The general solution of $\tan(x) = \tan(2x-π/2)$ The general solution given by the teacher is $$x = \fracπ2 + kπ.$$
But to me, this solution is wrong because $\tan(π/2)$ has an indeterminate value (infinite). How do I know that one infinity is equal to the other? It has no sense.
Therefore, I would put that it has no solution for the real values. Am I correct?
 A: The solution you were presented is surely wrong, or at least would require further explanation of context. It does not make sense to have a solution outside the domain of definition of the involved expressions. The two graphs never intersect and there is no solution to the equation.
A: If you set $t=\tan x$ and apply the duplication formula for the tangent, you  have
$$\tan\Bigl(2x-\frac \pi 2\Bigr)=-\frac 1{\tan 2x}=\frac{t^2-1}{2t},$$
so the equation becomes
$$\frac{t^2-1}{2t}=t\iff t^2-1=2t^2\iff t^2=-1,$$
which has no real solution.
A: $$\tan{(2x-\frac{\pi}{2})}=-\tan{(\frac{\pi}{2}-2x)}=-\frac{1}{\tan{2x}}$$
$$\tan{x}=-\frac{1}{\tan{2x}}\rightarrow\tan{x}\tan{2x}=-1$$
$$\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}=-1\rightarrow2\tan^2{x}=-1+\tan^2{x}\rightarrow\tan^2{x}=-1!$$
Somthing is wrong in the problem!
A: You are halfway there with your approach. Note that $\tan(x)$ has period $\pi$, so if $x + \pi = 2x - \pi/2$, then $\tan(x) = \tan(x + \pi) = \tan(2x - \pi/2)$ as well.
Extending this idea to shifting $\tan(x)$ by $k$ periods left/right, where $k$ is any integer, we need that $x + k \pi = 2x - \pi/2 \implies k \pi + \pi/2 = x$. But $\tan(k \pi + \pi/2) = \tan(\pi/2)$ by periodicity, which is undefined!
Hence there is no solution for $x$ (the empty set).
