If $\tan(\pi/12 -x),\tan(\pi/12), \tan(\pi/12 + x)$, are 3 consecutive terms of a GP then sum of the solutions in $[0, 314]$ is $k\pi$. What is $k$?

Question

if $\tan(\frac{\pi}{12} -x),\tan(\frac{\pi}{12}), \tan(\frac{\pi}{12} + x)$, in order are all three consecutive terms of a GP then sum of all the solutions in $[0, 314]$ is $k\pi$. find the value of $k$

My Approach: I applied the $b^2=ac$ formula and after basic trigonometric manipulations, I got the equation, $\frac{\cos(\frac{\pi}{6})}{\cos(2x)}=\cot^2(\frac{\pi}{12})$ and this is where

Answer 1

By your work $$\tan^2\frac{\pi}{12}=\frac{\tan^2\frac{\pi}{12}-\tan^2x}{1-\tan^2\frac{\pi}{12}\tan^2x},$$ which gives $\tan x=0.$

Can you end it now?