Solving Trigonometric Equations
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I'm sorry, I still don't understand!!!
Harry_the_cat
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I can see you are having trouble understanding the meaning of n in these equations.
Let's take a step back and consider finding the general solution for say \(\displaystyle tan(x) =\sqrt{3}\).
You know that \(\displaystyle tan( \frac{\pi}{3}) = \sqrt{3}\). Right?
You also know that \(\displaystyle tan (\frac{\pi}{3}+\pi) = \sqrt{3}\), because tan is positive in the first and third quadrant.
Drawing a diagram might help you here.
These two angles are marked above on a set of axes.
Note that the angle in the first quadrant (ie \(\displaystyle \frac{\pi}{3}\)) is coterminal with \(\displaystyle \frac{\pi}{3}+ 2\pi\) and \(\displaystyle \frac{\pi}{3}+4\pi\) etc.
The angle in the third quadrant (ie \(\displaystyle \frac{\pi}{3}+\pi\)) is coterminal with \(\displaystyle \frac{\pi}{3}+3\pi\) and \(\displaystyle \frac{\pi}{3}+5\pi\) etc.
You can also subtract \(\displaystyle 2\pi\) to arrive at coterminal angles.
So all of these angles will have a tan of \(\displaystyle \sqrt{3}\). There are an infinite number of solutions. It's impossible to list them all.
So, the easiest and simplest way to lump all of these solutions together is \(\displaystyle \frac{\pi}{3}\) + any integer number of \(\displaystyle \pi\), ie \(\displaystyle \frac{\pi}{3}+n\pi\) where n is an integer.
You can generate as many solutions as you like by choosing appropriate integer values for n.
Let's take a step back and consider finding the general solution for say \(\displaystyle tan(x) =\sqrt{3}\).
You know that \(\displaystyle tan( \frac{\pi}{3}) = \sqrt{3}\). Right?
You also know that \(\displaystyle tan (\frac{\pi}{3}+\pi) = \sqrt{3}\), because tan is positive in the first and third quadrant.
Drawing a diagram might help you here.
These two angles are marked above on a set of axes.
Note that the angle in the first quadrant (ie \(\displaystyle \frac{\pi}{3}\)) is coterminal with \(\displaystyle \frac{\pi}{3}+ 2\pi\) and \(\displaystyle \frac{\pi}{3}+4\pi\) etc.
The angle in the third quadrant (ie \(\displaystyle \frac{\pi}{3}+\pi\)) is coterminal with \(\displaystyle \frac{\pi}{3}+3\pi\) and \(\displaystyle \frac{\pi}{3}+5\pi\) etc.
You can also subtract \(\displaystyle 2\pi\) to arrive at coterminal angles.
So all of these angles will have a tan of \(\displaystyle \sqrt{3}\). There are an infinite number of solutions. It's impossible to list them all.
So, the easiest and simplest way to lump all of these solutions together is \(\displaystyle \frac{\pi}{3}\) + any integer number of \(\displaystyle \pi\), ie \(\displaystyle \frac{\pi}{3}+n\pi\) where n is an integer.
You can generate as many solutions as you like by choosing appropriate integer values for n.
But where do you get pi/3 from? And how do they go from -npi/3 to npi/3 ??
I also still don't understand what n = -1, n = 0, n = 1, etc means. Does n = -1 mean one cycle backwards; n = 0 means no cycle; n = 1 means 1 cycle after??
I must say that I find the suggested METHOD of solution unnecessarily obscure.
Replace the argument to prevent transcription errors and save time
[MATH]sin \left( \dfrac{\pi}{4} \ - 3x \right ) - cos \left ( \dfrac{\pi}{4} - 3x \right ) = 0.[/MATH]
Define [MATH]u = \dfrac{\pi}{4} - 3x \implies sin(u) - cos(u) = 0.[/MATH]
That is not hard and is sure to reduce errors and save time if any considerable work has to be done. It may also spark your intuition.
Solve the simplified equation in the interval [MATH]- \pi \le u \le \pi.[/MATH]
Notice that the length of that interval is 2 pi.
[MATH]sin(u) - cos(u) = 0 \implies sin(u) = cos(u) \implies \dfrac{sin(u)}{cos(u)} = 1 \implies tan(u) = 1.[/MATH]
[MATH]- \dfrac{\pi}{2} \le u \le \dfrac{\pi}{2} \text { and } tan(u) = 1 \implies arctan \{ tan(u) \} = arctan(1) \implies u = \dfrac{\pi}{4}.[/MATH]
But the tangent function is periodic with period pi.
