Time period of Sinusoidal Signal I have confusion in calculating the Time period of different signals.
1) $x(t) =3cos(4t + \pi/3)$
solution: 2$\pi$/4 = $\pi$/2 (It is periodic)
3) $x(t) =cos(t-\pi/9) +9sin(2\pi t + \pi/12)$
Solution: Time period for first sinusoid is $T_1 =2\pi$ and Time period for second sinusoid is $T_2 =1$. Since the ration is $\frac{T_1}{T_2} = 2\pi$ is Irrational. So it is not periodic.
My Question is in the first question we have $\pi$ term in the answer than how it is periodic?
 A: If you add together two sinusoids, the period of the sum is the least common multiple of the periods. For examples:
$\sin(\pi x)+\cos(2 \pi x/5)$ has period lcm(2,5)=10.
$\sin(\pi x)+\cos(\pi x/3)$ has period lcm(2,6)=6. 
$\sin(x)+\sin(2x)$ has period lcm($2 \pi$,$\pi$)=$2 \pi$.
Whenever the two periods are a rational multiple of one another, this least common multiple exists. When the two periods are irrational multiples of one another, the least common multiple does not exist. For suppose the two periods are $p_1$ and $p_2$. Suppose $p_3=k_1 p_1=k_2 p_2$ for nonzero integers $k_1,k_2$. Then $p_1=\frac{k_2}{k_1} p_2$, so $p_1$ is a rational multiple of $p_2$.
A: The first function is a simple sinusoidal with an irrational period length. That's nothing special.
Assuming that you forgot the argument t in the $9sin(2πt +π/12)$ 
The second one is the sum of sinusoidals with an irrational and a rational period.
That's quite a different beast. 
You correctly state that the second function is nonperiodic, but that is due to the ratio of the component periods being irrational, thus incompatible.
The first function has only one component, which is of course compatible with itself.
If a function $f(t)$ is periodic, there is a real value $P$ (the period) such that $f(t+n*P) = f(t) $ for all integers $n$. 
Given the sum of 2 periodic functions $f$ and $g$ with periods $P_1$ and $P_2$
we need to check, if the function $s(t) = f(t)+g(t)$ is periodic. 
If $s$ is periodic, its period $P_3$ will in general not be identical to $P_1$ or $P_2$. 
So, 
$s(t) = s(t + k*P_3) = f(t+n*P_1)+g(t+m*P_2)$  with $k,n,m$ being integers and $P_1,P_2,P_3$ reals. 
But $k*P_3 = n*P_1 = m*P_2$ must all be the same number as it is our displacement from the base case. 
That means the ratio $\frac{P_1}{P_2} = \frac{m}{n}$ must be a rational number.
In your cases, the first example is periodic being just one well-known periodic function. 
The second case is nonperiodic as the ratio in question is irrational.
