2. Find the real solution to the equation $\sqrt[3]{x(3+\sqrt{8 x-3})-1}+\sqrt[3]{x(3-\sqrt{8 x-3})-1}=1$.

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 translated text:
Find the real solution to the equation $\sqrt[3]{x(3+\sqrt{8 x-3})-1}+\sqrt[3]{x(3-\sqrt{8 x-3})-1}=1$.

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Let $a=\sqrt[3]{x(3+\sqrt{8 x-3})-1}, b=\sqrt[3]{x(3-\sqrt{8 x-3})-1} \Rightarrow\left\{\begin{array}{l}a^{3}+b^{3}=6 x-2, \\ a+b=1\end{array}\right.$ $\Rightarrow a^{2}+b^{2}-a b=(a+b)^{2}-3 a b=1-3 a b=6 x-2 \Rightarrow a b=1-2 x$.
Then let $t=a+b$, we have $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$\Rightarrow t^{3}+3(2 x-1) t-(6 x-2)=(t-1)\left(t^{2}+t+6 x-2\right)=0$. Since $x \geqslant \frac{3}{8}$, and when $x=\frac{3}{8}$, $t=1$;
thus $t^{2}+t+6 x-2=\left(t+\frac{1}{2}\right)^{2}+6\left(x-\frac{3}{8}\right)>0$, meaning for any $x \geqslant \frac{3}{8}$, we have $t=a+b=1$.
Therefore, the set of real solutions to the equation is $\left\{x \in \mathbf{R} \left\lvert\, x \geqslant \frac{3}{8}\right.\right\}$.