I am given a covering map $p: \mathbb{R}^+ \times \mathbb{R} \to \mathbb{R}^2 \setminus \{0,0\}$ defined by $p(r, \theta)=(r \cos 2 \theta,r \sin 2 \theta)$
Let $\alpha: [0,1] \to \mathbb{R}^2 \setminus \{0,0\}$ defined by $\alpha(t)=(2-t,0)$.
I want to find the lift of $\alpha$, namely $\tilde{\alpha}(t): [0,1] \to \mathbb{R}^+ \times \mathbb{R}$ such that $\alpha=p \circ \tilde{\alpha}$
$\textbf{My Attempt:}$ I need $\tilde{\alpha}$ to be such that $r \sin \theta =0$ and $r \cos \theta =2-t$
I am really not sure how to solve this. It is probably just algebra but I would really appreciate some hints in the right direction.
To find the lift, define $\tilde\alpha(t) = (r=f(t) , \theta=g(t))$
$$\alpha = ( 2-t,0) = p \circ \tilde \alpha = ( f(t) \cos( 2 g(t) ) , f(t) \sin( 2 g(t)) $$
From here it's easy to see if we take
$$f(t) = \frac{2-t}{ \cos ( 2 g(t) ) } = \pm( 2-t) \quad \& \quad g(t) = \frac{1}{2} (n \pi)\quad \quad n \in \mathbb{N}$$
This will lift will project down to the curve you want, the problem is your projection map doesn't give you much freedom.