2、The solution set of the equation $16 \sin \pi x \cos \pi x=16 x+\frac{1}{x}$ is
2. $\left\{\frac{1}{4},-\frac{1}{4}\right\}$

Analysis: The original equation is $8 \sin 2 \pi x=16 x+\frac{1}{x}$. When $x>0$, $16 x+\frac{1}{x} \geq 8 \geq 8 \sin 2 \pi x$, equality holds if and only if $x=\frac{1}{4}$; When $x<0$, $16 x+\frac{1}{x} \leq-8 \leq 8 \sin 2 \pi x$, equality holds if and only if $x=-\frac{1}{4}$. Therefore, the solution set of the original equation is $\left\{\frac{1}{4},-\frac{1}{4}\right\}$.