I'm looking for an algebraic solution to : $\cos(\frac{x}{2}-1) = \cos^2(1-\frac{x}{2})$. So I simplified the equation: first off, $\cos(\frac{x}{2}-1) = \cos(1-\frac{x}{2})$. Then I divided both sides by that. and so I'm left with two things to solve:
$\cos(\frac{x}{2}-1) = 0$ (because I divided both sides by that expression, I have to also include the $0 $ solution too). and $\cos(\frac{x}{2}-1) = 1$. And the general solution would be, I think, the union of those.
However, I'm kinda lost at this point. I've attempted to solve each equation. First off, I know that $\cos(x) = 0$ at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. So, the general solution for $\cos(x) = 0$ would be $x=\frac{\pi}{2} +2\pi k, \cup \ \frac{3\pi}{2}+2\pi k, k\in Z.$ I got up to this point, but don't know how to proceed.
The thing that's most confusing to me is I don't know how the $-1$ in the argument plays into the solution. Does it just change the graph to the right? Playing with desmos shows that graph is shifting by 2, but I thought that it'd shift by 1. More importantly: does it also affect the period of the function?
Additional question: In my book the answers are given in a different form. For example, the union I wrote would be written as: $x= (-1)^k\frac{\pi}{2} + \pi k, k \in Z.$ And in every case the period is "reduced" to $\pi k$. Why is that?