2. If $x, y \in\left[-\frac{\pi}{6}, \frac{\pi}{6}\right], a \in R$, and satisfy $\left\{\begin{array}{l}x^{3}+\sin x-3 a=0 \\ 9 y^{3}+\frac{1}{3} \sin 3 y+a=0\end{array} \quad\right.$, then $\cos (x+3 y)=$
Hui Gui 1 .
Analysis Transform the known conditions into $\left\{\begin{array}{l}x^{3}+\sin x-3 a=0 \\ (3 y)^{3}+\sin 3 y+3 a=0\end{array}\right.$, the function $f(t)=t^{3}+\sin t$ is an increasing function and an odd function in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and $x, 3 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $x=-3 y$, which means $x+3 y=0$, thus $\cos (x+3 y)=1$.