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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?

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    $\begingroup$ So if the solution for $\cos(x) =0$ is $(x) =\pi/2+2\pi k$ then the solution to $\cos(x/2-1)=0$ is $(x/2-1)=\pi/2+2\pi k$ similarly for the other solution. $\endgroup$ – kingW3 Jun 23 '20 at 23:49
  • $\begingroup$ well, $\cos (-u) = \cos u$ $\endgroup$ – Will Jagy Jun 23 '20 at 23:51
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Okay so, you noticed that ${\cos\left(\frac{x}{2}-1\right)=\cos\left(1-\frac{x}{2}\right)}$. And as you say, you end up with

$${\Leftrightarrow \cos^2\left(\frac{x}{2}-1\right)=\cos\left(\frac{x}{2}-1\right)}$$

And this implies

$${\cos\left(\frac{x}{2}-1\right)\left(\cos\left(\frac{x}{2}-1\right)-1\right)=0}$$

(as was pointed out by someone else - indeed it's probably bad practice to divide through here by ${\cos}$. Not that it's incorrect, since you did take into account the fact we would then miss the $0$ solution (which was awesome!!!) - but unnecessary).

For simplicity, we can replace ${\frac{x}{2}-1}$ with ${u}$ and just rearrange for ${x}$ at the end. So

$${\Rightarrow \cos(u)\left(\cos(u)-1\right)=0}$$

Now.... we want to solve ${\cos(u)=0}$. As you said, one solution is ${\frac{\pi}{2}}$... and if you take a look at the graph, you will notice that every other $0$ to the cosine function can be "reached" by hopping foward and backwards by multiplies of ${\pi}$... you are right in thinking indeed, the period of the cosine function is ${2\pi}$, but in fact ${0}$ reoccurs every ${\pi}$ radians. There is no problem with this. In order to have a full period (${2\pi}$ in this case), every value the function takes on must reoccur - only $0$ has. This actually means that

$${\cos(u)=0\Leftrightarrow u=\frac{\pi}{2}+n\pi, n \in \mathbb{Z}}$$

Now we can solve ${\cos(u)=1}$ (the other solution we need). Of course we have a ${1}$ at the point ${x=0}$, and you may notice that we can reach all the other 1's by jumping foward and backwards by multiples of ${2\pi}$ this time. Hence the solution is

$${\cos(u)=1\Leftrightarrow u=0 + 2n\pi =2n\pi, n \in \mathbb{Z}}$$

Now, you can simply plug back in the definition of ${u}$ in terms of ${x}$, adding ${1}$ and multiplying both sides by two and you will end up with the solutions being

$${x=\pi + 2n\pi + 2=(2n+1)\pi + 2, n \in \mathbb{Z}}$$

$${x=4n\pi + 2, n \in \mathbb{Z}}$$

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    $\begingroup$ This explained things very clearly and helped me solved another, similar problem too!. Thanks!!. $\endgroup$ – Ebrin Jun 24 '20 at 0:33
  • $\begingroup$ No problem at all! :) $\endgroup$ – Riemann'sPointyNose Jun 24 '20 at 0:35
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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 $$x\equiv2+\pi\bmod 2\pi \quad\text{ or }\quad x\equiv2\bmod 4\pi.$$

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  • $\begingroup$ Oh, the first part explains why the answer only contained the positive answers. So, the -1 in the argument just corresponds to the graph shifting to the right by two, right? It has no impact on the period. $\endgroup$ – Ebrin Jun 24 '20 at 0:07
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    $\begingroup$ Absolutely no impact. $\endgroup$ – Bernard Jun 24 '20 at 0:09
  • $\begingroup$ @kingW3: You're right. I shouldn't calculate directly on screen. Thank you for pointing it! $\endgroup$ – Bernard Jun 24 '20 at 0:17
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Approach to the problem I would like to discuss an approach and leave solving the equations to you. $\cos(y) = 0$ implies that y is an odd multiple of $\frac{\pi}{2}$ or $(2n + 1)\frac{\pi}{2}$, where n is an integer. You can even write $(2n - 1)$, which is fine too as it gives odd vales. Similarly, $\cos(\frac{x}{2} - 1) = 0$ implies that $\frac{x}{2} - 1 = (2n + 1)\frac{\pi}{2}.$ You can simplify it further. Then solve the next one which is $\cos(\frac{x}{2} - 1) = 1 $which implies that $\frac{x}{2} - 1 = 2n\pi$ (even multiple of pi), where $n$ is an integer. The $'n'$ in both solutions can be same because the first solution will always give an odd value of $\frac{\pi}{2}$ and the second solution will always give an even value of $\frac{\pi}{2},$ so we can use the same symbol 'n'. Now, to unify both solutions: Observe if there is any similarity between first and second solution. Is there any value(s) or range for which both solutions are same? If yes, we can get an even compact expression. If no, then write both solutions joined together with union symbol.

