The number of solutions of the equation $$\sin x = \cos^2 x$$ in the interval $$(0, 10)$$ is ______.

Given,

$$\sin x=\cos^2 x$$

Using

$$\cos^2 x=1-\sin^2 x$$

we get

$$\sin x=1-\sin^2 x$$

$$\sin^2 x+\sin x-1=0$$

Let

$$\sin x=t$$

Then,

$$t^2+t-1=0$$

$$t=\frac{-1\pm\sqrt5}{2}$$

Since

$$-1\le\sin x\le1$$

only

$$\sin x=\frac{\sqrt5-1}{2}$$

is valid.

Now,

$$0<\frac{\sqrt5-1}{2}<1$$

Hence, in each interval of length $$2\pi$$, there are two solutions.

Since

$$10<4\pi$$

the interval $$(0,10)$$ contains one complete interval $$[0,2\pi]$$ and part of the next interval.

The solutions in $$(0,2\pi)$$ are two.

Also, in $$(2\pi,10)$$, two more solutions occur because

$$2\pi+\sin^{-1}\left(\frac{\sqrt5-1}{2}\right)<10$$

and

$$3\pi-\sin^{-1}\left(\frac{\sqrt5-1}{2}\right)<10$$

Therefore, the total number of solutions is

$$\boxed{4}$$