One solution to Trig Function, but what about the other(s)?

[Ocata]

Homework Statement

[itex]7.5Sin(\frac{π}{10}x) = 5[/itex]Solving for x

The Attempt at a Solution

[/B]
[itex]7.5Sin(\frac{π}{10}x) = 5[/itex]

[itex]Sin^{-1}(\frac{5}{(7.5}) = \frac{π}{10}x[/itex]

[itex] \frac{10}{π}Sin^{-1}(\frac{5}{7.5}) = x = 2.32[/itex]


If this function were a parabola, there would be two answers based on quadratic equations algebraically producing two answers, which match the graph of the parabola.

But how come I only get one answer for a trigonometric function instead of two or more, so that it actually matches the graph?

Thank you

 

[ecastro]

I might be missing something, but your function doesn't seems to be quadratic in form.

 

[Ocata]

Exactly ecastro,

This function is trigonometric, not parabolic. And this trigonometric function produces an arch, just like that of a parabolic, so it too should have more than one solution as well.

However, I can find both solutions for a parabola but solving for this trigonometric function yields only one solution, which I am sure there should be more solutions based on the nature of what a sine function looks like graphically.

 

[HallsofIvy]

You say "If this function were a parabola" but it isn't! In fact, instead of having two solutions, it will have an infinite number of them because the sine function is periodic.

Yes, from [itex]7.5 sin\left(\frac{\pi}{10}x\right)= 5[/itex], you get [itex]sin\left(\frac{\pi}{10}x\right)= \frac{5}{7.5}= \frac{2}{3}[/itex]. And now use your knowledge of the sine function! We know, for example, that [itex]sin(x)= sin(\pi- x)[/itex] so while [itex]\frac{10}{\pi}sin^{-1}\left(\frac{2}{3}\right)[/itex] is a solution, so is [itex]\pi- \frac{10}{\pi}sin^{-1}\left(\frac{2}{3}\right)[/itex]. If the problem is to find solutions to the equation in [itex][0, 2\pi][/itex], that would give both. Further, sin(x) is periodic with period [itex]2\pi[/itex] so adding any integer multiple of [itex]2\pi[/itex] will also give a solution.

 

[Mark44]

[itex]7.5Sin(\frac{π}{10}x) = 5[/itex]

[itex]Sin^{-1}(\frac{5}{(7.5}) = \frac{π}{10}x[/itex]

When you're working with trig equations, your second step above limits your possible solutions to just one; namely, a solution in the interval ##[-\pi/2, \pi/2]##. The reason for this is that the inverse sine function is one-to-one. Very often, trig equations have multiple solutions, not just the single solution that the inverse trig functions produce. The post by HallsOfIvy shows how to find the solutions that aren't produced by the inverse trig functions.

 

[Ocata]

You say "If this function were a parabola" but it isn't! In fact, instead of having two solutions, it will have an infinite number of them because the sine function is periodic.

Yes, from [itex]7.5 sin\left(\frac{\pi}{10}x\right)= 5[/itex], you get [itex]sin\left(\frac{\pi}{10}x\right)= \frac{5}{7.5}= \frac{2}{3}[/itex]. And now use your knowledge of the sine function! We know, for example, that [itex]sin(x)= sin(\pi- x)[/itex] so while [itex]\frac{10}{\pi}sin^{-1}\left(\frac{2}{3}\right)[/itex] is a solution, so is [itex]\pi- \frac{10}{\pi}sin^{-1}\left(\frac{2}{3}\right)[/itex]. If the problem is to find solutions to the equation in [itex][0, 2\pi][/itex], that would give both. Further, sin(x) is periodic with period [itex]2\pi[/itex] so adding any integer multiple of [itex]2\pi[/itex] will also give a solution.

Thank you HallsofIvy,

I did not know of this: [itex]sin(x)= sin(\pi- x)[/itex] I had not seen it in the my textbook I was referencing, Sullivan 7th edition, so thank you so much for advising me about this property.

[itex] \frac{10}{π}sin^{-1}(\frac{5}{7.5})= x = 2.32 [/itex]

the interval of the first arch is at x = 0 to x = 10.

so [itex]sin(x)= sin(\pi- x)[/itex] = [itex]sin(2.32)= sin(10- 2.32)[/itex] = [itex]sin(2.32)= sin(7.67)[/itex]

so at y = 5, x = 2.32 and 7.67 between x intervals (0,π) => (0,10)

Thank you, very, very much.

 

[Ocata]

When you're working with trig equations, your second step above limits your possible solutions to just one; namely, a solution in the interval ##[-\pi/2, \pi/2]##. The reason for this is that the inverse sine function is one-to-one. Very often, trig equations have multiple solutions, not just the single solution that the inverse trig functions produce. The post by HallsOfIvy shows how to find the solutions that aren't produced by the inverse trig functions.

Ah yes, this makes wonderful sense. Thank you, Mark44.

 

[insightful]

Just to add a little humor: