Finding general solution for a trigonometrical equation

[pope4]

I was doing this question, and I got the right answer for the solutions themselves. However, generally, our teacher always told us that a general solution is something along the lines of x+n2pi, n E I (where x is one solution). However, when I went to check whether I got the right answer or not, I realized that the 2npi pattern doesn't work here, and the solution is x+6n (where x is one solution). Why is that? Why is the pattern different for this question than the ones my teacher was talking about? How can you know when to use 2npi and when to use something else?

 

[Dr.Peterson]

I was doing this question, and I got the right answer for the solutions themselves. However, generally, our teacher always told us that a general solution is something along the lines of x+n2pi, n E I (where x is one solution). However, when I went to check whether I got the right answer or not, I realized that the 2npi pattern doesn't work here, and the solution is x+6n (where x is one solution). Why is that? Why is the pattern different for this question than the ones my teacher was talking about? How can you know when to use 2npi and when to use something else?

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The reason for the 2pi n is that the period of the function is 2pi. But this function's period is 6, namely 2pi/(pi/3). For the second, similarly, the period is not 360, but 360/15.

What reason did your teacher give for adding 2pi n? You have to think, not just write something you memorize.

 

[Dr.Peterson]

I was doing this question, and I got the right answer for the solutions themselves. However, generally, our teacher always told us that a general solution is something along the lines of x+n2pi, n E I (where x is one solution). However, when I went to check whether I got the right answer or not, I realized that the 2npi pattern doesn't work here, and the solution is x+6n (where x is one solution). Why is that? Why is the pattern different for this question than the ones my teacher was talking about? How can you know when to use 2npi and when to use something else?

View attachment 34477View attachment 34476

Since the work shown is not actual work, but lets the calculator do the thinking, perhaps you have not seen how to solve these problems yourself. Here is how I would do it, which will show the reason for the result:
[math]\sin\left(\frac{\pi}{3}\left(x-1\right)\right)=\frac{1}{2}[/math][math]\frac{\pi}{3}\left(x-1\right)=\frac{\pi}{6}{\color{Red}+2\pi n}\text{ or }\frac{5\pi}{6}{\color{Red}+2\pi n}[/math][math]x-1=\frac{3}{\pi}\left(\frac{\pi}{6}+2\pi n\right)=\frac{1}{2}+6n\text{ or }x-1=\frac{3}{\pi}\left(\frac{5\pi}{6}+2\pi n\right)=\frac{5}{2}+6n[/math][math]x=\frac{3}{2}+6n\text{ or }x=\frac{7}{2}+6n[/math]So we do use [imath]2\pi[/imath], because that is the period of the basic sine function, but subsequently change it to 6.

If you're letting the calculator do all the work, you have to separately think about the period of the function.