Tricky trigonometry question Find all solutions of the equations $\tan(3\arccos{x})=11/2$


I got $x = \cos(\frac{1}{3}\arctan(11/2) + n\pi/3)$  

Which gives $6$ unique values of $x$, but only $3$ of them seem to work when I sub them in using my calculator. How do I know which $3$ values of $n$ to pick?
HINT:

Let $\arccos x=\theta\implies 0\le \theta\le\pi$  and as the Principal value of $\cos$ lies in $\in[0,\pi]$

So, the problem becomes $$\tan3\theta=\frac{11}2$$ and as $$\tan3\theta=\frac{3t-t^3}{1-3t^2}$$

$$\implies 2t^3-33t^2-6t+11=0\  \  \  \  (1)$$ where $\displaystyle t=\tan\theta$

Solve $(1)$ for $\displaystyle t$

If $\displaystyle t=\tan\theta\ge0, \theta=\arccos x$ will lie in $\in[0,\frac\pi2]$

If $\displaystyle t=\tan\theta&lt;0, \theta=\arccos x$ will lie in $\in(\frac\pi2,\pi)$
Have you covered the general solutions to trigonometric equations? For example, consider the general solution to $tan A = 11/12$. Tan is a positive number (in this case $+11/12$) in the first and third quadrants.

So, the general solution is $A = tan^-1(11/12) +k\pi = 0.74 radians + k\pi$. k is any integer.

But $A = 3cos^-1 (x)$.

So, $cos^-1(x) = 0.18 radians + k\pi/3$

Can you go on from here?