3. Let in triangle $A B C$

$$
\cos (2 \angle A-\angle B)+\sin (\angle A+\angle B)=2 \text {. }
$$

Find the side $B C$, if $A B=4$.
Answer: 2.

Solution. Each term in the left part of the equation does not exceed 1. Therefore, the equality will hold only if each of them equals 1. We solve the corresponding equations, denoting $\alpha=\angle A, \beta=\angle B:$

$$
2 \alpha-\beta=2 \pi n, n \in \mathbb{Z} ; \quad \alpha+\beta=\frac{\pi}{2}+2 \pi k, k \in \mathbb{Z}
$$

Considering that $0<\alpha<\pi, 0<\beta<\pi$, we get

$$
2 \alpha-\beta=0 ; \quad \alpha+\beta=\frac{\pi}{2} .
$$

From this,

$$
\beta=2 \alpha, \quad 3 \alpha=\frac{\pi}{2}, \quad \alpha=\frac{\pi}{6}, \quad \beta=\frac{\pi}{3} .
$$

Thus, $A B C$ is a right triangle with hypotenuse $A B$ and an angle of $30^{\circ}$ opposite side $B C$. Therefore, $B C=A B / 2=2$.

Evaluation. 12 points for a correct solution.