Difficult problem involving a percentage of the period of a sinusoid Im having difficulty intuitively understanding how to solve this problem:
$x(t) = A\cos(\omega t + \phi)$
$A > 0$
$\phi\in(−\pi,\pi]$.
$x(t) ≥ 2.4$ for $18$% of each period
takes $0.123$ seconds for $x(t)$ to drop from $2.4$ to the next minimum
the ﬁrst zero for $t>0$ occurs at $t = 0.040$ seconds. 
How does one solve this type of problem for $x(t)$? I am not sure how to find the period at all.
 A: This isn't a complete solution, but hopefully it will help you on your way.
If $x(t) = 0$, then you know that $\omega t + \phi = \pm\pi/2$. Let $\omega > 0$. (You could also say $\omega < 0$ and get an equivalent answer.) At $t=0.040$ seconds, $x(t)=0$ for the first time, which means that the cosine argument must be $\pi/2$, so
$$
\omega \cdot 0.040 + \phi = \pi/2
$$
Something else you know is that the minimum will occur when the cosine argument is equal to $\pi$. Also, say that at $x(t_0)=2.4$ and picture a line at $2.4$ cutting the top off the sinusoid. If $x$ is greater than $2.4$ for $18$ percent of the period, then you know that $x(t_0 + 0.18T) = 2.4$, where $T=2\pi/\omega$ is the period. Furthermore, you know that at $t_0 + 0.09T$, $x$ reaches its peak. If it takes $0.123$ seconds to reach the minimum from $2.4$ and it takes $0.09T$ seconds to go from its peak to $2.4$, then that is half a period:
$$
0.09T + 0.123 = T/2.
$$
With the period, you can get the frequency. With the frequency you can get $\phi$. Finaly, with both $\omega$ and $\phi$, you can get $A$.
I realize the above is kind of rambling and disjoint, but I think it will help you find the answer. 
