We need to find solution set for equation $\sin(x/4)=3/8$.
There is video with process of finding said solution set on Khan Academy, but I have problems with it. Here is the video: https://www.youtube.com/watch?v=ES15mSuqHM8&t=51s
According to the video, one part of solution set can be found this way:
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$\sin(\pi-\frac{x}{4}+2\pi n)=\frac{3}{8}$, where $n$ is any integer. We put both sides of the equation inside arcsine
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$\arcsin(\sin(\pi-\frac{x}{4}+2\pi n))=\arcsin(\frac{3}{8})$. Here we apply theorem $\arcsin(\sin(θ))=θ$
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$\pi-\frac{x}{4}+2\pi n=\arcsin(\frac{3}{8})$. Now let's rearrange letters to get expression equal to $x$.
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$x=-4\arcsin(\frac{3}{8})+4\pi+8\pi n$
I see transition from step #2 to step #3 as problematic, because it is based on incorrect use of theorem $\arcsin(\sin(θ))=θ$. In order to get the same $θ$ both inside and the outside of $\sin()$, $θ$ must belong to range of $\arcsin()$, namely to closed interval $[-\frac{\pi}{2};\frac{\pi}{2}]$. But is it the case with "$\pi-\frac{x}{4}+2\pi n$"? I want to argue that no. This transformation must work for all values of $n$, thus it's enough to show single counterexample with any $n$ of our choosing, in order to show that the transformation is not correct. For sake of simplicity let's take $n=0$, thus we are left with "$\pi-\frac{x}{4}$".
We know that $\frac{x}{4}$ is an acute angle because $\arcsin(\sin(\frac{x}{4}))=\arcsin(\frac{3}{8})$ leads to
$x/4=\arcsin(\frac{3}{8})$. And here $\arcsin(\frac{3}{8})$ is equal to about $22$ degrees angle (I use degrees here only once, for sake of easy illustration of acuteness). So we have pi minus acute angle, which gives as the second quarter on the unit circle, which means that the angle "$\pi-x/4$" is more than $\pi/2$, which means that it's incorrect to use the theorem $\arcsin(\sin(θ))=θ$.
Best Answer
Yes, it's incorrect. And uselessly complicated.
From $\sin(x/4)=3/8$ you simply get $$ \frac{x}{4}=\arcsin(3/8)+2k\pi \qquad\text{or}\qquad \frac{x}{4}=\pi-\arcsin(3/8)+2k\pi $$ that yields $$ x=4\arcsin(3/8)+8k\pi \qquad\text{or}\qquad x=4\pi-4\arcsin(3/8)+8k\pi $$ No need to use $\arcsin(\sin(\theta))=\theta$, which is actually false unless, as you note, $\theta\in[-\pi/2,\pi/2]$.
If you have $a$ such that $-1< a< 1$, there are two angles in the interval $(-\pi,\pi)$ having sine equal to $a$: just intersect the unit circle with the line $y=a$. It's clear, geometrically, that one of the angles is in the interval $(-\pi/2,\pi/2)$ and that the other one is the “supplementary” angle.
For $a=1$ or $a=-1$, the two families of solutions degenerate into one.