Finding common points of $\cos x$ and $(1/2)^x$ This might be easy math for you guys but I'm just not getting it.
Question says,graph:
$y= \cos x$
$y=(\frac{1}{2})^x$
On the same axis. Easy enough.
Next part says, state $4$ points where the two intersect.
I think they mean algebraically, because other way it would be too easy.
So I did $\cos x=(\frac{1}{2})^x$ and blanked out.
I tried a learning resource for Grade $12$ math to no avail.
Am I misunderstanding the question? If not, how do I solve this?
Hints are just as good as answers, thanks for your help! 
 A: I am going to gamble on approximate solutions being the best solutions. The equation $\cos(x)=\left(\frac{1}{2}\right)^x$ clearly has integer solution $x=0$. There will be no other rational solutions, and that is a discussion probably deserving its own page. There is certainly no algebraic solution. However, we can find infinitely many approximate solutions, each becoming better by the following logic:
As $x$ approaches infinity (becomes larger), the function $\left(\frac{1}{2}\right)^x$ monotonically converges to zero. Formally,
$$\displaystyle\lim_{x \rightarrow \infty}\left(\frac{1}{2}\right)^x =0.$$
The $\cos(x)$ function however has zeroes wherever $x=\frac{(2n+1)\pi}{2}$, where $n$ is an integer ($n\in\mathbb{Z}$).
What this specifically means in regards to finding approximate solutions to the equation $\cos(x)=\left(\frac{1}{2}\right)^x$, is that as $n$ becomes larger, the solution $x=\frac{(2n+1)\pi}{2}, \, n \in \mathbb{N}$ becomes more exact.
It should be said that the $x$ intersections will never be rational for $x>0$.
Just for fun you might declare that, other than $x=0$, you can declare a solution that is more precise than any other in the class. For example, other than $x=0$, I doubt that anyone will declare a more exact solution than $x=\dfrac{\left(2 \cdot 10^{100}+1\right) \pi}{2}$. I usually save googols for when I really need them though in circumstances like these.  
