Q:

Solving $f\left(x\right)+f\left(1/x\right)=K$

I would like to know if there's a way to solve
$$
\forall x \in \mathbb{R}^{*+}, \ K \in \mathbb{R}, \ f\left(x\right)+f\left(\frac{1}{x}\right)=K
$$
We know that for $\displaystyle K=\frac{\pi}{2}$ we have ( $x \in \mathbb{R}^{*+}$ )
$$
\text{arctan}\left(x\right)+\text{arctan}\left(\frac{1}{x}\right)=\frac{\pi}{2}
$$
And for $K=0$
$$
\ln\left(x\right)+\ln\left(\frac{1}{x}\right)=0
$$
Only thing i've found is that it implies
$$
-f'\left(\frac{1}{x}\right)+x^2f'\left(x\right)=0 \text{ and }f\left(1\right)=\frac{K}{2}
$$

A:

Sure, take $f(x) = \log(x)+\frac{K}{2}$.

Edit: I'm interpreting the problem as
$$ \forall K\in\mathbb{R}, \ \exists f:\mathbb{R}^{*+}\to\mathbb{R} \text{ s.t. } 
\forall x \in\mathbb{R}^{*+}, \ f(x)+f\left(\frac{1}{x}\right) = K$$

A:

$$f(x)+f(1/x)=K$$
Change of function :  $\quad f(x)=F\left(\ln|x|\right)+\frac{K}{2}$
$f(1/x)=F\left(\ln|1/x|\right)+\frac{K}{2}=F\left(-\ln|x|\right)+\frac{K}{2}$
$$F\left(\ln|x|\right)+F\left(-\ln|x|\right)=0$$
Change of variable : $\quad X=\ln|x|$
$$F(X)+F(-X)=0$$
$$F(X)=-F(-X)$$
Thus $F(X)$ is any odd function.
The solution is :
$$f(x)=F(\ln|x|)+\frac{K}{2}\quad\text{with any odd function }F.$$
EXAMPLES :
With $F(X)=X$ : 
$\begin{cases}
f(x)=\ln|x|+\frac{K}{2}\\
f(1/x)=-\ln|x|+\frac{K}{2}
\end{cases}
\quad f(x)+f(1/x)=\ln|x|+\frac{K}{2}+\left(-\ln|x|+\frac{K}{2}\right)=K$
With $F(X)=\sin(X)$ :
$\begin{cases}
f(x)=\sin(\ln|x|)+\frac{K}{2}\\
f(1/x)= -\sin(\ln|x|)+\frac{K}{2}
\end{cases}
\quad f(x)+f(1/x)=\sin(\ln|x|)+\frac{K}{2}+\left( -\sin(\ln|x|)+\frac{K}{2}
\right)=K$
Etc.

