We shall consider the matrix-valued differential operator


$$(L u)(x) :=u'(x) - \begin{pmatrix} 0 & \sin(2\pi x-\frac{\pi}{6})\\ - 2\sin(2\pi x+\frac{\pi}{6}) & 0 \end{pmatrix} u(x).$$


This is a $1$-periodic operator. Thus, does there exist a $\lambda \in \mathbb C$ and a $1$-periodic solution to this ODE such that


$$ (L - \lambda)u = 0.$$


Probably there is no explicit solution, but can we show the existence of such a solution?



Existence of periodic solution to ODE
The solutions of $(L-\lambda)u=0$ are the functions $u(x)=e^{i\lambda x}v(x)$, where $v$ satisfies $Lv=0$. The periodicity amounts to $e^{i\lambda}v(1)=v(0)$. Thus your problem does admit infinitely many solutions. Just consider the monodromy matrix $M:v(0)\mapsto v(1)$, whose determinant equals $1$ (by the Wronskian). Take an eigenvalue $\mu$ of $M$, which is therefore non-zero. Then any $\lambda\in{\mathbb C}$ such that $e^{i\lambda}=\mu$ solves your problem.


