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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.

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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*}

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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: enter image description here

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.

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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 $\arccos,$

$\arccos(\cos y)=\begin{cases} y &\mbox{if } 0\le y\le\pi \\ 2\pi-y & \mbox{if } \pi<y<2\pi \end{cases}$

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  • $\begingroup$ Then $x+A=n\pi$? $\endgroup$ – Shemafied May 27 '15 at 16:30
  • $\begingroup$ @Shemafied, So, $x\pm A=n\pi,$ right? $\endgroup$ – lab bhattacharjee May 27 '15 at 16:31
  • $\begingroup$ And that's where the two possible solutions come from. $\endgroup$ – Shemafied May 27 '15 at 16:33
  • $\begingroup$ Making "Prosthaphaeresis" my word of the day. ;) $\endgroup$ – user137731 May 27 '15 at 17:00

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