guringpw

2021-12-26

If I have a such equation to solve:

Philip Williams

Beginner2021-12-27Added 39 answers

Within the range $[0,2\pi ]\text{}\text{},\mathrm{sin}\frac{\pi}{6}=\mathrm{sin}\frac{5\pi}{6}=\frac{12}{}$

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 $\mathrm{sin}(2n\pi +\theta )=\mathrm{sin}\theta$ where n is an integer.

So,

$\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{\pi}{6}=\mathrm{sin}(2n\pi +\frac{\pi}{6})\Rightarrow 3x=2n\pi +\frac{\pi}{6}$

Similarly,

$\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{5\pi}{6}=\mathrm{sin}(2n\pi +\frac{5\pi}{6})\Rightarrow 3x=2n\pi +\frac{5\pi}{6}$

So,

Also sine has periodicity of

So,

Similarly,

Jim Hunt

Beginner2021-12-28Added 45 answers

When

$\mathrm{sin}x=\mathrm{sin}\alpha \Rightarrow x=n\pi +{(-1)}^{n}\alpha$ ,

where$n=0,\pm \pm 1,\pm 2,\pm 3,\dots$ So here

$\mathrm{sin}3x=\mathrm{sin}\frac{\pi}{6}\Rightarrow 3x=n\pi +{(-1)}^{n}\frac{\pi}{3}\Rightarrow x=\frac{n\pi}{3}+{(-1)}^{n}\frac{\pi}{9},\text{}\text{}\text{}n=0,\pm 1,\pm 3,\dots$

where

nick1337

Expert2022-01-08Added 573 answers

Here's the kicker:

But we want to know the solutions for the entire domain

And now by dividing by 3, you can see why your period is scaled by