1. Use the principal-value identity

For real $x \in [-1,1]$, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}.$

Let $\sin^{-1}x = y.$ Then $\cos^{-1}x = \frac{\pi}{2} - y.$

2. Substitute into the given equation

Given $2\sin^{-1}x + 3\cos^{-1}x = \frac{2\pi}{5},$ so $2y + 3\left(\frac{\pi}{2} - y\right) = \frac{2\pi}{5}.$

Now simplify: $2y + \frac{3\pi}{2} - 3y = \frac{2\pi}{5}$ $-y + \frac{3\pi}{2} = \frac{2\pi}{5}$ $y = \frac{3\pi}{2} - \frac{2\pi}{5} = \frac{15\pi - 4\pi}{10} = \frac{11\pi}{10}.$

So we get $\sin^{-1}x = \frac{11\pi}{10}.$

3. Check principal value range

The principal value range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$

But $\frac{11\pi}{10} > \frac{\pi}{2},$ so this is impossible for any real $x$.

Hence, there is no real solution.

4. Number of real solutions

Therefore, the number of real solutions is $0.$