Therefore, the general solution in terms of u is
[MATH]u = \dfrac{\pi}{4} + n \pi, \text { for any integer n}.[/MATH]
But
[MATH]u = \dfrac{\pi}{4} - 3x \implies \dfrac{\pi}{4} + n \pi = \dfrac{\pi}{4} - 3x \implies x = - \dfrac{n \pi}{3}.[/MATH]
Straight forward. Now I don’t like their explanation. We have said n can be ANY integer. So we can get a simpler answer by multiplying by minus 1. Still an integer.
[MATH]\therefore x = \dfrac{n\pi}{3}, \ n \in \mathbb Z.[/MATH]
We have a simple but general solution in x.
Replace the argument to prevent transcription errors and save time
[MATH]sin \left( \dfrac{\pi}{4} \ - 3x \right ) - cos \left ( \dfrac{\pi}{4} - 3x \right ) = 0.[/MATH]
Define [MATH]u = \dfrac{\pi}{4} - 3x \implies sin(u) - cos(u) = 0.[/MATH]
That is not hard and is sure to reduce errors and save time if any considerable work has to be done. It may also spark your intuition.
Solve the simplified equation in the interval [MATH]- \pi \le u \le \pi.[/MATH]
Notice that the length of that interval is 2 pi.
[MATH]sin(u) - cos(u) = 0 \implies sin(u) = cos(u) \implies \dfrac{sin(u)}{cos(u)} = 1 \implies tan(u) = 1.[/MATH]
[MATH]- \dfrac{\pi}{2} \le u \le \dfrac{\pi}{2} \text { and } tan(u) = 1 \implies arctan \{ tan(u) \} = arctan(1) \implies u = \dfrac{\pi}{4}.[/MATH]
But the tangent function is periodic with period pi.
Therefore, the general solution in terms of u is
[MATH]u = \dfrac{\pi}{4} + n \pi, \text { for any integer n}.[/MATH]
But
[MATH]u = \dfrac{\pi}{4} - 3x \implies \dfrac{\pi}{4} + n \pi = \dfrac{\pi}{4} - 3x \implies x = - \dfrac{n \pi}{3}.[/MATH]
Straight forward. Now I don’t like their explanation. We have said n can be ANY integer. So we can get a simpler answer by multiplying by minus 1. Still an integer.
[MATH]\therefore x = \dfrac{n\pi}{3}, \ n \in \mathbb Z.[/MATH]
We have a simple but general solution in x.
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For (a):
I think I understand now - so x = npi/3 is the simplified answer?
For (b):
I am still confused... How can 'n' be the number of solutions, but also the number of cycles after?? And what does it mean when n = 0? There are 0 solutions??????
Also for the last part:
I don't understand why you are making it equal to zero... We only learnt that tan(x) = sin(x)/cos(x) = 1
I'm sorry, in all honestly this is too complex for me I don't really know what's going on
You do not understand replacing a complex expression with a simple variable????
[MATH]sin \left (\dfrac{\pi}{4} - 3x \right ) - cos \left (\dfrac{\pi}{4} - 3x \right ) = 0 \text { and } u = \dfrac{\pi}{4} - 3x \implies[/MATH]
[MATH]sin(u) - cos(u) = 0.[/MATH]
[MATH]sin \left (\dfrac{\pi}{4} - 3x \right ) - cos \left (\dfrac{\pi}{4} - 3x \right ) = 0 \text { and } u = \dfrac{\pi}{4} - 3x \implies[/MATH]
[MATH]sin(u) - cos(u) = 0.[/MATH]
(a) As I said to you in a reply to another thread, the 'answer' is the list of numbers:
...-2pi/3, -pi/3, 0, pi/3, 2pi/3...
but you can generate lists of the same numbers using different formulas:
e.g. the formula npi/3 or the formula -npi/3, substituting in n=0, n=1, n=-1, n=2, n=-2 etc...
I don't see one of those formulas being much 'simpler' than the other, except I suppose the missing '-' sign makes npi/3 slightly simpler. However, either would do nicely.
(b) 'n' has nothing to do with the number of solutions, nor I guess 'the number of cycles after' - I'm afraid I don't really know what you mean by this phrase. 'n' is a 'dummy variable' with no meaning itself for the question. It is only used as a means of generating the solutions to the equation, as I have just described in (a).
If I say to you that the solutions to the equation are of the form: npi/3
I mean that you can make a list of the solutions by sticking n=0 (the 0 has no meaning), n=1 etc... and generate the list:
0, pi/3, -pi/3, 2pi/3, -2pi/3 etc...
For the last part: by making 'it' equal to 0 I assume you mean cosine equal to 0.
tanx = sinx/cosx for every value of x except values of x, where cosx=0. E.g. tan(x) does not exist when x=pi/2. However I wouldn't worry about the last part - it's the least of your worries (and the rest of the world seems to ignore it quite happily anyway).
