Substituting cot(x) with an identity in equation Assume:
$$\cos(x) / \cot(x) = 0\tag A$$
I rewrite it as
$$\cos(x)/ \left(\cos(x)/\sin(x)\right) = 0\tag B$$
and get
$$\sin(x) = 0\tag C$$
This implies $x = 180^\circ\times k$
But, under (B), if $\sin(x) = 0$ I get undefined for $\cos(x)/\sin(x)$ and (C) would not be possible.  
Also, $\cot(0)$ is undefined.
 A: This happens sometimes.  Since $\cot$ has a domain that excludes multiples of $180^\circ$, any "solutions" to an equation involving $\cot$ that are themselves mutliples of $180^\circ$ don't really exist.  These are called "Extraneous (or Missing or Spurious) solutions".
Similarly, the equation $\cos(x) / \cot(x) = 1$ doesn't have any solutions either, because in order for it to happen you get $\sin(x)=1$ but that means $\cos(x)=\cot(x)=0$ and we're again dividing by zero.
In general, when trying to solve equations, you must beware of situations where the solution ends up not being in the valid domain of the original expressions.
Usually when this happens, it happens because naive simplification -- that is, simplification without keeping the domain restrictions in mind -- makes it look like it should work.  This also means that a graphical inspection looks like it should work too, if not done carefully.  Here's the correct graph of $\cos(x)/\cot(x)$:

Note the discontinuities every $90^\circ$.  These come from the fact that $\cot(x)$ is undefined for $x = 180^\circ \times k$ and zero (thus forcing division by zero) for $x = 90^\circ + 180^\circ \times k$, for integer $k$.  Most computer graphing systems will not show these, because it will only show if it happens to exactly pick an $x$ value that lands there.
