
Trig Derivative Problem
Hello everyone,
Could anyone please offer hints or tips as to how to solve this problem? Some of my work is shown below.
Thank you.

5. Consider the function $\displaystyle f(x) = x^2 + \cos(kx) $, where k is the real parameter. For what values of k does the given function have no inflection points?

I found these:
$\displaystyle f'(x) = 2x  k\sin(kx) $
$\displaystyle f''(x) = 2  k^2 \cos(kx)$
To determine the inflection points, I set $\displaystyle f''(x) = 0 $.
Therefore:
$\displaystyle k^2 \cos(kx) = 2 $
$\displaystyle cos(kx) = \frac{2}{k^2} $
$\displaystyle x = \frac{\cos^{1} \frac{2}{k^2}}{k} $
However, I am not sure how to proceed from here.

The condition is $\displaystyle \frac{2}{k^2}>1$, (because $\displaystyle \cos\alpha\in[1,1], \ \forall\alpha\in\mathbf{R}$)
Then $\displaystyle k^2<2\Rightarrow k\in(\sqrt{2},\sqrt{2})$