Problem 861

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., $x \in[0,4]$.