30-60-90 triangle. In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\). Given that \(AB = 4\), which is the hypotenuse, we can determine the lengths of the other sides.

6. The side opposite the \(30^\circ\) angle (which is \(BC\)) is half the hypotenuse:
   \[
   BC = \frac{AB}{2} = \frac{4}{2} = 2
   \]

The final answer is \(\boxed{2}\)