How to numerically or algebraically (preferably algebraically) solve the equation $\frac{x}{5} + \cos2x = 2$

Question

As the title suggests, I would like to solve the equation $x/5 + \cos2x = 2$ numerically or algebraically. I choose to have both formulations numerically and algebraically as I do not know which one fits this problem best. I use the word "preferably" as I would like an exact answer but I still want to have a more open question. Anyways, I do not know how to do what I want to do (I had a thought about using a summation formula for arccos among things). I will now give you some background.

Here we have a very similiar question, but I already understand that in order to solve the problem digitally, you "split" up the equation into two functions and then look at where they are equal to each other, in order to get the correct answer(s). What I am interested in is how to numerically or preferably algebraically solve the equation. Maybe that I could get some kind of answer from questions like this, but I am uncertain if all of these methods would be good for this particular problem? I want the most simplistic way to solve this problem numerically or preferably algebraically.

I would appreciate to be enlightened!

Answer 1

You can get an idea of how many solutions there are and approximately where to look for them by sketching the graphs of the cosine curve $y=\cos2x$ and the line $y=2-{1\over5}x$. Note that the line crosses the $x$-axis at $x=10$, and has $y$ values between $1$ and $-1$ only between $x=5$ and $x=15$, so all the solutions must occur in the interval $[5,15]$. The cosine curve $y=\cos2x$ takes the peak value $1$ at multiples of $\pi$, of which there are three in the interval $[5,15]$ (namely $2\pi, 3\pi$, and $4\pi$), and the trough value $-1$ halfway between each peak. If you draw things carefully enough (especially near $x=15$), you should be able to see the two graphs intersect precisely $7$ times, as noted in Dr. Sonnhard Graubner's answer.

To find the solutions numerically, bisection may be your best bet. You can try Newton's method on $f(x)={1\over5}x+\cos2x-2$, but it's not clear (to me, at least) which roots you can converge to in that way. Maybe someone else can elucidate or offer a better alternative.

Answer 2

Hint: define $$f(x)=\frac{x}{5}+\cos(2x)-2$$ and start with an Image of $f(x)$ I Count seven intersection points with the $x$-axes

Answer 3

After Barry Cipra's answer, let us consider that we would use Newton method to find the roots.

As, usual, the problem is the initial guess. So, let us use Taylor series limited to second order built at $x=k \pi$ to get by the end the quadratic $$-2 (x_k-k\pi )^2+\frac{1}{5} (x_k-k\pi )+\frac{k\pi }{5}-1=0$$ the solutions of which being $$x_k=\frac{1}{20}+k\pi \pm\frac{1}{20} \sqrt{40 k\pi -199}$$ Applying to the different roots $(k=2,3,4,5)$, we shall have

$$\left( \begin{array}{cccccccc} n & x_{2-}^{(n)} &x_{2+}^{(n)} &x_{3-}^{(n)} &x_{3+}^{(n)} &x_{4-}^{(n)} &x_{4+}^{(n)} &x_{5-}^{(n)}\\ 0 & 5.97150 & 6.69487 & 8.80771 & 10.1418 & 11.7451 & 13.4877 & 14.7220 \\ 1 & 5.96696 & 6.70965 & 8.76370 & 10.2343 & 11.6187 & 13.7362 & 14.4506 \\ 2 & 5.96698 & 6.70943 & 8.76425 & 10.2335 & 11.6164 & 13.7782 & 14.3922 \\ 3 & & & & & & 13.7804 & 14.3873 \end{array} \right)$$ which are the seven solutions for six significant figures.