## anonymous one year ago Friends please if you can help me with this exercise, this came into my calculus test 1: parameters found for "w" and "n" the equation cos (wx) - x / n has a real solution and calculate how many second solutions has exactly ..... please still do not know how to solve it thanks

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1. perl

The directions are somewhat unclear. Are you quoting word for word?

2. perl

Is this the equation : $$\Large \cos (wx) - \frac xn = 0$$

3. anonymous

yes that is the equation

4. anonymous

search parameters of "w" and "n" for the equation has a real solution

5. anonymous

but also sought to find the exact number of solutions of the equation

6. perl

That's harder :) We can try using calculus

7. anonymous

That confused me

8. anonymous

This exercise was in my calculus test 1

9. perl

this problem is within the subject of calculus 1

10. perl

we can see that as n gets large $$\Large \cos (wx) - \frac xn \to \cos(wx)$$

11. perl

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

12. perl

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.

13. perl

I don't see anything that pops out. @Michele_Laino @ParthKohli

14. ParthKohli

They call these things transcendental equations, don't they? Graphing is the way to go about these.

15. perl

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

16. anonymous

theorem of Bolzano or intermediate value theorem that thought but how to use it with the values of "w" and "n"

17. anonymous

but forget to mention $w >0,n \ge1$

18. perl

we can assume n are integers. https://www.desmos.com/calculator/0pdovzb6kn

19. anonymous

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