Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $$x$$ which satisfy $$\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1} x$$ is equal to:

We need to find the number of real solutions of $$\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1}(x)$$.

For the equation to be well-defined, we need $$\left|\frac{3x}{5}\right| \leq 1$$, $$\left|\frac{4x}{5}\right| \leq 1$$, and $$|x| \leq 1$$. The most restrictive condition is $$|x| \leq 1$$.

Clearly, $$x = 0$$ satisfies the equation since all three terms become 0.

For $$x \neq 0$$, we rearrange: $$\sin^{-1}\left(\frac{3x}{5}\right) = \sin^{-1}(x) - \sin^{-1}\left(\frac{4x}{5}\right)$$. Taking the sine of both sides and applying the identity $$\sin(A - B) = \sin A \cos B - \cos A \sin B$$:

$$\frac{3x}{5} = x\sqrt{1 - \frac{16x^2}{25}} - \frac{4x}{5}\sqrt{1 - x^2}$$

Since $$x \neq 0$$, dividing both sides by $$x$$ (noting the sign is preserved since sine inverse is odd):

$$\frac{3}{5} = \frac{1}{5}\sqrt{25 - 16x^2} - \frac{4}{5}\sqrt{1 - x^2}$$

Multiplying by 5: $$3 = \sqrt{25 - 16x^2} - 4\sqrt{1 - x^2}$$.

Let $$u = \sqrt{1 - x^2} \geq 0$$. Then $$25 - 16x^2 = 25 - 16(1 - u^2) = 9 + 16u^2$$. The equation becomes:

$$\sqrt{9 + 16u^2} = 3 + 4u$$

The right side must be non-negative, which holds since $$u \geq 0$$. Squaring both sides (valid since both sides are non-negative):

$$9 + 16u^2 = 9 + 24u + 16u^2$$

This simplifies to $$24u = 0$$, so $$u = 0$$, which gives $$1 - x^2 = 0$$, i.e., $$x = \pm 1$$.

Verification for $$x = 1$$: $$\sin^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{4}{5}\right)$$. Since $$\sin^{-1}\left(\frac{3}{5}\right)$$ and $$\sin^{-1}\left(\frac{4}{5}\right)$$ are complementary angles (as $$3, 4, 5$$ form a Pythagorean triple, $$\cos\left(\sin^{-1}\frac{3}{5}\right) = \frac{4}{5}$$), their sum equals $$\frac{\pi}{2} = \sin^{-1}(1)$$. Valid.

Verification for $$x = -1$$: By the odd symmetry of $$\sin^{-1}$$, $$\sin^{-1}\left(-\frac{3}{5}\right) + \sin^{-1}\left(-\frac{4}{5}\right) = -\frac{\pi}{2} = \sin^{-1}(-1)$$. Valid.

Therefore, there are 3 real values of $$x$$: $$x \in \{-1, 0, 1\}$$.