Trigonometric integral inequality Let us consider the integral equation
\begin{equation}
    f(x)=\lambda \int_{0}^{\pi} \cos (x-y) f(y) \hspace{1mm}dy+g(x), \quad x \in[0, \pi],
\end{equation}
where $f$ is an unknown function on $[0, \pi]$, $g(x)$ is a given continuous function on $[0, \pi]$ and $\lambda$ is a given real constant. Prove that the equation has a unique solution $f \in C[0, \pi]$ for each $\lambda \neq \frac{2}{\pi}$.

The function $y\mapsto |\cos(x-y)|$ is $\pi$-periodic. Therefore, its integral over an interval of length $\pi$ does not depend on the endpoints. In particular :
\begin{align}
\int_0^\pi|\cos(x-y)|\text dy &= \int_{x-\pi/2}^{x+\pi/2} |\cos(y-x)|\text dy\\
&= \int_{-\pi/2}^{\pi/2} \cos(y)\text dy \\
&= 2
\end{align}
