Basic trigonometry intuition I have already posted a question regarding the same function here
However, now I simply can not grasp why the function has to have two solution sets:$$\cos y=\cos \Bigl(\frac{\pi }{2}-4x\Bigr)\iff\begin{cases}y= \dfrac{\pi }{2}-4x + 2k\pi \\\text{or}\\y= 4x- \dfrac{\pi }{2}+2k\pi\end{cases}$$
Is it because we can write $\cos(y)=\cos\begin{pmatrix}
\left | \frac{\pi}{2}-4x+2k\pi \right |
\end{pmatrix}$, therefore having two possible cases  ($\cos$ being an even function)? How would one deal with a function something like $\arcsin \left ( \cos 4x \right )$? If someone could explain it graphically, that would be totally awesome.
 A: For simplicity's sake, write $u = \frac{\pi}2 - 4x$; then the problem you are trying to solve becomes
$$\cos y = \cos u$$
Now let's think about when it's possible for $y$ and $u$ to have the same cosine.  If you interpret $y$ as an angle (measured in radians) that locates a point on a unit circle, then $\cos y$ tells you the $x$-coordinate of that point.  Then $\cos y = \cos u$ means that the angles $y$ and $u$ locate points with the same $x$-coordinate.  There are two ways that can happen, illustrated by the following diagrams:

The left-hand picture shows $y$ and $u$ locating the same point on the unit circle.  This happens when $y=u$ (obviously) or when $y=u+2\pi k$ (slightly less obvious).  This is the part of the solution you are apparently comfortable with.
The right-hand picture shows $y$ and $u$ locating points that are directly above/below one another on the unit circle.  This is the part of the solution you are having trouble understanding.  It happens when $y = -u$ (obvious, once you have the picture) or when $y = -u + 2 \pi k$ (slightly less obvious).
Now go back and substitute in $u = \frac{\pi}2 - 4x$ and the full solution should make sense.
A: It is a basic fact about trigonometric equations that:
\begin{align*}
\cos x=\cos\theta &\iff \begin{cases}x\equiv\theta\\x\equiv-\theta\end{cases} \mod 2\pi \\
\sin x=\sin\theta &\iff \begin{cases}x\equiv\theta\\x\equiv\pi-\theta\end{cases} \mod 2\pi \\
\tan x=\tan\theta &\iff\quad x\equiv\theta \mod \pi \\
 \end{align*}



A: If $\cos2x=\cos2A$
using Prosthaphaeresis Formula, $2\sin(x+A)\sin(x-A)=0$
If $\sin(x-A)=0,x-A=n\pi$ where $n$ is any integer
What if $\sin(x+A)=0?$

$\arcsin(\cos4x)=\dfrac\pi2-\arccos(\cos4x)$
Find suitable integer value of $m$ for $y=2m\pi+4x$ i.e.,$y\equiv4x\pmod{2\pi}$ such that $0\le y<2\pi$
Use definition of the principal value of $\arccos,$
$\arccos(\cos y)=\begin{cases} y &\mbox{if } 0\le y\le\pi \\
2\pi-y & \mbox{if } \pi<y<2\pi \end{cases}$
