Solve the equation sin2x = cotx

We know that $\sin(2x) = 2\sin(x)\cos(x)$ and $\cot(x) = \frac{\cos(x)}{\sin(x)}$.
Substitute these into the given equation:

Note that for $\cot(x)$ to be defined, $\sin(x) \neq 0$. This means $x \neq n\pi$ for any integer $n$.

Multiply both sides by $\sin(x)$ (which is non-zero based on the restriction):

Move all terms to one side:

This gives two separate cases:
Case 1: $\cos(x) = 0$
Case 2: $2\sin^2(x) - 1 = 0$

If $\cos(x) = 0$, then $x$ is an odd multiple of $\frac{\pi}{2}$.

where $n$ is an integer.

If $2\sin^2(x) - 1 = 0$, then:

We can use the identity $\cos(2x) = 1 - 2\sin^2(x)$.
So, $2\sin^2(x) - 1 = -(1 - 2\sin^2(x)) = -\cos(2x)$.
Therefore, the equation becomes $-\cos(2x) = 0$, or $\cos(2x) = 0$.
If $\cos(2x) = 0$, then $2x$ is an odd multiple of $\frac{\pi}{2}$.

Divide by 2:

where $n$ is an integer.

For $x = \frac{\pi}{2} + n\pi$, $\sin(x) = \pm 1$, which is not zero. These solutions are valid.
For $x = \frac{\pi}{4} + \frac{n\pi}{2}$, $\sin(x) = \pm \frac{\sqrt{2}}{2}$, which is not zero. These solutions are valid.

The general solutions are the union of the solutions from Case 1 and Case 2.
The solutions are:

where $n \in \mathbb{Z}$.

The final answer is $\boxed{x = \frac{\pi}{2} + n\pi \text{ or } x = \frac{\pi}{4} + \frac{n\pi}{2}, \text{ where } n \in \mathbb{Z}}$.