Let $r$ be an irrational real number. For real $x>0$, let $U_x$ be a random variable (r.v.) uniformly distributed on the interval $[0,x]$, and then let
$$C_x:=\cos rU_x+2\cos U_x.$$ Then the problem can be restated as follows: Is it true that
\begin{equation*}
P(C_x>0)\to1/2\,\text{?} \tag{1}
\end{equation*}
Everywhere here, the limits are taken for $x\to\infty$.

The answer to this question is yes. Indeed,
for each $(k,n)\in\{0,1,\dots\}^2$,
\begin{align*}
&E\cos^k rU_x\,\cos^n U_x \\
&=2^{-k-n}\,\frac1x\,\int_0^x du\,(e^{iru}+e^{-iru})^k \,(e^{iu}+e^{-iu})^n \\
&=2^{-k-n}\sum_{p=0}^k\sum_{q=0}^n\binom kp\binom nq
\frac1x\,\int_0^x du\,
\exp\{iu[(2p-k)r+2q-n]\} \\
&\to2^{-k-n}\sum_{p=0}^k\sum_{q=0}^n\binom kp\binom nq
1(2p=k,2q=n),
\end{align*}
since $r$ is irrational. So,
\begin{align*}
&E\cos^k rU_x\,\cos^n U_x\to m_k m_n,
\end{align*}
where
\begin{equation*}
m_k:=1(k\text{ is even})2^{-k}\binom k{k/2}=E\cos^k U_{2\pi}.
\end{equation*}
So, by dominated convergence, for the joint characteristic function (c.f.) $f_{r,x}$ of the pair $(\cos rU_x,\cos U_x)$ of r.v's and all real $s,t$ we have
\begin{align*}
f_{r,x}(s,t)&=E\exp\{i(s\cos rU_x+t\cos U_x)\} \\
&=\sum_{n=0}^\infty \frac{i^n}{n!}\,E(s\cos rU_x+t\cos U_x)^n \\
&=\sum_{n=0}^\infty \frac{i^n}{n!}\,
\sum_{k=0}^n\binom nk s^k t^{n-k}E\cos^k rU_x\,\cos^{n-k} U_x \\
&\to\sum_{n=0}^\infty \frac{i^n}{n!}\,
\sum_{k=0}^n\binom nk s^k t^{n-k}m_k m_{n-k} \\
&=h(s)h(t),
\end{align*}
where
\begin{equation*}
h(s):=\sum_{k=0}^\infty \frac{i^k s^k m_k}{k!}=E\exp\{is\cos U_{2\pi}\},
\end{equation*}
so that $h$ is the c.f. of the (symmetric absolutely continuous) r.v. $\cos U_{2\pi}$.

So, the pair $(\cos rU_x,\cos U_x)$ of r.v's converges in distribution to a pair $(A,B)$ of independent copies of the r.v. $\cos U_{2\pi}$. So, for any real $b$, the r.v. $\cos rU_x+b\cos U_x$ converges in distribution to the symmetric absolutely continuous r.v. $A+bB$.

Thus, (1) follows.