Period of a continuous signal So I have 
$$x_1=2 \cos(.6\sqrt\pi x+\pi/6)$$ and $$x_2= \sin(1.2\sqrt\pi x-\pi/3)$$ and need to find the period of $(x_1+x_2)^2$.

Let $$a=.6\sqrt\pi x+\pi/6 ~~~\text{and} ~~~ b=1.2\sqrt\pi x-\pi/3$$ so that 
$$x_1=2 \cos(a) ~~ \text{and} ~~ x_2=\sin(b).$$ 

And finally we get: 
$$(x_1+x_2)^2=1.5+2\cos(2a)+2\sin(a+b)+2\sin(a-b)-.5\cos(2b)$$

Now, the how do I calculate the time period of it?

Edit:

$$\text{period of the individul sinusoids in the square}~~T_1=5\sqrt\pi/3 ~~ T_2=10\sqrt\pi/9~~T_3=10\sqrt\pi/3 ~~ T_4=5\sqrt\pi/6 $$

$$\text{The ratios are}~~T_4/T_3=1/4~~T_4/T_2=3/4~~T_4/T_1=1/2$$

$$\text{LCM of denominators is 4 } $$

$$\text{Hence period }~~T=4T_4=10\sqrt\pi/3$$
is that correct?
Let $T_0$ be the period of the signal  $s(t) = x_1(t) + x_2(t)$.
$$s(t) = s(t+T_0) \tag{1}$$

Then the period for $s^2(t) = ( x_1 + x_2)^2$ will be either $T_0$ or $T_0/2$ :

First, it's obvious that $T_0$ is also a period for $s(t)^2$ as it satisfies $$s^2(t) = s^2(t+T_0) \tag{2}$$ 

However, $T_0 / 2$ may also be period for $s^2(t)$. Since we are looking for the smallest number to satisfy eq(2).

Hence do the following:


Find the period $T_0$ of $s(t) = x_1(t)+x_2(t)$.
Check if $T_0/2$ satisfies $s^2(t) = s^2(t+T_0)$.
If yes; then period of $s^2(t)$ is $T_0/2$; otherwise $T_0$.