I doubt that this vector ODE be explicitly integrable. However you can do the following.
Write it as $\frac{d\phi}{dx}=\lambda A(x)\phi$, where
$$A(x)=\begin{pmatrix} \sin2\pi x & 1- \cos2\pi x \\ -1- \cos2\pi x & - \sin2\pi x \end{pmatrix}.$$
Notice ${\rm Tr} A\equiv0$.
The general solution of the Cauchy problem is given in the form $\phi(x)=M(\lambda;x)\phi(0)$. Your question amounts to whether $\mu=1$ is an eigenvalue of $M(\lambda;1)$, or not.
First of all $\frac{d\det M}{dx}=\lambda ({\rm Tr}A(x))\det M\equiv0$, hence $\det M\equiv1$ (because $M(\lambda;0)=I_2$). Thus $1$ is an eigenvalue if and only if the trace of $M(\lambda;1)$ equals $2$.
Thus we are led to the study of $\lambda\mapsto{\rm Tr}M(\lambda;1)$. To this end, we notice that
$$\frac{dM}{dx}=\lambda A(x)M,\qquad M(\lambda;0)=I_2.$$
The map $\lambda\to M$ is analytic. Expanding $M=I_2+\lambda M_1+\lambda^2M_2+\cdots$, we have $M_1'=A$ and $M_2'=AM_1$ with $M_j(\lambda;0)=0_2$. This can be integrated explicitly. Obviously ${\rm Tr}M_1\equiv0$. If my calculation is correct, we have ${\rm Tr}M_2(1)=-1$. Thus $|{\rm Tr}M(\lambda;1)|<2$ for small $\lambda\ne0$.
This is a useful information: when $\lambda\ne0$ is small, then the eigenvalues of $M(\lambda;1)$ are $e^{\pm i\theta}$, where $\theta(\lambda)$ varies analytically, with $\theta'(0)\ne0$. For a dense subset near zero, the frequency has the form $\theta(\lambda)=\frac{2\pi p}q$ where $\frac pq$ is rational (irreducible). Then every solution is $q$-periodic.
To answer properly and positively your question, there remains to check whether $\lambda\to{\rm Tr}M(\lambda;1)$ takes values above $2$. If so, there is a parameter $\lambda$ for which this trace equals $2$, and thus $\mu=1$ is a double eigenvalue. Then the ODE admits a $1$-periodic solution.