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?
Best Answer
You're making things more complicated than they are
First, cosine is an even function, so the equation can be written $$\cos\Bigl(\frac{x}{2}-1\Bigr) =\cos^2\Bigl(\frac{x}{2}-1\Bigr)$$ as well. Note the equation implies $\cos\bigl(\frac{x}{2}-1\bigr)\ge 0$.
Also, if $0<c<1$, note that $c<c^2$ Therefore the equation is equivalent to $$\cos\Bigl(\frac{x}{2}-1\Bigr)=0\:\text{ or }\:1.$$ Now this is easy to solve in terms of congruences: \begin{cases} \cos\Bigl(\frac{x}{2}-1\Bigr)=0\iff \frac{x}{2}-1\equiv \frac\pi 2 \mod\pi \iff\frac x2\equiv 1+\frac\pi 2\mod\pi \\ \cos\Bigl(\frac{x}{2}-1\Bigr)=1\iff \frac{x}{2}-1\equiv 0\mod2\pi \iff\frac x2\equiv 1\mod2\pi \end{cases} and ultimately $$x\equiv2+\pi\bmod 2\pi \quad\text{ or }\quad x\equiv2\bmod 4\pi.$$