How can one solve this system of equations: $$ \begin{align} x(t) = a \cos(\omega t + \phi) + \dfrac{\beta a^3}{32} \cos(3(\omega t + \phi)), \\ p(t) = -a \omega \sin(\omega t + \phi) - 3 \omega \dfrac{\beta a^3}{32} \sin(3(\omega t + \phi)),\\ \omega = \omega_0 + \dfrac{3}{8} a^2, \end{align}, \\ \text{if } x(0) = x_0, p(0) = p_0 $$ where $\omega_0, \beta$ are constants?
I can write:
$$ \begin{align} x_0 = a \cos(\phi) + \dfrac{\beta a^3}{32} \cos(3\phi), \\ p_0 = -a \omega \sin(\phi) - 3 \omega \dfrac{\beta a^3}{32} \sin(3\phi)),\\ \omega = \omega_0 + \dfrac{3}{8} a^2, \end{align} $$ But i can't figure out how to solve it for $a$ and $\phi$. Could you help?
Not a solution but possibly helpful. Make use of the triple angle formulas \begin{align} \sin(3\theta) &=3\sin\theta−4\sin^3\theta,\\ \cos(3\theta) &= 4 \cos^3\theta - 3 \cos \theta. \end{align} The first equation ($x_0 = \cdots$) will end up being third order in $\cos\phi$ and the other ($p_0 = \cdots$) third order in $\sin\phi$. With a bit of luck you can solve the first for $\cos\phi$ and the second for $\sin\phi$, then take it from there knowing that $\cos^2\phi + \sin^2\phi = 1$.