# Solve this equation $\cos{\left(\frac{\pi}{3}-\frac{\pi}{3r}\right)}=\sqrt{\frac{11}{r^2}-2}$

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?

• Plot these two functions and you'll see that there are exactly 2 real solutions. One (the one you found) is positive. The other is negative and I doubt there's an analytic form for it. – John Barber May 14 at 4:18

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 $$r\approx -2.34179$$.