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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 $$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;\pi/3<x<\pi +2 \arcsin \left(\frac{3^{1/4}}{\sqrt{2}}\right).$$ Numerically, this interval is $1.05<x<5.53$. It can be extended by switching the branch of the square root.

The solution can be generalized to larger $k$; put all $f_n(x)=x$ for $n=1,2,\ldots, k-1$ and then solve for $f_k(x)$. The expression is lengthy and I omit it here.

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 $$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;\pi/3<x<\pi +2 \arcsin \left(\frac{3^{1/4}}{\sqrt{2}}\right).$$ Numerically, this interval is $1.05<x<5.53$. It can be extended by switching the branch of the square root.

The solution can be generalized to larger $k$; put all $f_n(x)=x$ for $n=1,2,\ldots, k-1$ and then solve for $f_k(x)$. The expression is lengthy and I omit it here. The interval in which a real solution holds shrinks with increasing $k$.

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 $$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;\pi/3<x<\pi +2 \arcsin \left(\frac{3^{1/4}}{\sqrt{2}}\right).$$ Numerically, this interval is $1.05<x<5.53$.

This can be generalized to larger $k$; put all $f_n(x)=x$ for $n=1,2,\ldots, k-1$ and then solve for $f_k(x)$.

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 $$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;\pi/3<x<\pi +2 \arcsin \left(\frac{3^{1/4}}{\sqrt{2}}\right).$$ Numerically, this interval is $1.05<x<5.53$. It can be extended by switching the branch of the square root.

The solution can be generalized to larger $k$; put all $f_n(x)=x$ for $n=1,2,\ldots, k-1$ and then solve for $f_k(x)$. The expression is lengthy and I omit it here.

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 $$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;\pi/3<x<\pi +2 \arcsin \left(\frac{3^{1/4}}{\sqrt{2}}\right).$$ Numerically, this interval is $1.05<x<5.53$.

This can be generalized to larger $k$; put all $f_n(x)=x$ for $n=1,2,\ldots k-1$ and then solve for $f_k(x)$.

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 $$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;\pi/3<x<\pi +2 \arcsin \left(\frac{3^{1/4}}{\sqrt{2}}\right).$$ Numerically, this interval is $1.05<x<5.53$.

This can be generalized to larger $k$; put all $f_n(x)=x$ for $n=1,2,\ldots, k-1$ and then solve for $f_k(x)$.