Let us consider the following operator $$ Af(x)=\sin(x)f(x), $$ in $L^2(\mathbb R)$.
This operator is self-adjoint, so $\sigma(A)\subset \mathbb R.$
Solving equation $(A-\lambda I)f=g,$ we obtain $(A-\lambda I)^{-1}g=\frac{g}{\sin x-\lambda}$, and this formula defines resolvent if $\lambda\in\mathbb R\setminus[-1,1].$
Let $\lambda \in [-1,1].$ From the fact, that $\parallel \frac{1}{\sin x- \lambda}\parallel=\infty$, we get, that $[-1,1]$ is continuous spectrum.
Now I am trying to check, if this continuous spectrum is (or is not)a absolutely continuous spectrum. I would by very grateful if anyone can give me a hint, how to do that.