We know that the two functions $\{\;\cos (ax),\;2\cos (b x)\;\}$ where $\frac ab \notin \mathbb{Q}$ are independently positive (and negative) over $\frac 12$ of the domain.


Is it possible to estimate the fraction of the domain in which $\;\cos (ax)+2\cos (b x) \;$ is positive (in this case, the function is not periodic)?


In other words, can we estimate
$$ \lim\_{m\to\infty} \frac 1m \int\_{0}^{m} \Big(\cos (ax)+2\cos (b x) >0\Big) \,dx?$$


Any hints and comments are greatly appreciated.



Are we able to estimate the fraction of the domain where $\cos (ax)+2\cos (b x)$ with $\frac ab \notin\mathbb{Q}$ is positive?
Since $a,b$ are incommensurable, $(ax,bx)$ is asymptotically equidistributed
in the torus $({\bf R} / 2\pi{\bf Z})^2$.
[One proof is via a continuous version of Weyl's equidistribution criterion:
for any integers $r,s$ with $(r,s) \neq (0,0)$ we have
$$
\frac1m \int\_0^m e^{i(rax+sbx)}\, dx = O\_{r,s}(1/m) \to 0
$$
as $m \to \infty$.] Therefore $\{x > 0 \mid \cos ax + 2 \cos bx > 0 \}$
has the same density in the positive reals as
$\{ (\theta,\phi) \in ({\bf R} / 2\pi{\bf Z})^2 \mid 
 \cos \theta + 2 \cos \phi > 0\}$, which is $1/2$ by symmetry.


