$x$-intercept of cosine graph I am having problems understanding how to find the $x$-intercept of a cosine graph. 
Example:
$10\cos(x/2)$
Answer:$((2n + 1)\pi , 0 )$
I have the answer just need help understanding the steps, thanks
 A: The cosine function (i.e. $\cos(x)$) is zero at $x=\text{some odd integer multiple of}\;\frac{π}{2}$. You can see that in the unit circle. $\cos$ is 0 at $\frac{π}{2}, \frac{3π}{2},\frac{5π}{2}\;\text{or going clockwise at}\;\frac{-π}{2}, \frac{-3π}{2}, \frac{-5π}{2}$, etc.

So, it makes sense that the zeros of $\cos(x/2)$ would be at odd integer multiples of $π$ (rather than odd integer multiples of $\frac{π}{2}$ because the $2$ in the denominator is already there).
In the solution you provided, $((2n+1)π,0), n$ is any integer. Because of this, $2n+1$ is any odd integer. 
The 10 doesn't matter because $10$ times $0$ is still $0$. 
A: $$y=10 \cos{ \left (\frac{x}{2} \right )}$$
$y-$intercept:
$$x=0 \Rightarrow y=\cos{ 0 }=1$$
$\left ( 0,1 \right )$
$$\\ \\ \\$$
$x-$intercept:
$$y=0 \Rightarrow 10 \cos{ \left (\frac{x}{2} \right )}=0 \Rightarrow \cos{ \left (\frac{x}{2} \right )} =0 \Rightarrow \frac{x}{2}=\pi n+\frac{\pi}{2} \Rightarrow x=2 \pi n +\pi=(2n+1) \pi$$
$\left ( (2n+1) \pi, 0 \right )$
A: 1) The x-intercept of a  graph of a function $f(x)$ is in the form $(x_0, 0)$, where $x_0$ is solution of the equation $f(x)=0$
2) $10\cos{\frac{x}{2}}=0$
$$10\cos{\frac{x}{2}}=0$$
$$\frac{x}{2}=\frac{\pi}{2}+\pi n$$
$$x=\pi+2\pi n=\pi(2n+1)$$
I am sorry for my bad English
