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(calculus) finding the number of zeros/solutions

  1. Jan 5, 2016 #1
    • Member warned about posting problem statement as an image
    1. The problem statement, all variables and given/known data
    Determine, for each real value of k, the number of solutions in [0,2pi] of:
    x+sqrt(2)cos(x)=k
    And there is a hint: pi<4(1+sqrt(2))/3
    2. Relevant equations
    none

    3. The attempt at a solution
    I tried to define a function f(x)=x+sqrt(2)cos(x)-k and to find the number of zeros for the derivative that will give me(using Rolle's Theorem) the maximum number of zeros to the function. That really didn't help since I got that there are 2 zeros to the derivative, which mean at most 4 zeros to the function.
    I also thought to use intermediate value theorem but it's no useful since k is unknown.

    Thank you,
    Thomas
     

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    Last edited: Jan 5, 2016
  2. jcsd
  3. Jan 5, 2016 #2

    SteamKing

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    You should make a plot of x + √2 * cos (x) and see how picking different values of k would determine the max. number of roots on the interval [0, 2π].
     
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