1. The solution to the equation $\frac{16}{3} x^{4}+\frac{1}{6 x^{2}}=\sin \pi x$ is
$-1 . \frac{1}{2}$.
Notice,
$$
\begin{array}{l}
\frac{16}{3} x^{4}+\frac{1}{12 x^{2}}+\frac{1}{12 x^{2}} \\
\geqslant 3 \sqrt[3]{\frac{16}{3} x^{4} \cdot \frac{1}{12 x^{2}} \cdot \frac{1}{12 x^{2}}}=1 \\
\geqslant \sin \pi x .
\end{array}
$$

The equality holds if and only if $x=\frac{1}{2}$.
Thus, $x=\frac{1}{2}$.