First question
Find distinct values of $\sin x$ given that $\sin 5x =0.5$
my method
There will be $5$ unique values of $\sin x$
working in first quadrant
$5x = \pi/6 + 2\pi m$ for $m=0,\pm 1,\pm2,...$ (I use values of $m$ that keep $-\pi \le x\le \pi$)
$x = \pi/30, 13\pi/30, -11\pi/30, ... $
working in second quadrant
$5x = 5\pi/6 + 2\pi m$ for $m=0,\pm 1,\pm2,...$ (I use values of $m$ that keep $-\pi \le x\le \pi$)
$x = 5\pi/30, 17\pi/30, -7\pi/30,...$
Eliminate duplicates using the identity $\sin(\pi-x)=\sin x$
This gives $\sin x = \sin (\pi/30), \sin(5\pi/30), \sin (-7\pi/30), \sin (-11\pi/30),\sin(13\pi/30)$
My method is quite time-consuming because it generates duplicate solutions.
Is there a way to know immediately what values of $m$ will give the unique values of $\sin x$ ? Or is there a another way to approach this question?
Second question
If the question was instead, :
Find distinct values of $\cos x$ given that $\cos 4x=0.5$
, the values of $x$ that give unique values of $\cos x$ are simply :
First quadrant : $4x=\pi/3+2\pi m$ for $m=0,1$
Fourth quadrant : $4x=5\pi/3+2\pi m$ for $m=0,1$
Here there was no need to use negative values of $m$. How do I know when to use positive/negative values of $m$ ?
Thank you.
The intuitive shortcut to identifying all values $x$, within a modulus of $2\pi$ such that $\sin(5x) = 0.5$ is as follows.
First of all, you want to visualize the problem against the backdrop of the sine function superimposed on the unit circle. Here, the sine function represents the height (or $y$ coordinate) as $\theta$ goes from $0$ through $2\pi$.
So, it should be immediately apparent that for any value within the $(-1,1)$ boundary, there will be $2$ angles that satisfy the condition. Then, for the specific values $1,-1$, there is only one angle for +1, and one such angle for $-1$.
The 2nd portion of the intuitive shortcut involves being exposed to the $n$ roots of the complex equation $z^n = 1.$
Let $f(k)$ denote $\cos(2k\pi/n) + i\sin(2k\pi/n) ~: ~k \in \Bbb{Z_{\geq 0}}.$
You learn that for $k \in \{0,1,2,\cdots,n-1\}$, the function $f(k)$ yields $n$ distinct values. Further, you realize, that by DeMoivre's rule, $[f(k)]^n = 1,$ for each of those values of $k$.
So, then your intuition is stretched.
When you see a function like $\sin(5x)$, you know that the sine function is periodic, with period $(2\pi)$. So, based on the concepts in the last few paragraph, your intuition naturally considers what happens if you take any value $\theta$, and compare $\sin(5\theta)$ against $\sin(5[\theta + 2k\pi/5]) ~: ~k \in \{0,1,2,\cdots, 4\}.$
So, once your intuition has been stretched, when you see a problem like:
it becomes 2nd nature to think that
$5\theta = \pi/6 = 30^\circ$ or $5\theta = 5\pi/6 = 150^\circ.$
One pair of solutions is therefore $\theta = (30^\circ)/5$ and $\theta = (150^\circ/5)$.
You instinctively know that the single pair of solutions immediately above can be converted into $5$ distinct pairs of solutions, that involve incorporating $2k\pi/5, ~: ~k \in \{0,1,2,3,4\}.$
Again, the key here is to have your intuition so developed that Mathematical analysis is virtually unnecessary. You know that you want $(4x) \in \{\pi/3, -\pi/3 = 5\pi/3\}.$
You also know that the single pair of solutions generated by $x = \pm \frac{\pi}{3 \times 4}$ will be converted into $4$ distinct pairs of solutions, by consideration of $2k\pi/4 ~: k \in \{0,1,2,3\}.$
Addendum
I may have introduced a point of confusion.
The problems that you posed referred to finding the distinct values of $\sin(x)$ (for the 1st question) and $\cos(x)$ (for the 2nd question).
I re-interpreted these questions to signify finding the distinct values of $x$ within a modulus of $(2\pi)$, for each of these questions. This is how similar questions are more typically posed.
So, for the specific questions asked in the original posting, you should adjust my responses, based on the idea that you are looking for distinct values of (for example) $\sin(x)$, rather than distinct values of $(x)$.