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?
 A: When you reach at:  $θ=\displaystyle \frac {\pi} 8 +\frac{k \pi} 2 ,k\in\mathbb{Z}\text{ } (1) $
We are given that:  $θ\in[-\pi, \pi] \text{ } (2)$
So:  $(1),(2)\implies -\pi<\displaystyle \frac {\pi} 8 +\frac{k \pi} 2<\pi\iff -\frac {\pi} 8-\pi<\frac{k \pi} 2< -\frac {\pi} 8+\pi\iff $
$$ \displaystyle -\frac {9\pi} 8<\frac{k \pi} 2<\frac {7\pi} 8 \iff -\frac {9\pi} 4<{k \pi} <\frac {7\pi} 4 \iff -\frac{9} 4<{k } <\frac {7} 4$$ But, $k\in\mathbb{Z}$  so we have to find all the $k$ that belong to:  $\begin{bmatrix}\displaystyle-\frac{9} 4,\frac {7} 4 \end{bmatrix}$  such as:  $k\in \mathbb{Z}. $
After you find the $k$ replace it to the equation $(1)$ and you find all the $θ$ 
So all the possible $k$ that meet our conditions are:  $k=-2,-1,0,1$
So we conclude that, all the possible solutions are:
$$ \begin{Bmatrix}\displaystyle -\frac {7\pi} 8 ,-\frac {3\pi} 8,\frac {\pi} 8 ,\frac {5\pi} 8\end{Bmatrix}$$
A: 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
$$\left\{~\frac{\pi}{8}, ~\frac{5\pi}{8}, ~\frac{-7\pi}{8}, ~\frac{-3\pi}{8} ~\right\}. \tag4 $$
A: Take a look at the plot of $\sin(4 \theta)$.  Does this help you?

A: In the equality $\theta=\frac\pi8+\frac{k\pi}2$:

*

*if $k=1$, you get $\frac{5\pi}8$;

*if $k=-1$, you get $-\frac{3\pi}8$;

*if $k=-2$, you get $-\frac{7\pi}8$.

A: Your question can be rewritten as:

Solve the equation for the angle $\theta$, that satisfies $$\sin(4\theta)=1$$ with $−\pi \leq \theta \leq \pi$

At first, solving the equation gives:
$$\arcsin(4x) = \arcsin(1)$$
$$4x=\frac{\pi}{2}+k\cdot2\pi$$
$$x=\frac{\pi}{8}+\frac{k\pi}{2}$$
Of the part where we put $k2\pi$ into it, remember that by adding any number $k$, that $k \in \mathbb{Z}$ (Whole number, if you may wonder), it is the same act as rotating the point on a circle $360^{\circ}$, and go back to the same position. Now considering that the value limit which was given is $−\pi \leq \theta \leq \pi$. First, solve for half of the region we are bounded, $\pi \leq \theta \leq 0$, with the solution, $x=\frac{\pi}{8}+\frac{k\pi}{2}$, you may realize that you have to make the angle $\theta$ stay inside the region, which correlates with the I and II quadrant.
$\hspace{4.5cm}$
Thus, as you might see, the number $k \in \mathbb{Z}$ can be such value like 1 for the angle to stay inside the boundaries. For example, with $k = 1 $ , the angle $\theta$ is:
$$\theta = \frac{5\pi}{8}$$ which stays inside the I and II quadrant.
For the negative, or the other half, remember that negative angles start counter-clockwise from the starting point, that is, now it is all angles stay inside the III and IV quadrant, accordingly.
A: $$ \cos 4 \theta = \sqrt{1-\sin^2 4 \theta }=0 $$
$$ \theta = (2k-1) \frac{\pi}{2}\cdot \frac{1}{4} $$
so the first halt of radius vector is at the principal value $+\dfrac{\pi}{8}$ and next it continues around the origin halting at  multiples of $\dfrac{\pi}{2}$ along a unit circle either in the clockwise or in the anti- clockwise direction.

