Trig and Triangle Math Club Question: Finding Side Length

Question

I recently had a math club competition, and I was unsure of how to approach one of the problems on the test:

In $\triangle ABC$,
$\ \ \ \ \ \ \ \ \cos(2A-B) + \sin(A + B) = 2$
$\ \ \ \ \ \ \ \ \overline{AB} = 4$.
What is the length of $\overline{BC}$?

My immediate reaction was to simplify down all the trig functions, which resulted in: $$(\cos^2 A - \sin^2 A)\cos B + 2\sin A \cos A \sin B + \sin A \cos B + \sin B \cos A = 2$$

And I don't see any way to further simplify it. I think my original method is wrong, or I just can't see the next step forward, so I would really appreciate any tips on how to solve the problem.

Answer 1

Since $$\cos(2A-B)\le 1 \ \text{and} \ \sin(A + B)\le 1;\\ \cos(2A-B) = \sin(A + B) = 1 \Rightarrow \\ \begin{cases}2A-B=0 \\ A+B=\frac{\pi}{2} \end{cases}\Rightarrow \begin{cases}A=\frac{\pi}{6} \\ B=\frac{\pi}{3}\end{cases} \Rightarrow C=\frac{\pi}{2}.$$ Hence: $$BC=AB\sin \frac{\pi}{6}=2.$$

Answer 2

Hint:

For real $x,$ $$\cos x,\sin x\le1$$

$$\cos(2A-B) + \sin(A + B) = 2\implies\cos(2A-B)=\sin(A+B)=1$$

and $\sin(A+B)=\sin(\pi-C)=\sin C$

Can you take it from here?