Show that the group $H_{\omega}^{ev}(X)$ depends only on the cohomology de-Rhan class $[\omega]$.

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Let $\Omega^{\bullet}(X)$ be the space of differential forms on a smooth manifold $X$. Let $\omega$ be an odd degree form. Consider

$$d_{\omega}:\Omega^{\bullet}(X) \to \Omega^{\bullet}(X) $$

such that $d_{\omega}(\eta)= d\eta + \omega\wedge\eta$. Now consider the twisted cohomology group

$$H_{\omega}^{ev}(X):= \ker(d^{ev}_{\omega})/ {\rm Im}(d^{odd}_{\omega})$$ where $d_{\omega}^{ev}:\Omega^{2k}(X) \to \Omega^{2k}(X)$, and $d^{odd}_{\omega}$ is defined in a similar way.

Show that the group $H_{\omega}^{ev}(X)$ depends only on the cohomology de-Rhan class $[\omega]$.

Can anyone provide me an idea of what this question is asking for?