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The directions are somewhat unclear. Are you quoting word for word?
Is this the equation : \( \Large \cos (wx) - \frac xn = 0 \)
yes that is the equation
search parameters of "w" and "n" for the equation has a real solution
but also sought to find the exact number of solutions of the equation
That's harder :) We can try using calculus
That confused me
This exercise was in my calculus test 1
this problem is within the subject of calculus 1
we can see that as n gets large \(\Large \cos (wx) - \frac xn \to \cos(wx) \)
We could modify the equation \( \Large n\cdot \cos (wx) -x = 0 \) The derivative is \( \Large -n \sin(wx)w -1 = 0\) However I don't see how this helps
Again just to be clear you want to know exactly how many real solutions are there to the equation: $$\Large \cos (wx) - \frac xn = 0$$ Does n have to be an integer? Clearly n=0 won't work.
They call these things transcendental equations, don't they? Graphing is the way to go about these.
yes . not an algebraic equation. i dont see any other way than by checking for sign change, using intermediate value theorem . but this does not give you a nice result
theorem of Bolzano or intermediate value theorem that thought but how to use it with the values of "w" and "n"
but forget to mention \[w >0,n \ge1\]
we can assume n are integers. https://www.desmos.com/calculator/0pdovzb6kn
I think it has infinitely many solutions. As n goes larger, the continuity of the cosine (regarding W values) will ensure the existence of number k= cos (c) for some c in R such that k between (-1,1). Now we can vary n in such a way that x/n is between (-1,1) for any value of x (since n is independent). That is to say, for any x in the domain, there exist some n such that their ratio is between (-1,1). on the other hand, we can change w as we want such that the equality still holds . in short, infinitely many solutions That's just what I think