Functional analysis exercise - how to use Fredholm theory to solve integral equation I need to find for which $\lambda\in\mathbb{R},1<p<\infty$, the following equation: $$f(x)-\lambda\int_0^{\pi/2}\cos(x-y)f(y)dy=1$$ has a solution in $L^p[0,\pi/2]$ over $\mathbb{R}$. The direction given is to use Fredholm theory for spectrum of compact operators and to take derivatives inside integrals. I found a solution that uses other tools, and I'm interested in finding out how the direction given can be used to solve it in another way.
 A: Here is how I would do it: The kernel is separable since $\cos(x-y) = \cos(x)\cos(y) + \sin(x) \sin(y)$. Thus we can write $$f(x) - \lambda \cos(x) \int^{\pi/2}_0 \cos(y) f(y) dy - \lambda \sin(x) \int^{\pi/2}_0 \sin(y) f(y) dy = 1.$$ Put $f_c = \int^{\pi/2}_0 \cos(y)f(y) dy, \,\,\, f_s = \int^{\pi/2}_0 \sin(y) f(y) dy$. Then $$f(x) -\lambda f_c \cos(x) - \lambda f_s \sin(x) = 1.$$ If we can solve for $f_c, f_s$ then we are done to do so, multiply the equation by $\cos(x)$ and integrate. This gives $$f_c - \lambda f_c \int^{\pi/2}_0 \cos(x)^2 dx - \lambda f_s \int^{\pi/2}_0 \cos(x) \sin(x) dx = \int^{\pi/2}_0 \cos(x) dx$$ or $$f_c - \frac{\pi\lambda}{4} f_c - \frac{\lambda}{2} f_x = 1.$$ Likewise, multiplying the above equation by $\sin(x)$ and integrating gives $$f_s - \frac{\lambda}{2} f_c- \frac{\pi \lambda}{4} f_s = 1.$$ Together these equations form the system $$\left( \begin{matrix} 1 - \frac{\pi \lambda}{4} & -\frac{ \lambda}{2}\\ -\frac{\lambda}{2} &1 - \frac{\pi \lambda}{4}\end{matrix} \right)\binom{f_c}{f_s} = \binom 1 1.$$ Thus  the integral equation has a solution whenever this linear system has a solution; I leave the analysis of this linear system to you.
EDIT: Here is another method. We see that $$f(x) = 1 + \lambda \int^{\pi/2}_0 \cos(x-y) f(y) dy.$$ Now if $f \in L^p[0,\pi/2]$, then the right hand side is continuous and differentiable to any order (since it is a convolution with a smooth function) so that means the left hand side is also differentiable and we see by differentiating under the integral $$f'(x) = -\lambda\int^{\pi/2}_0 \sin(x-y) f(y) dy$$ and thus $$f''(x) = -\lambda\int^{\pi/2}_0 \cos(x-y) f(y)dy.$$ Adding this  last equation to the first gives $$f''(x) + f(x) = 1$$ which has solution $$f(x) = 1 + A \cos(x) + B \sin(x)$$ as before. Of course, out ability to solve for $A,B$ will depend on $\lambda$. 
