2010 AMC 12B Problem 13

A cosine plus a sine equals $2$ only when each equals $1.$ So $\cos(2A-B)=1$ and $\sin(A+B)=1,$ giving $2A-B=0^\circ$ and $A+B=90^\circ.$ Solving, $A=30^\circ$ and $B=60^\circ,$ so $\triangle ABC$ is a $30\text{-}60\text{-}90$ right triangle with the right angle at $C.$ With hypotenuse $AB=4,$ the side $BC$ opposite the $30^\circ$ angle is half the hypotenuse, so $BC=2.$ Thus, the correct answer is C.