4. The solution to the equation $\sqrt{\sqrt{3}-\sqrt{\sqrt{3}+x}}=x$ is
4. $x=\frac{\sqrt{4 \sqrt{3}-3}-1}{2}$
Let $\sqrt{a+x}=y$, where $a=\sqrt{3}$. Then
$$
\left\{\begin{array} { l } 
{ \sqrt { a + x } = y , } \\
{ \sqrt { a - y } = x }
\end{array} \Rightarrow \left\{\begin{array}{l}
a+x=y^{2}, \\
a-y=x^{2} .
\end{array}\right.\right.
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
x+y=y^{2}-x^{2}=(y-x)(y+x) \\
\Rightarrow(x+y)(x-y+1)=0 .
\end{array}
$$

Thus, $x+y=0$ or $y=x+1$.
(1) If $x+y=0$, since $x \geqslant 0, y \geqslant 0$, it follows that $x=y=0$.
From $\sqrt{a+x}=y$, we know $a=0$, which contradicts $a=\sqrt{3}$.
(2) If $y=x+1$, i.e., $\sqrt{a+x}=x+1$, then, $x^{2}+x+1-a=0$,
i.e., $x^{2}+x+1-\sqrt{3}=0$.
Solving this, we get $x=\frac{\sqrt{4 \sqrt{3}-3}-1}{2}$.