Find the spectrum of $(Ff)(x)=\int_{-\pi}^{\pi} (1 + \cos(x-y))f(y)dy$

@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.