3. Let in triangle $A B C$

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

Find the side $B C$, if $A B=4$.
Answer: $2 \sqrt{2}$.

Solution. Each term in the left part of the equation is no more than 1. Therefore, the equality will hold only if each of them equals 1. We solve the corresponding equations, denoting $\alpha=\angle A, \beta=\angle B:$

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

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

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

From this,

$$
\beta=\alpha, \quad 2 \alpha=\frac{\pi}{2}, \quad \alpha=\frac{\pi}{4} .
$$

Thus, $A B C$ is an isosceles right triangle with hypotenuse $A B$. Therefore, $B C=A B \cdot \frac{\sqrt{2}}{2}=2 \sqrt{2}$.

Evaluation. 12 points for a correct solution.