The number of solutions of the equation $$|\cot x| = \cot x + \frac{1}{\sin x}$$ in the interval $$[0, 2\pi]$$ is ________.

We need to solve $$|\cot x| = \cot x + \frac{1}{\sin x}$$ in $$[0, 2\pi]$$.

Note that $$\frac{1}{\sin x} = \csc x$$, so the equation is $$|\cot x| = \cot x + \csc x$$.

Rearranging: $$|\cot x| - \cot x = \csc x$$.

We consider two cases based on the sign of $$\cot x$$.

Case 1: $$\cot x \geq 0$$. Then $$|\cot x| = \cot x$$, so the equation becomes $$\cot x - \cot x = \csc x$$, i.e., $$0 = \csc x$$. Since $$\csc x = \frac{1}{\sin x}$$ is never zero, there are no solutions in this case.

Case 2: $$\cot x < 0$$. Then $$|\cot x| = -\cot x$$, so the equation becomes $$-\cot x - \cot x = \csc x$$, i.e., $$-2\cot x = \csc x$$.

Writing in terms of sine and cosine: $$-2 \cdot \frac{\cos x}{\sin x} = \frac{1}{\sin x}$$.

Multiplying both sides by $$\sin x$$ (noting $$\sin x \neq 0$$): $$-2\cos x = 1$$, so $$\cos x = -\frac{1}{2}$$.

In $$[0, 2\pi]$$, $$\cos x = -\frac{1}{2}$$ gives $$x = \frac{2\pi}{3}$$ and $$x = \frac{4\pi}{3}$$.

We need to check that $$\cot x < 0$$ at these points. At $$x = \frac{2\pi}{3}$$: $$\sin x > 0$$ and $$\cos x < 0$$, so $$\cot x < 0$$. This works. At $$x = \frac{4\pi}{3}$$: $$\sin x < 0$$ and $$\cos x < 0$$, so $$\cot x > 0$$. This does not satisfy $$\cot x < 0$$, so it is rejected.

Therefore, the number of solutions is $$1$$.