Trigonometric equation

One question though, the lower limit is ≈ -10,6. However, if I use this value in 3*sin(k(x−10)) = 1, I get more than 3 solutions.
 
Sweden_01 said:
One question though, the lower limit is ≈ -10,6. However, if I use this value in 3*sin(k(x−10)) = 1, I get more than 3 solutions.
Click to expand...

If I use numeric approximations for the interval I provided, it would be:

[MATH][3.06193399626770,4.349830709614651)[/MATH]
Do you perhaps have your calculator in degree mode rather than radians?
 
Yep, my calculator was on degrees-mode. But could you please explain why the period must be smaller than 170 degrees? I am really trying to understand this problem step by step, your answers as awesome, however, it's the mathematics that I haven't understand completely yet.
 
Sweden_01 said:
Yep, my calculator was on degrees-mode. But could you please explain why the period must be smaller than 170 degrees? I am really trying to understand this problem step by step, your answers as awesome, however, it's the mathematics that I haven't understand completely yet.
Click to expand...

After writing the equation in terms of \(\theta\), we have:

[MATH]20^{\circ}\le\theta\le170^{\circ}[/MATH]
So, we know at the very least, since a sine function is going to reach some positive value inside its range only twice during one period, that the period must be smaller than the upper bound of the domain so that we would have three solutions.

As it turned out, the period needed to be even smaller than that, but this was just a preliminary observation on the period. :)
 
Okay, so if > 170 degrees, we get more solutions since the sine function begins to decrease, right?
 
Sweden_01 said:
Okay, so if > 170 degrees, we get more solutions since the sine function begins to decrease, right?
Click to expand...

As the period increases, we will get fewer oscillations within a finite domain. Recall, the period describes the angular distance required to make one complete oscillation. So, if we find an insufficient number of oscillations to generate the number of solutions we want, then the period must be decreased until we get the number of solutions we want, but not decreased too much, lest we get too many.

In the form:

[MATH]f(x)=sin(kx)[/MATH] where \(0<k\)

We can think of \(k\) as the angular velocity. As \(k\) increases, so does the frequency, and the period shares an inverse relationship with the frequency, so as \(k\) increases, the period decreases.

Knowing this initial restriction on the period, helped us to determine that we were not going to be able to get a solution on the first increase of the sine function. :)
 
Okay, I think I am now understanding this problem. Honestly, I didn't have a clue, in the beginning, however, thanks to your help and patience I now understand this. Honestly, I don't know how to thank you, you can't imagine how much I value the energy and effort you put down in helping, thanks :)
 
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