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

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.
