(calculus) finding the number of zeros/solutions

[mr.tea]

Homework Statement

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

Homework Equations

none

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

 

[SteamKing]

Homework Statement

Attached as file

Homework Equations

none

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

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].