Let $r>0$, solve this equation $$\cos{\left(\dfrac{\pi}{3}-\dfrac{\pi}{3r}\right)}=\sqrt{\dfrac{11}{r^2}-2}$$
I have found that $r=2$ is a solution, as $$LHS=\cos{\dfrac{\pi}{6}}=\dfrac{\sqrt{3}}{2}$$ and $$RHS=\sqrt{\dfrac{11}{4}-2}=\dfrac{\sqrt{3}}{2},$$ so $r=2$ is a root.
How can other solutions be found?
If you consider the function$$f(r)=\cos{\left(\dfrac{\pi}{3}-\dfrac{\pi}{3r}\right)}-\sqrt{\dfrac{11}{r^2}-2}$$ its domain is restrited to $-\sqrt{\frac{11}{2}} \leq r \leq \sqrt{\frac{11}{2}}$.
If you did what @John Barber commented, you must have noticed that the negative solution is close to the left bound where the function value is $\approx 0.077$; this means that the root is quite close to the bound (just above).
So, let $r=t-\sqrt{\frac{11}{2}} $ and expand the function a very truncated series around $t=0$; this should give $$f(t)=\cos \left(\frac{11+\sqrt{22}}{33} \pi \right)-2 \sqrt[4]{\frac{2}{11}} \sqrt{t}+O\left(t\right)$$ Ignoring the higher order terms, then $$t\sim \frac{1}{4} \sqrt{\frac{11}{2}} \cos ^2\left(\frac{11+\sqrt{22}}{33} \pi \right)\approx 0.00347582\implies r\approx -2.34173$$ while the exact solution is $-2.34180$.
Using the expansion up to $O\left(t^{3/2}\right)$ and solving the quadratic in $\sqrt t$ would give $t\approx 0.00341749$ which implies $r\approx -2.34179$.
I should precise here that, using Newton method without any step size control, we face serious problem since, at least for a time, many ieterates are complex numbers.