Hammerstein integral equation with inverse of the solution

In signal processing theory I found this integral equation that I recognized to be of Hammerstein type: $$u(t)-\int_{0}^{1}d\phi cos(\omega t+\phi)\frac{1}{u(\phi)}=0$$ Unfortunately the solution function $u(t)$ appears in the denominator, so I am unable to solve it. Can someone help me? Thanks.

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If you differentiate the equation twice in t, you can get $u′′(t)+\omega^2 u(t)=0$ so $u(t)=A\cos(\omega t+\beta)$,which can then be solved by plugging it back into the original equation and plugging in something like $t=0$. Something about this feels quite strange though because the units don′t make physical sense. u(t) looks like some sort of signal that depends on time but $\phi$ is a phase shift of radians. Are you sure this equation is correct? –  Alex R. Jul 5 '12 at 14:19
It's seems strange also to me. In any case I found something about an equation of this kind here: www-users.mat.uni.torun.pl/~tmna/files/v15n2-02.pdf but my feeling is that maybe a simpler treatment can be found. –  Riccardo.Alestra Jul 5 '12 at 14:43
The other way is to iterating integration by parts. In this way you easily recognize that the solution can be cast in a form similar to the one given in the previous comment. –  Jon Jul 13 '12 at 12:28
What if in the kernel the term is of the form $cos(\phi t)$. I do know, in signal theory such a kernel could not arise, but i faced a similar problem in quantum field theory and i have to take a long route to solve it. –  Immo Ali Nov 9 '14 at 20:45
Still another possibility is to write $\cos(\omega t+\phi)=\cos\omega t\cos\phi-\sin\omega t\sin\phi$, which immediately displays $u$ as a linear combination of $\sin\omega t$, $\cos\omega t$. –  Christian Remling Nov 10 '14 at 1:18