2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $\left\{\begin{array}{l}x^{3}+\sin x-2 a=0 \\ 4 y^{3}+\sin y \cos y+a=0\end{array}\right.$, then $\cos (x+2 y)=$
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2.【Analysis and Solution】From the given, we have $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$. 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 $\cos (x+2 y)=1$