Regarding graph To interpret a function $f(x - 1) $interms of $f(x)$: The value which you get on putting $x$ in a function $f(x)$ will now [in $f(x - 1)]$ be obtained by setting $x$ as $(x+1)$ in $f(x - 1)$. Put $x+1$ in place of $x$ in $f(x-1)$, you will get $f(x)$ again. What does it mean? It means that you will have to shift your graph of $f(x)$ towards right by $1$ unit. Practice by drawing the graph of $f(x) = x^2.$ Then draw $f(x) = (x - 1)^2.$ But your problem has $\cos(\frac{x}{2} - 1)$ which is similar to $f(ax + b)$. We can now interpret $'b'$ as we did in last para. What about $'a'$?

Think like this. Every value of function $f(x)$ is now obtained at $\frac{x}{a}$ when the function is transformed to $f(ax)$. To show this, set $x$ as $\frac{x}{a}$ in $f(ax)$, you will get f(x) again. It also means that the frequency of the function $f(ax)$ is now decreased by $a$ factor of $'a'$.

In case of $f(\frac{x}{a})$, it can be said that every old value of $f(x)$ is obtained at $'ax'$ (show). Here the frequency is increased by a factor of $'a'$. Every old value of $f(x)$ is obtained quickly with $f(\frac{x}{a}).$ You can relate this with period since frequency and period are related. Or, you could interpret it in terms of period in the first place itself. If the frequency has increased, it means the value of the function will be repeated quickly, thereby decreasing the period. Period of $\cos(x)$ is $2\pi$. Period of $\cos(ax+b)$ will be $\frac{2\pi}{a}.$ The constant b only shifts the values, it does not account for frequency or period.

Thus, your interpretation of graph of $\cos(\frac{x}{2} - 1)$ is incomplete. To obtain its graph, we have to expand the graph of $\cos(x)$ by a factor of $2$, then shift the graph towards right by $1$ unit.

Regarding your last doubt In academics it is completely fine to represent the solution as simple as possible according to you. Generally, we want to make it more compact. The solution which you have wrote there. Put integral values of $k$. You will find that you have actually written odd multiples of $\frac{\pi}{2}$ in any case. Try to put several odd and even (positive integers for ease) integral values for $k$ and interpret the result yourself.

I thank @Ebrin for his help in formatting this answer.

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    $\begingroup$ Sorry for the late reply, I've read your answer thoroughly and found it very helpful. Thanks for taking the time and explaining things! $\endgroup$ – Ebrin Jun 28 '20 at 15:59
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So if $\cos W = 0$ then $W = \pm \frac \pi 2 + 2k\pi$. Notice $-\frac \pi 2 + 2k \pi = \frac \pi 2 + (2k-1)\pi$. And $\frac \pi 2 + 2k \pi = -\frac \pi 2 + (2k+1) \pi$. So to simplify $W = \frac \pi 2 + m\pi$ would be the simplest way to state this.

(Alternativily as $0 = -0$ and $\cos(W \pm \pi) = -\cos W$ we'd note that for an $W = \pm \frac \pi 2 +2k \pi$ then $\mp \frac \pi 2+ (2k+1)\pi$ is a solution).

So $\frac x2 -1 = \frac \pi 2 + m\pi$ then $x = (2m+1)\pi + 2$

And if $\cos W = 1$ then .... well... do I need to point out that solution is $W=m\pi$?

So $\frac x2 -1= m\pi$ so $x=2m\pi +1$.

So the solutions are $x = k \pi + 2$. If $k$ is odd then $\cos(\frac x2 -1)=\cos(\frac k2 \pi) = 1$. And if $k$ is odd then $\cos(\frac x2 -1)=\cos(\frac {k-1}2\pi + \frac \pi 2) = 0$.

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