As an exercise, I was told to find the spectrum of the bounded operator $K\in B(L^2[-\pi,\pi])$ defined by $$K\varphi (t)=t\int_{-\pi}^\pi\varphi (x)\cos (x)dx+\cos t\int_{-\pi}^\pi x\varphi(x)dx.$$
This seems pretty horrifying, so I thought of looking at one eigenspace at a time i.e at the span of one of $\cos(nt),\sin(nt),1$. If $\varphi(t)$ is in the span of $1,\cos(nt)$, the second integral dies, and we get something proportional to $t$, so no eigenvectors.
If $\varphi$ is in the span of $\sin(nt)$ the first integral dies and $K\varphi(t)$ is proportional to $\cos(t)$ which is orthogonal to $\sin (nt)$. Hence I conclude there are no eigenvectors and so no eigenvalues.
Is my solution correct? Moreover, since the spectrum should actually check the values for which $K-\lambda I$ is not invertible, did I miss anything?
Note that the range of $K$ is in a finite dimensional subspace
$$\text{span} \{x, \cos x\},$$
thus $K$ is a compact operator and has only point spectrum ($0$ is a eigenvalue too).
To look for eigenfunction, we need only restrict ourselve to the above subspace. Note that
$$ K (ax+b \cos x) = b\pi x + a\frac{2\pi^3}{3} \cos x,$$
This implies that
$$ K( x \pm \sqrt{\frac 23} \pi \cos x) = \pm \sqrt{\frac 23} \pi^2( x \pm \sqrt{\frac 23} \pi \cos x)$$
Thus $K$ has three eigenvalues $0, \pm \sqrt{\frac 23} \pi^2$.