I have got the 4th order coefficient of $\epsilon$ from the equation (9) from the paper. I got : $$\phi_4+ \ddot \phi_4+\omega_2 \ddot \phi_2 - \Delta \phi_2+g_2 \phi_2^2+2 \phi_1 \phi_3+ 3g_3 \phi_1^2\phi_2+g_4 \phi_1^4$$ Putting the values from equation (paper13), paper(14), paper (22) we get, $$\phi_4+ \ddot \phi_4+\omega_2 \left[-p_2 \cos\tau- q_2 \sin\tau- \frac{4g_2}{6} \right]- [\Delta p_2 \cos\tau + \Delta q_2 \sin\tau \frac{g_2}{3}(\cos2\tau-3)(p_1 \Delta p_1+(\nabla p_1)^2)] +g_2 \left[ p_2^2 \cos^2 \tau+ q_2^2 \sin^2 \tau+ \frac{g_2^2}{36} p_1^4 (\cos2\tau-3)^2+2p_2q_2 \sin\tau \cos\tau+ \frac{g_2p_1^2 q_2}{3}\sin\tau(\cos2\tau-3)+\frac{g_2 p_1^2 p_2}{3} \cos\tau (\cos2\tau-3)\right]+2\left[ p_1q_3 \sin\tau\cos\tau+p_1 p_3 \cos^2\tau+\frac{p_1^2}{3}\cos \tau\{\frac{1}{8}\left(\frac{4}{3}g_2^2- \lambda\right) p_1^2\cos3\tau +g_2[q_2 \sin2\tau+p_2(\cos2\tau-3)]\} \right]+3g_3 \left[p_1^2 p_2 \cos\tau \cos3\tau+ p_1^2q_2\sin\tau \cos^2\tau+ \frac{g_2}{6} p_1^4 \cos^2 \tau (\cos2\tau-3)\right]+g_4 p_1^4 \cos^4\tau$$
Now the writer of the paper wrote that equations (23) and (24) are the conditions for vanishing of the terms in $\sin\tau$ and $\cos\tau$. My problem is how they got equations (23) and (24)?
What the writer is implying is that the left-hand-side of equation (23) is the coefficient of $\sin\tau$ in equation (22). Similarly, the left-hand-side of equation (24) is the coefficient of $\cos\tau$ in equation (22).
To check this, you need to write equation (22) entirely in terms of $\sin\tau$ and $\cos\tau$. In other words, you need to get rid of things like $\cos 3\tau$ and $\sin 2\tau$. Conceptually, this not difficult to do -- for example, $\sin 2\tau = 2\sin\tau\cos\tau$, and $\cos 3\tau = \cos^3\tau - 3\cos\tau\sin^2\tau$, and so on.
But, the work is tedious and error-prone, so better to do it with help from a symbolic math program like Maple or Mathematica.
I haven't checked the derivation myself. I'm just making a guess about what the author means.