I'm to prove that the following equation has a unique solution:
$$f(x) = \int_0^1 e^{-sx} \cos(\alpha f(s)) ds.$$
(Here, $\alpha \in (0,1)$.) The form of the exercise screams to apply the contraction mapping principle, so I want to show that the operator is a contraction.
Let $f,g \in C[0,1]$, then we want
$$\left|\int_0^1 e^{-sx} \cos(\alpha f(s))ds - \int_0^1 e^{-sx} \cos (\alpha g(s)) ds\right| < c||f-g||$$
for some constant $c < 1$. The left-hand side is equal to
$$\int_0^1 e^{-sx} [\cos(\alpha f(s)) - \cos(\alpha g(s)) ]ds$$
which itself is less than or equal to
$$\int_0^1[\cos(\alpha f(s)) - \cos(\alpha g(s))]ds,$$
but I don't see an easy way to work with this integrand.