<s>[INST] 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?
 [/INST] 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:
$$ \begin{Bmatrix}\displaystyle -\frac {7\pi} 8 ,-\frac {3\pi} 8,\frac {\pi} 8 ,\frac {5\pi} 8\end{Bmatrix}$$
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