3. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\sin y \cdot \cos y+a=0 .
\end{array}\right.
$$

Then the value of $\cos (x+2 y)$ is ( ).
(A) 1
(B) -1
(C) 0
(D) $\frac{\sqrt{2}}{2}$
3.A.

From the problem, we have $\left\{\begin{array}{l}x^{3}+\sin x=2 a, \\ (-2 y)^{3}+\sin (-2 y)=2 a \text {. }\end{array}\right.$ Let $f(t)=t^{3}+\sin t, t \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then $f(t)$ is an increasing function on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, from $f(x)=f(-2 y)$, we get $x=-2 y$, i.e., $x+2 y=0$. Hence, $\cos (x+2 y)=1$.