Solving $2\sin(x)= \cot(5x)$ without finding the intersections of the graphs of the two sides

Question

I have solved the equation $$2\sin(x)= \cot(5x)$$ by finding the intersections of the graphs of the two sides.

Is there a method to solve this equation without using graphs?

Answer 1

It is not so bad. Since $x=0$ cannot be a solution, multiply everything by $\sin(5x)$ and consider that you look for the zero's of function $$f(x)=2 \sin (x) \sin (5 x)-\cos (5 x)$$

There are very few roots to be computed since $$f(x)=f(-x) \qquad f(\pi-x)=f(\pi+x)\qquad f(x+2\pi)=f(x)$$ This means that we have to only find five roots for $x \in (0,\pi)$, all other being deduced from the previous identities.

The first one can be easily approximated using a Taylor series around $x=0$. This gives $$f(x)=-1+\frac{45 x^2}{2}-\frac{555 x^4}{8}+O\left(x^6\right)$$ gives $$x_1^{(0)}=\sqrt{\frac{2\left(45-\sqrt{915}\right)}{555} }$$

Similarly, around $x=\pi$ $$f(x)=1-\frac{5}{2} (x-\pi )^2-\frac{415}{24} (x-\pi )^4+O\left((x-\pi )^6\right)$$ gives $$x_5^{(0)}=\pi -\sqrt{\frac{2\left(\sqrt{2715}-15\right)}{415} }$$ Being lazy, I should simply use as starting estimate $$x_n^{(0)}=\frac{5x_1^{(0)}-x_5^{(0)}}4+\frac{x_5^{(0)}-x_1^{(0)}}4 n \qquad \qquad n=2,3,4$$

Let us see what this gives $$\left( \begin{array}{cc} 0 & 0.230558 \\ 1 & 0.228943 \\ 2 & 0.228942 \end{array} \right)$$ $$\left( \begin{array}{cc} 0 & 0.852598 \\ 1 & 0.749694 \\ 2 & 0.754455 \\ 3 & 0.754441 \end{array} \right)$$ $$\left( \begin{array}{cc} 0 & 1.474638 \\ 1 & 1.333158 \\ 2 & 1.351424 \\ 3 & 1.351323 \end{array} \right)$$ $$\left( \begin{array}{cc} 0 & 2.096678 \\ 1 & 1.964882 \\ 2 & 1.984847 \\ 3 & 1.984935 \end{array} \right)$$ $$\left( \begin{array}{cc} 0 & 2.718719 \\ 1 & 2.680447 \\ 2 & 2.682368 \\ 3 & 2.682372 \end{array} \right)$$ All of that seems to be quite fast.

Answer 2

We have that $\cos 5x-2\sin x\sin 5x=0,$ or linearising the second term, $$\cos 4x-\cos 5x-\cos 6x=0.$$ Now using the identity $$\cos z=\frac{e^{iz}+e^{-iz}}{2},$$ and with $y=e^{ix}$ we obtain $$y^4+1/y^4+y^5+1/y^5+y^6+1/y^6=0.$$ There are many ways to deal with this. You may explore the substitution $$u=y+1/y.$$ However, clear the fractions to obtain $$y^{12}+y^{11}+y^{10}+y^2+y+1=0,$$ and we can factor the polynomial to get $$y^{10}(y^2+y+1)+y^2+y+1=0,$$ or $$(y^{10}+1)(y^2+y+1)=0,$$ or $$((y^2)^5+1)(y^2+y+1)=0,$$ which becomes $$(y^2+1)((y^2)^4-(y^2)^3+(y^2)^2-y^2+1)(y^2+y+1)=0,$$ and since the quartic in $y^2=z$ gives $$z^4-z^3+z^2-z+1=0,$$ which may be dealt with by dividing by $z^2$ and using the substitution $w=z+1/z,$ the problem is essentially finished. The rest is only a matter of endurance.