Finding the period of this periodic function How do you calculate the period of the following:
$$x(t)=\dfrac{\sin(2t)+\sin(3t)}{2\sin(t)}$$
 A: As some other answers remarked, the three basic components of $x(t)$, namely $sin(t)$, $sin(2t)$ and $sin(3t)$ have $2\pi$ has a period, thus $x(t)$ also has $2\pi$ a period. 
[Recall that the definition for a function $f(x)$ defined over $\mathbb{R}$ to have $\ell\neq 0$ as a period is that for all $x$ we have $f(x+\ell)=f(x)$.
If $\ell$ is a period of $f$, then any multiple of $\ell$ is also a period of $f$. By definition, the period is the smallest period of $f$ (if it exists).
]
Now, to show that $2\pi$ is the period of $x(t)$, we need to prove that $2\pi$ is not only a period of $x$ but the smallest of those. For this, you need to show that no strict divisor of $2\pi$ is a period of $x(t)$. This can be done, for example, by studying the table of variation of $x$ which shows that the only candidate period would be $\pi$ and then verifying that since $x(0)=5/2$ and $x(\pi)=1/2$, $\pi$ cannot be a period of $x(t)$.
This final step of verifying that no shorter period exists is essential, as shown by the following example:
$$
x'(t)=\frac{sin(3t)+sin(5t)}{sin(t)},
$$
whose period is $\pi$ (and not $2\pi$, despite the fact that $2\pi$ is the lcm of $2\pi$, $2\pi/3$ and $2\pi/5$).

Final remark on the definition of periods:
The definition given between brackets only applies to function defined all over the real line. If the set of definition of $f$ is a strict subset of $\mathbb{R}$, then we need to add the condition that this set of definition is invariant by translation by $\pm \ell$.
A: $$\sin(2t)=2\sin t\cos t$$
$$\sin(3t)=3\sin t-4\sin^3t$$
Thus
$$x(t)=\cos t+\frac{3}{2}-2\sin^2t=2(\cos t+\frac{1}{4})^2-\frac{5}{8}$$
The period of $x(t)$ is $2\pi$
A: Notice how $x \left( t \right)$ is composed of a combination of $\sin{t}$, $\sin{2t}$ and $\sin{3t}$ terms.  The period of $x \left( t \right)$ is simply the least common multiple (lcm) of each of the sine term's periods.  Therefore the answer is $2 \pi$.
Be careful how you use this 'theorem' however, as it can easily be abused.  For instance, the period of $y \left( t \right) = \left( \sin{2t}+\sin{t}-\sin{t} \right)$ is $\pi$.  If we tried to calculate the lcm without first simplifying $y \left( t \right)$, we might arrive at the erroneous answer $2 \pi$.
A: You could start by graphing the function

But it helps to look at the periods of the separate parts of the function
$\sin(2t)$ has period $\pi$
$\sin(3t)$ has period $\frac{2 \ \pi}{3}$
$\sin(t)$ has period $2 \pi$
We are looking for the shortest period into which each of these divides and the answer is $2 \pi$ 
