2. The solution set of the inequality $\sqrt{x-x^{2}}+\sqrt{4 x-x^{2}} \geqslant 2$ is $\qquad$ .
2. $\left\{\frac{4}{5}\right\}$.

From the domain, we get $0 \leqslant x \leqslant 1$.
$$
\begin{array}{l}
\text { By } \sqrt{x-x^{2}}+\sqrt{4 x-x^{2}} \geqslant 2 \\
\Rightarrow \sqrt{4 x-x^{2}} \geqslant 2-\sqrt{x-x^{2}} \\
\Rightarrow 4 \sqrt{x-x^{2}} \geqslant 4-3 x \\
\Rightarrow 25 x^{2}-48 x+16 \leqslant 0 \\
\Rightarrow(5 x-4)^{2} \leqslant 0 \Rightarrow x=\frac{4}{5} .
\end{array}
$$

Therefore, the solution set of the original inequality is $\left\{\frac{4}{5}\right\}$.