Solve it if you can!!

For $\arccos(x)$ I used what is typically considered to be the principal arc cosine. This returns a value between $0$ and $\pi$ . I'll divide the domain into four equal intervals, and examine them. As $x$ goes from $0$ to $\pi$ , the range of $\arccos(x)$ is the same as the domain of $\cos(x)$ , so they exactly invert each other. Hence, $f(x)=x$ for $0\leq x\leq \pi$ . Next, from $x=\pi$ to $x=2\pi$ , I used the identity $\cos(2\pi-x)=\cos(x)$ to adjust the angle to always be between $0$ and $\pi$ so that the domain of cos matches the range of arccos, so we can again cancel the cos and arccos. Hence, $f(x)=2\pi - x$ for $\pi \leq x \leq 2\pi$ . The next two relevant identities are $\cos(x-2\pi)=\cos(x)$ and $\cos(4\pi-x)=\cos(x)$ . This leads to four equations corresponding to each interval: $x=\frac{10-x}{x}$ for $0\leq x \leq \pi$ , $2\pi-x=\frac{10-x}{x}$ for $\pi \leq x \leq 2\pi$ , $x-2\pi =\frac{10-x}{x}$ for $2\pi \leq x \leq 3\pi$ , $4\pi-x=\frac{10-x}{x}$ for $3\pi \leq x \leq 4\pi$ . Only the first three have solutions inside their respective intervals.