2. $[0,4]$.
(Method 1) It is easy to see that
$$
\begin{array}{l}
\sqrt{2 x+2-2 \sqrt{2 x+1}}+\sqrt{2 x+10-6 \sqrt{2 x+1}} \\
=\sqrt{(\sqrt{2 x+1})^{2}-2 \sqrt{2 x+1}+1^{2}}+\sqrt{(\sqrt{2 x+1})^{2}-6 \sqrt{2 x+1}+3^{2}} \\
=|\sqrt{2 x+1}-1|+|\sqrt{2 x+1}-3| \\
\geqslant 2,
\end{array}
$$

where the equality holds if and only if $1 \leqslant \sqrt{2 x+1} \leqslant 3$, i.e., $x \in[0,4]$.
(Method 2) According to the problem, we can set
$$
\begin{array}{c}
\sqrt{2 x+2-2 \sqrt{2 x+1}}=1-d, \\
\sqrt{2 x+10-6 \sqrt{2 x+1}}=1+d,
\end{array}
$$

From
$$
\left\{\begin{array}{l}
1-d \geqslant 0 \\
1+d \geqslant 0
\end{array}\right.
$$

we get $-1 \leqslant d \leqslant 1$. By $(2)^{2}-(1)^{2}$, we get
$$
\sqrt{2 x+1}=2-d,
$$

Substituting into equations (1) and (2), we get
$$
\left\{\begin{array}{l}
\sqrt{1+(2-d)^{2}+2(2-d)}=1-d, \\
\sqrt{9+(2-d)^{2}+6(2-d)}=1+d
\end{array}\right.
$$

which holds for all $-1 \leqslant d \leqslant 1$.
Therefore, the original equation holds as long as $\sqrt{2 x+1}=2-d \in[1,3]$, i.e., $x \in[0,4]$.
2. The solution set of the equation $\sqrt{2 x+2-2 \sqrt{2 x+1}}+\sqrt{2 x+10-6 \sqrt{2 x+1}}=2$ is