Let $$S = \left\{x : \cos^{-1} x = \pi + \sin^{-1} x+\sin^{-1}(2x+1)\right\}$$. Then $$\sum_{x \in S}^{}(2x-1)^{2}$$ is equal to_________.

For $$\cos^{-1} x$$: we need $$-1 \le x \le 1$$.

For $$\sin^{-1} x$$: we need $$-1 \le x \le 1$$.

For $$\sin^{-1}(2x+1)$$: we need $$-1 \le 2x+1 \le 1$$, i.e., $$-1 \le x \le 0$$.

So the domain is $$x \in [-1, 0]$$.

Substituting $$\cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x$$:

$$\frac{\pi}{2} - \sin^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x+1)$$

$$-\frac{\pi}{2} = 2\sin^{-1} x + \sin^{-1}(2x+1)$$

$$\sin^{-1}(2x+1) = -\frac{\pi}{2} - 2\sin^{-1} x$$

$$2x+1 = \sin\left(-\frac{\pi}{2} - 2\sin^{-1} x\right) = -\sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right) = -\cos(2\sin^{-1} x)$$

Let $$\theta = \sin^{-1} x$$, so $$\sin\theta = x$$.

$$\cos(2\theta) = 1 - 2\sin^2\theta = 1 - 2x^2$$

$$2x + 1 = -(1 - 2x^2) = 2x^2 - 1$$

$$2x^2 - 2x - 2 = 0$$

$$x^2 - x - 1 = 0$$

$$x = \frac{1 \pm \sqrt{5}}{2}$$

$$x = \dfrac{1 + \sqrt{5}}{2} \approx 1.618$$ — outside the domain.

$$x = \dfrac{1 - \sqrt{5}}{2} \approx -0.618$$ — inside $$[-1, 0]$$. Valid.

For $$x = \dfrac{1-\sqrt{5}}{2}$$: we check that $$\sin^{-1}(2x+1) = \sin^{-1}(2-\sqrt{5})$$, and $$2-\sqrt{5} $$, which is in $$[-1, 1]$$, so the inverse sine is defined.

The RHS value $$-\frac{\pi}{2} - 2\sin^{-1}x $$ and $$\sin^{-1}(-0.236) $$

$$S = \left\{\dfrac{1-\sqrt{5}}{2}\right\}$$, so:

$$\sum_{x \in S}(2x-1)^2 = \left(2 \cdot \frac{1-\sqrt{5}}{2} - 1\right)^2 = (1 - \sqrt{5} - 1)^2 = (-\sqrt{5})^2 = 5$$

The answer is $$\boxed{5}$$.