2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, 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 $\cos (x+2 y)=$ . $\qquad$
2. From the given, $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$, let $f(t)=$ $t^{3}+\sin t$, then $f(x)=(-2 y)$. Since the function $f(t)=t^{3}+\sin t$ is increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\therefore x=-2 y, x+2 y=0$, hence $\cos (x+2 y)=1$.