Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $$\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$$ is equal to

We need to find the sum of all solutions of $$\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$$.

Step 1: Use the identity $$\cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x)$$.

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

$$\frac{\pi}{2} - 3\sin^{-1}(x) = \cos^{-1}(2x)$$

Step 2: Take cosine of both sides.

$$\cos\left(\frac{\pi}{2} - 3\sin^{-1}(x)\right) = 2x$$

$$\sin(3\sin^{-1}(x)) = 2x$$

Step 3: Expand using the triple angle formula.

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

$$\sin(3\theta) = 3\sin\theta - 4\sin^3\theta = 3x - 4x^3$$

Setting equal to $$2x$$:

$$3x - 4x^3 = 2x$$

$$x - 4x^3 = 0$$

$$x(1 - 4x^2) = 0$$

$$x = 0$$ or $$x^2 = \frac{1}{4}$$, i.e., $$x = \pm\frac{1}{2}$$

Step 4: Check domain constraints.

For $$\cos^{-1}(2x)$$ to be defined: $$-1 \leq 2x \leq 1 \Rightarrow -\frac{1}{2} \leq x \leq \frac{1}{2}$$.

Also, $$\cos^{-1}(x)$$ requires $$-1 \leq x \leq 1$$, which is already satisfied.

Additionally, LHS = $$\frac{\pi}{2} - 3\sin^{-1}(x)$$ must be in $$[0, \pi]$$ (range of $$\cos^{-1}$$).

Step 5: Verify each solution.

$$x = 0$$: LHS = $$\cos^{-1}(0) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$. RHS = $$\cos^{-1}(0) = \frac{\pi}{2}$$. Valid. ✓

$$x = \frac{1}{2}$$: LHS = $$\cos^{-1}\left(\frac{1}{2}\right) - 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} - 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} - \frac{\pi}{3} = 0$$. RHS = $$\cos^{-1}(1) = 0$$. Valid. ✓

$$x = -\frac{1}{2}$$: LHS = $$\cos^{-1}\left(-\frac{1}{2}\right) - 2\sin^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} - 2\left(-\frac{\pi}{6}\right) = \frac{2\pi}{3} + \frac{\pi}{3} = \pi$$. RHS = $$\cos^{-1}(-1) = \pi$$. Valid. ✓

Step 6: Find the sum.

$$0 + \frac{1}{2} + \left(-\frac{1}{2}\right) = 0$$

The correct answer is Option A: $$0$$