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

When $x>0$,
$$
16 x+\frac{1}{x} \geqslant 8 \text {, }
$$

with equality holding if and only if $x=\frac{1}{4}$.
Also, $16 \sin \pi x \cdot \cos \pi x=8 \sin 2 \pi x \leqslant 8$, with equality holding if and only if $x=\frac{1}{4}+k(k \in \mathbf{Z})$. Therefore, when $x>0$, the equation has only the solution $x=\frac{1}{4}$. By the property of odd functions, we know that $x=-\frac{1}{4}$ is another solution. Hence, the solution set of the equation is $\left\{\frac{1}{4},-\frac{1}{4}\right\}$.