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.

  • 1
    $\begingroup$ I'm not sure there is a closed form for this solution $\endgroup$ – Astyx Nov 7 '16 at 22:18
  • 2
    $\begingroup$ Related question here: math.stackexchange.com/questions/46934/… $\endgroup$ – Von Kar Nov 7 '16 at 22:28
  • $\begingroup$ No closed form solution is known, or is likely to exist. You must use numerical methods such as Newton's, or iterate $x_{n+1} = \cos(x_n)/5$ starting from, say, $x_1 = 0$. $\endgroup$ – Robert Israel Nov 7 '16 at 22:35

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 $3\sqrt(3)-5$


$\cos(3\sqrt(3)-5) = 0.9808237171904401982886706009894$

$5*(3\sqrt(3)-5) = 0.98076211353315940291169512258809$

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.

  • $\begingroup$ Lol, I use this method all the time. Especially since quadratics are easy to solve. $\endgroup$ – Simply Beautiful Art Nov 8 '16 at 22:14

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.

  • $\begingroup$ $(\cos x)'=-\sin x$. $\endgroup$ – Nyfiken Nov 7 '16 at 22:27
  • $\begingroup$ Yes of course, I edited, thanks ! $\endgroup$ – Astyx Nov 7 '16 at 22:28
  • $\begingroup$ Of course we assume $\cos x$ is the cosine in radians, not degrees... $\endgroup$ – GEdgar Nov 7 '16 at 22:45

Here are graphs for $\cos x$ and $5x$. We can see there is only one crossing point where $\cos x = 5x$, right?



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.


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