Find the spectrum of $(Ff)(x)=\int_{-\pi}^{\pi} (1 + \cos(x-y))f(y)dy$ As the title suggests we have a function $T: L^2 [-\pi,\pi] \to L^2[-\pi,\pi]$ that sends $f \to \int_{-\pi}^{\pi} (1 + \cos(x-y))f(y)dy$.
I have read a couple of related posts, but unfortunately I am not sure how to proceed. Most of what I read either worked with Fourier transforms (I see that if $K(x)=1+cos(x)$, then the expression we have is $K$ convoluted with $f$) or directly by finding point spectrum and invoking some property where doing this sufficed to compute the total spectrum.
I did not succed with the first one, and after that went to try to compute the point spectrum. What I get is that $\pi$ and $2\pi$ are the (point) spectrum, and any function of the form $ b_1 \cos(t) +  b_2 \sin(t)$ and $b_0$ (constant one) are their corresponding eigenfunctions, respectively. To do this I just worked out the expression, using the fact that we can rewrite $\cos(x-y)$ as $\cos(x)\cos(y)+\sin(x)\sin(y)$. From here I got that the image of our map can be written as $a_1 + a_2\cos(t) +  a_3 \sin(t)$, from where I deduced that these are the unique eigenvalues after some computations.
What approach do you guys suggest? Does my function have any special properties (like compactness or self adjointess) to conlude this is it? Is the Fourier transformation easier here? Thanks a lot!

@cmk is right. Look at that post. But if you want a more specific answer for your specific convolution kernel, here goes. (But please then look at the link that cmk suggested afterwards.)
Take $\psi_k(x) = e^{i k x}$. Then $F \psi_k = \lambda_k \psi_k$ where $\lambda_0=2\pi$, $\lambda_1=\lambda_{-1}=\pi$ and $\lambda_k=0$ for $k \not\in \{-1,0,1\}$. That is a complete orthonormal basis. So the spectrum is pure-point. It is a rank-3 self-adjoint operator. Just integrate to get these results: multiplying by $(2\pi)^{-1}$ you get
$$
\frac{1}{2\pi} \int_{-\pi}^{\pi} \left(1+\frac{1}{2} e^{i x} e^{-iy} + \frac{1}{2} e^{-ix} e^{iy}\right) e^{i k y}\, dy\, ,
$$
which is equal to the expression utilizing $L^2$-inner products
$$
\langle \psi_0\, ,\ \psi_k\rangle\, \psi_0(x) 
+ \frac{1}{2} \langle \psi_1\, ,\ \psi_k\rangle\, \psi_1(x)
+ \frac{1}{2} \langle \psi_{-1}\, ,\ \psi_k\rangle\, \psi_{-1}(x)\, ,
$$
using the mathematicians' convention for sesquilinearity.
