We have a one-variable function for $x>0$ where $f_i(x)$ are some periodic trigonometric functions
$F(x)=f_1(x)\cdot f_2(x)\cdot f_3(x)$
We know that
- $f_1(x)$ is positive for $\frac23$ of its domain;
- $f_2(x)$ is positive for $\frac12$ of its domain;
- $f_3(x)$ is positive for $\frac12$ of its domain;
Using this data, is it possible to conclude that $F(x)$ is positive for what portion of its domain?
The way the question is worded, it doesn't clearly state that the three domains are the same, but if they aren't, then no conclusions can be drawn, so I'll assume they are. We can assume without loss of generality that the domain is $(0,1)$, and we can further assume that $f_2$ is negative for $(0,0.5)$ and positive for $(0.5,1)$. We can then assume that $f_3$ is negative for $(x,x+0.5)$ for some $0<x<0.5$. Then $f_2f_3$ is positive in $(x,0.5)$ and $(x+0.5,1)$, and those ranges have size $0.5-x$ each, for a total of $1-2x$. As $x$ can range from $0$ to $0.5$, $f_2f_3$ can be positive anywhere from $0$ to $1$ length.
If $f_2f_3$ is positive for two thirds of the domain, then we can line up its negative portion with $f_1$'s and get the product to be positive (almost[1]) everywhere. If $f_2f_3$ is positive for one third of the domain, we can line its positive portion up with the negative portion of $f_1$, and make the product negative (almost) everywhere. So there is no upper or lower limit to what percentage of the domain $F$ is positive (other than not more than 100% and not less than 0%, of course).
[1] The function will have go between positive and negative, and trigonometric functions are continuous, so they have to hit zero at some point. So while $F$ can't be positive everywhere, it can be positive everywhere except for a measure-zero set.