Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $$2\sin^{-1} x + 3\cos^{-1} x = \frac{2\pi}{5}$$, is ________

We need to solve: $$2\sin^{-1}x + 3\cos^{-1}x = \frac{2\pi}{5}$$.

Using $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$, so $$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$$.

Substituting: $$2\sin^{-1}x + 3\left(\frac{\pi}{2} - \sin^{-1}x\right) = \frac{2\pi}{5}$$

$$2\sin^{-1}x + \frac{3\pi}{2} - 3\sin^{-1}x = \frac{2\pi}{5}$$

$$-\sin^{-1}x = \frac{2\pi}{5} - \frac{3\pi}{2} = \frac{4\pi - 15\pi}{10} = -\frac{11\pi}{10}$$

$$\sin^{-1}x = \frac{11\pi}{10}$$

But the principal value of $$\sin^{-1}x$$ lies in $$[-\pi/2, \pi/2]$$, and $$\frac{11\pi}{10} > \frac{\pi}{2}$$. So there is no solution.

The number of real solutions is 0.

The answer is 0.