You need to specify $\lambda_k$ more clearly. More precisely, what is the spectrum of this operator? The equation

$$ \cos x\cosh x =1 $$

seems to have a unique solution $\mu_k$ on any interval of the form $(k\pi/2, k\pi/2+\pi)$, $k\in\mathbb{Z}$, so I assume the spectrum might be $\mu_k^4$, $k\in\mathbb{Z}$?!? (Please edit you question to remove this ambiguity.) In any case if the boundary value problem is elliptic (please *check that*) then the spectrum is discrete. In particular it can be determined by finding the eigenfunctions which means solving some ode's. My guess is that you found all the eigenfunctions, i.e., the system you found is complete.

**Update.** You need to check two things: 1) the boundary value problem is elliptic 2) it is symmetric. I'll deal with the 2nd issue first because it is faster. Denote by $A$ the operator

$$A=\frac{d^4}{dx^4}. $$

A simple integration by parts shows that for any $u,v\in C^4([0,1])$ we have

$$ \int_0^1 \bigl(\; v(Au) -u(Av)\;\bigr) dx=\sum_{j=0}^3(-1)^j\bigl( u^{(3-j)}(1)v^{(j)}(1)- u^{(3-j)}(0) v^{(j)}(0)\;\bigr). $$

If the function $u$ satisfies your boundary conditions $u^{(k)}(x)=0$ for $k=2,3$, $x=0,1$ the above equality simplifies a bit

$$ \int_0^1 \bigl(\; v(Au) -u(Av)\;\bigr) dx= \sum_{j=2}^3(-1)^j\bigl( u^{(3-j)}(1)v^{(j)}(1)- u^{(3-j)}(0) v^{(j)}(0)\;\bigr). $$

If the function $v$ satisfies the same boundary conditions as $u$, then the last equality takes the very simple form

$$ \int_0^1 \bigl(\; v(Au) -u(Av)\;\bigr) dx= 0. $$

This says that the boundary value problem is symmetric, or formally selfadjoint.

The ellipticity of this problem is another issue. The most readable account I could find is in Chap. 20 vol.3 of the book *The Analysis of Linear Partial Differential Operators* by the late great Lars Hormander.

The ellipticity of the boundary value problem requires that the symbol of your operator $A$ be elliptic (which it is) and that the boundary value conditions should satisfy the so called *Lopatinskii-Schapiro conditions*.

In your case they are trivially satisfied because you work on a one-dimensional space $[0,1]$. The upshot is that in your case the boundary conditions are elliptic. We can form the unbounded operator $\newcommand{\bD}{\boldsymbol{D}}$

$$ \hat{A}: \bD(\hat{A})\subset L^2(0,1)\to L^2(0,1), u\mapsto \frac{d^4 u}{dx^4} $$

Where the domain $\bD(\hat{A})$ of $\hat{A}$ consists of functions in the Sobolev space $L^{4,2}(0,1)$ (four weak derivatives in $L^2$) such that $u^{(j)}(x)=0$ for $x=0,1$, $j=2,3$.

The results in the above monograph show that $\hat{A}$ viewed as an unbounded operator on the Hilbert space $L^2(0,1)$, is closed, densely defined, selfadjoint and has compact resolvent. This is all you need. Arguably, the above argument is a bit heavy, and it feels like hunting a mosquito using a bazooka.

There is a direct, more elementary approach to proving that $\hat{A}$ has compact resolvent. Observe first that the above integration by parts formulae show that the operator $\hat{A}+1$ is positive, i.e.,

$$ (\hat{A}u,u)_2+(u,u)_2>0,\;\;\forall u\in \bD(\hat{A})\setminus 0, $$

where $(-,-)_2$ denotes the $L^2$-inner product. Hence $\hat{A}+1$ is injective. Then follow the strategy in the proof of Theorem 8.22 in Brezis' book *Functional Analysis, Sobolev Spaces and Partial Differential Equations* to prove that $\hat{A}+1$ is invertible and it's inverse is compact as an operator $L^2(0,1)\to L^2(0,1)$.