Multiple solutions for trig function with period Find all angles with limits $−\pi \leq \theta \leq \pi$ which satisfy $\sin 4 \theta = 1$
The working out is given as:
If $\sin 4 \theta = 1$ then $4 \theta = \pi/2 + 2 k \pi$, so $\theta = \pi/8 + k \pi/2$
For $−\pi \leq \theta \leq \pi$ we have $\theta = \pi/8, 5 \pi/8, −3 \pi/8, −7\pi/8$
How did they find the other values of $5 \pi/8, −3 \pi/8$ and $−7 \pi/8$?
Was it just using the unit circle?

Alternative approach:
You have that $4\theta$ must be congruent to $\pi/2$, within a modulus of $(2\pi).$
So, form the following sequence

*

*$4\theta = \pi/2 \implies $ 
$\theta = \pi/8.$

*$4\theta = 2\pi + \pi/2 = 5\pi/2 \implies $ 
$\theta = 5\pi/8.$

*$4\theta = 9\pi/2\implies $ 
$\theta = 9\pi/8.$

*$4\theta = (13)\pi/2\implies $ 
$\theta = (13)\pi/8.$

*$4\theta = (17)\pi/2\implies $ 
$\theta = (17)\pi/8.$

*$4\theta = (21)\pi/2\implies $ 
$\theta = (21)\pi/8.$

*$\cdots$
At this point, you stop and take stock:
the candidate values for distinct solutions are the elements in the following set:
$$\left\{~\frac{\pi}{8}, ~\frac{5\pi}{8}, ~\frac{9\pi}{8}, ~\frac{13\pi}{8}, ~\frac{17\pi}{8}, ~\frac{21\pi}{8}, \cdots ~\right\}. \tag1 $$
Now, you (again) stop and consider.
If you examine the values in (1) above, considering that you are only interested in values that are distinct, with respect to a modulus of $(2\pi)$, you realize that
$$\frac{\pi}{8} \equiv \frac{17\pi}{8} \pmod{2\pi}, ~~~\frac{5\pi}{8} \equiv \frac{21\pi}{8} \pmod{2\pi}. \tag2 $$
Further, you should also realize at this time, that the ongoing pattern will recur.  That is, you should realize that the following infinite sequence of angles
$$\frac{25\pi}{8}, ~\frac{29\pi}{8},  ~\frac{33\pi}{8},  ~\frac{37\pi}{8}, \cdots $$
will not be yielding any distinct angles, with respect to the $(2\pi)$ modulus.
Therefore, you realize that there are only $(4)$ distinct solutions, within a modulus of $(2\pi)$.  These solutions are given by the set
$$\left\{~\frac{\pi}{8}, ~\frac{5\pi}{8}, ~\frac{9\pi}{8}, ~\frac{13\pi}{8} ~\right\}. \tag3 $$
At this point, you are close to your goal.
(3) above represents all distinct satisfying values $\theta$ such that $0 \leq \theta \leq 2\pi.$
However, this isn't good enough.  The problem requires you to identify all distinct values $\theta$ such that $-\pi \leq \theta \leq \pi.$
If you examine (3) above, you realize that the first two values (from the left) are okay, but that the next two values need adjustment.  What this means is that the two values of $~\dfrac{9\pi}{8}~$ and $~\dfrac{13\pi}{8}~$ need to be adjusted.
The adjustment needed is that these two values must be re-expressed in their modulus $(2\pi)$ equivalents that are within the range $-\pi \leq \theta \leq \pi.$
The easiest way to do this is to subtract $(2\pi)$ from each of the two out of range angles.
So:

*

*$\dfrac{9\pi}{8} - 2\pi = \dfrac{-7\pi}{8}.$

*$\dfrac{13\pi}{8} - 2\pi = \dfrac{-3\pi}{8}.$
Therefore, the refined expression of the 4 distinct solutions, from (3) above, are expressed as
$$\left\{~\frac{\pi}{8}, ~\frac{5\pi}{8}, ~\frac{-7\pi}{8}, ~\frac{-3\pi}{8} ~\right\}. \tag4 $$
