Solution of the equation $\cot \theta = 2\cot 2\theta$ I've tried to solve the equation $\cot \theta = 2\cot 2\theta$ with the  command 'Reduce' of Mathematica and obtained $\theta = n\pi$ as the solution with n an integer. But $\theta=n\pi$ is clearly a singularity in a cotangent function so this is puzzling.
I've realized that the above equation can be simplified to:
$\tan \theta=0$, and that is probably what Mathematica does to obtain the solution, but again, how $\theta=n\pi$ can be the solution if it is not in the domain of a $\cot$ function?
Is that an inconsistency in Mathematica or am I missing something?
 A: Reciprocate both sides:
$$\tan(\theta)=\frac12\tan(2\theta)=\frac{\sin(2\theta)}{2\cos(2\theta)}$$
$$=\frac{2\sin(\theta)\cos(\theta)}{2\cos(2\theta)}=\frac{\sin(\theta)\cos(\theta)}{2\cos^2(\theta)-1}$$
$$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$
Which allows us to cancel the sines and move all the cosines to the right.
$$1=\frac{\cos^2(\theta)}{2\cos^2(\theta)-1}$$
$$2\cos^2(\theta)-1=\cos^2(\theta)$$
$$\cos^2(\theta)-1=0$$
$$\cos(\theta)=\pm1$$
$$\theta=\pm n\pi$$
$$n=0,1,2,3,\dots$$
As I just noticed, along with the comments, is that the solution $\theta=\pm n\pi$ does not work for the original problem.
So, perhaps, I may note:
$$\lim_{\theta\to\pm n\pi}\cot(\theta)=\lim_{\theta\to\pm n\pi}2\cot(\theta)$$
Since the limit tends to be undefined, one notes that in a calculus sense, the problem is equivalent to:
$$\frac{2\cot(2\theta)}{\cot(\theta)}=1$$
And since the limit holds true, it is somewhat safe to say that is the answer.
However, take it as you will, accept the answer as correct if you want, there are no other "answers" other than maybe undefined.
A: If you expand $\cot2\theta$, you get
$$
\cot\theta=\frac{\cot^2\theta-1}{\cot\theta}
$$
that's clearly an inconsistent equation.
On the other hand, if you rewrite it as
$$
\tan2\theta=2\tan\theta
$$
and expand, you get
$$
\frac{\tan\theta}{1-\tan^2\theta}=\tan\theta
$$
that has the solutions $\tan\theta=0$, but they aren't solutions of the original equation, unless you allow $\infty$ as a value.
Depending on its implementation and conventions, a piece of software can give different results from “inconsistent”.
A: HINT: your equation is equivalent to $$\frac{1}{\cot(\theta)}=0$$
