At what value(s) of $x$ does $\cos x = 5x$? I've been given the following problem:
At what value(s) of $ x $ does $\cos x=5x $?
I'm not sure how to go about solving this problem.  If someone could please solve it step by step, that'd be great.
 A: It's not sophisticated, but you can approximate the answer using a series expansion
$$
\cos(x) = 1-(x^2/2)+x^4/24-x^6/720... = 5x
$$
Then, ignore the really small terms...
$$
\cos(x) \approx 1-(x^2/2) = 5x
$$
Then, rearrange to
$$
-x^2 - 10x + 2 = 0
$$
... giving a positive root of $3\sqrt(3)-5$
Checking:
$\cos(3\sqrt(3)-5) = 0.9808237171904401982886706009894$
$5*(3\sqrt(3)-5) = 0.98076211353315940291169512258809$
You could get more accurate by including more terms of the series expansion, if you know how to compute the roots of higher order equations like
$$
-x^6+30 x^4-360 x^2-3600 x + 720 = 0
$$
which is the equation you get from including one more term of the series expansion.
A: Study $f:x\mapsto \cos x - 5x$ :
Its derivative is $f':x\mapsto -\sin x - 5 \lt0$ thus $f$ is decreasing over $\Bbb R$, $f(0) = 1\gt0$ and $f({\pi\over2}) = -5{\pi\over2}\lt0$ thus there is only one solution over $\Bbb R$ and it is located in $[0, {\pi\over2}]$.
However there is no closed form of this root.
A: Here are graphs for $\cos x$ and $5x$.  We can see there is only one crossing point where $\cos x = 5x$, right?

A: A typical transcendent equation for which there is not elementary closed form of solution. Approximating, we know that $\cos 0=1$ and $5\cdot 0.2=1$ so the solution is near of $0.2$ which is near to $0$. We have  $\cos 0.2=0.999993907$ and $5\cdot 0.2=1$ so $\cos 0.2\lt5\cdot 0.2$ Proving with $x=0.199$ we find a better approximation and with $x=0.197$ we get $\cos 0.197=0.999994089$ and $5\cdot 0.197=0.985$ 
For many calculations this value $\color{red}{x=0.197}$ could be a good enough approximation.
