Finding function for X ~within~ cosine?

[VinZ]

Hi!

First off, i am not sure if i posted this in right category, cause i honestly do not know which category this question belongs to.

Basically i am trying to figure out how to make function for X from this equation:
a=x*b+c*cos(x*d)
10=x*10+2*cos(x*10) (same with some numbers)
I am quite clueless how to get X out of under cosine, so i will appreciate any help.

Also here is a link to wolfram, it has graph and answer for those specific numbers, but it doesn't show how it got the answer.
http://www.wolframalpha.com/input/?i=what+is+x+in+10=x*10+2*cos(x*10)

Thanks in advance.

EDIT:I guess for the equation not to be too complicated i might as well take out all the numbers and make it shorter:
a=cos(x*b) where a and b is known.

 

[Deleted member 4993]

Hi!

First off, i am not sure if i posted this in right category, cause i honestly do not know which category this question belongs to.

Basically i am trying to figure out how to make function for X from this equation:
a=x*b+c*cos(x*d)
10=x*10+2*cos(x*10) (same with some numbers)
I am quite clueless how to get X out of under cosine, so i will appreciate any help.

Also here is a link to wolfram, it has graph and answer for those specific numbers, but it doesn't show how it got the answer.
http://www.wolframalpha.com/input/?i=what+is+x+in+10=x*10+2*cos(x*10)

Thanks in advance.

EDIT:I guess for the equation not to be too complicated i might as well take out all the numbers and make it shorter:
a=cos(x*b) where a and b is known. - No you cannot do this (that is, reduce your first equation to this form) - in general.

10=x*10+2*cos(x*10) and you want to find x = ??

This is a non-linear trigonometric equation - it does not have closed-form solution/s.

It needs to be solved numerically, using one of the many methods such as Newton's algorithm.

 

[HallsofIvy]

If you mean "solve the equation for x" then, as Subhotosh Khan said, that cannot be done in terms of elementary functions.

As for your simplified version, thats very easy. From a= cos(bx), start by applying the inverse or "arc" cosine to both sides: arccos(a)= arccos(cos(bx))= bx. Then, of course, divide both sides by "b" (the inverse of multiplying by b): x= arccos(a)/b.


(Note that cosine is NOT a "one to one" function (many different values of x will give the same value for cos(x)) so does not have a "true" inverse. arccosine gives only the between 0 and \(\displaystyle 2\pi\) that has that cosine value.)

 

[VinZ]

Well I am not from English speaking country so I am not familiar with most Math terms, and on top of that I have not studied Calculus... so maybe my wording is off.

The way I see it, Wolfram can get results from that equation and i figured that there has to be a way to find solution(s). :roll:

Well thanks anyway. I might rethink the problem and see if i can come up with something else to solve it.

 

[JeffM]

Well I am not from English speaking country so I am not familiar with most Math terms, and on top of that I have not studied Calculus... so maybe my wording is off.

The way I see it, Wolfram can get results from that equation and i figured that there has to be a way to find solution(s). :roll:

Well thanks anyway. I might rethink the problem and see if i can come up with something else to solve it.

No one said it can't be APPROXIMATELY solved. It cannot be solved through a neat equation.

 

[VinZ]

No one said it can't be APPROXIMATELY solved. It cannot be solved through a neat equation.

So what CAN (approximately) be solved? How accurate and how?
I know that the equation can have multiple results when other inputs have specific values, but that isn't really a problem or is it?.
Would really like to learn how to get on with this problem.

Also my apologies for leaving the thread for so long.

 

[daon2]

So what CAN (approximately) be solved? How accurate and how?
I know that the equation can have multiple results when other inputs have specific values, but that isn't really a problem or is it?.
Would really like to learn how to get on with this problem.

Also my apologies for leaving the thread for so long.

You would need to use a numerical method, such as Newton's Method. You can get any desired accuracy you'd like with some amount of error, but the process can be tedius. Algorithms like this are best suited for computers when a certain amount of accuracy is required.

 

[VinZ]

You would need to use a numerical method, such as Newton's Method. You can get any desired accuracy you'd like with some amount of error, but the process can be tedius. Algorithms like this are best suited for computers when a certain amount of accuracy is required.

Well Newtons Method seems to have some flaws. If my graph would be jagged like 'saw blade', it would sometimes take too many iterations and sometimes fail altogether based on the initial guess.
I guess i would have to make the initial guess for Newtons Method in the specific 'saw tooth' for it to be successful and as i am indeed using this for computer program it would be necessary that it always works.

I did however made my own 'method' which is somewhat similar to Newtons Method except that it always works and is fairly accurate depending on how many iterations i run it trough.
I guess i could use my method to get closer to the approximation and then Newtons method to get even closer.
But the reason i made this thread was to find a better way to do it, because i feel that such methods gives me too many loops/calculations.
But as Subhotosh Khan said there is no neat function, so i guess this thread is dead end.

Maybe i should post the whole problem, but i am afraid it would be more of a physics problem.