I went over it again and wrote it out - I get it now! However, I don't understand the very last part...
If you make n = -1, then x will be pi/3 ?
No. It's a 'dummy variable'. The values of 'n' have no particular 'meaning'. 'n' is just used as a tool to generate the solutions.
Why 'n' generates solutions, is demonstrated in my pdf above (re. the tan graph).
I'm sorry this hasn't been much help. It is quite difficult to do this online.
I have to say that @JeffM 's post is particularly clear and concise and is worth studying. Perhaps it is worth you going back and redoing this topic again because you seemed not to have grasped the basic concept of what is going on. I think it may be very difficult for you to pick it up like this without a thorough study of the topic again from the start. (Only a suggestion). I have no doubt you'll get it.
OK Let me try this.
We first introduce the trigonometric functions in terms of the angles of right triangles other than the right angle. Using radian measure, that gives us a domain of [MATH](0.\ \pi /2).[/MATH]
We then extend the functions to a circle.
This gives us a domain of [MATH][0,\ 2 \pi].[/MATH]
We then extend the functions to the domain of [MATH](- \infty,\ \infty)[/MATH]
The basic trig functions are then periodic over an infinite domain.
[MATH]sin(x) = sin \left ( x + 2n \pi \right ) \text { for any integer } n.[/MATH]
Once we are dealing with an infinite domain, we may get an infinite number of answers to a trigonometric equation. How we number the periods is arbitrary. There is no first integer or last integer. It frequently makes sense, however, to say that when n = 0 is the first period. But that is totally arbitrary. The n has meaning only relative to where we defined n = 0.
As lex has said, it is a dummy variable to reflect that there are an infinite number of answers. You choose n to be relevant.
We first introduce the trigonometric functions in terms of the angles of right triangles other than the right angle. Using radian measure, that gives us a domain of [MATH](0.\ \pi /2).[/MATH]
We then extend the functions to a circle.
This gives us a domain of [MATH][0,\ 2 \pi].[/MATH]
We then extend the functions to the domain of [MATH](- \infty,\ \infty)[/MATH]
The basic trig functions are then periodic over an infinite domain.
[MATH]sin(x) = sin \left ( x + 2n \pi \right ) \text { for any integer } n.[/MATH]
Once we are dealing with an infinite domain, we may get an infinite number of answers to a trigonometric equation. How we number the periods is arbitrary. There is no first integer or last integer. It frequently makes sense, however, to say that when n = 0 is the first period. But that is totally arbitrary. The n has meaning only relative to where we defined n = 0.
As lex has said, it is a dummy variable to reflect that there are an infinite number of answers. You choose n to be relevant.
I think this is where I'm getting confused:
"sin(x) = sin(x + 2nπ) for any integer n"
^^ my understanding is that you add 2nπ because one full cycle is 2π for sin (and cos) graphs, and then for the n, I thought n represented the NUMBER of cycles that can occur, which is infinite. So if n is negative, then it's going clockwise on the unit circle however many cycles; if n is positive, then it's going anti-clockwise on the unit circle however many cycles.
Now I think we are getting somewhere.
If you think about x as an angle, then yes, positive means rotations counter-clock-wise and negative means rotations clock-wise. Thinking that way, obviously negative integers represent a different physical reality. And n = 0 means no rotation at all.
But then if I rotate a full circle clockwise, I end up with the same angle as I started with.And if I rotate a full circle anti-clock-wise, I end up with the same angle I started with.
[MATH]sin(x + 2 \pi) = sin(x) = sin(x - 2 \pi) \implies sin(x + 2 \pi) = sin(x - 2 \pi).[/MATH]
In terms of the value of the sine function, whether you add 2 pi to or subtract 2 pi from x makes absolutely no difference. And that is true for 4pi, 6 pi, 8 pi. We do not care in terms of the value of the sine whether we multiply 2 pi by a positive or negative integer. We get the same value over and over again.
This makes the trig functions useful for studying any kind of periodic phenomenon. We can forget about angles. We now consider the trig functions as being a way to describe any repetitive behavior. Once we do that, we really do not care about a specific value of n until we label one period as the zero period. Plus and minus do not then correspond to the direction of rotation. They correspond to how many periods before or after the arbitrary period selected as the zero period.
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Hmm, I think I get it now. So are we supposed to decide the arbitrary period selected as the zero period - or are you just saying that is the only way in which the specific value of n would matter? And would that ever occur?
Also, how are we supposed to know whether the solutions are positive or negative? --> in reference to one of the questions I came across where it asked for the first three positive solutions. Is it just trial and error every time, where you just plug in negative and positive values for n until you get the first three positive solutions for example?