guringpw

2021-12-26

If I have a such equation to solve:
$\mathrm{sin}3x=\frac{1}{2}$

Philip Williams

Within the range
So, $\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{\pi }{6}=\mathrm{sin}\frac{5\pi }{6}$
Also sine has periodicity of $2\pi$. Thus $\mathrm{sin}\left(2n\pi +\theta \right)=\mathrm{sin}\theta$ where n is an integer.
So,
$\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{\pi }{6}=\mathrm{sin}\left(2n\pi +\frac{\pi }{6}\right)⇒3x=2n\pi +\frac{\pi }{6}$
Similarly,
$\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{5\pi }{6}=\mathrm{sin}\left(2n\pi +\frac{5\pi }{6}\right)⇒3x=2n\pi +\frac{5\pi }{6}$

Jim Hunt

When
$\mathrm{sin}x=\mathrm{sin}\alpha ⇒x=n\pi +{\left(-1\right)}^{n}\alpha$,
where $n=0,±±1,±2,±3,\dots$ So here

nick1337

$\mathrm{sin}\left(x\right)$ is periodic, as in it has a period of $2\pi$. Therefore, it will intersect the same y-value every period, or every $2n\pi$ where n is an integer. By scaling $\theta$ by 3, you are actually scaling the period of $\mathrm{sin}\theta$ by 13. Thus the new period is $\frac{2\pi }{3}$. You can see this in action with your solution you gave, which was
$\mathrm{sin}\theta =\frac{1}{2}$
$\theta ={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)$
Here's the kicker: $\mathrm{sin}\theta$ intersects 12 twice every period, so there are two solutions. By inspecting the unit circle, find why this must be true. The two solutions are
$3\theta =\frac{\pi }{6}$
$3\theta =\frac{5\pi }{6}$
But we want to know the solutions for the entire domain $\left(-\mathrm{\infty },\mathrm{\infty }\right)$, so we want
$3\theta =\frac{\pi }{6}+2n\pi$
$3\theta =\frac{5\pi }{6}+2n\pi$
And now by dividing by 3, you can see why your period is scaled by $\frac{1}{3}$ (as well as the solution):
$\theta =\frac{\pi }{18}+\frac{2n\pi }{3}$
$\theta =\frac{5\pi }{18}+\frac{2n\pi }{3}$

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