Here is a *real* solution for $k=2$: Take $f_1(x)=x$ and
$$f_2(x)=\arccos\left( \sqrt{\tfrac{3}{4}- \cos ^4(x/2)} \operatorname{cotan} (x/2)-\tfrac{1}{2}\cos x\right).$$
Then
$$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;x_{\rm min}<x<x_{\rm max}.$$
Numerically, I find $x_{\rm min}=1.05$ and $x_{\rm max}=5.53$.