Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve $4x^3 - 9x - 1 = 0$, correct to 3 decimal places. I am trying to solve the following problem: 

Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve the equation $4x^3 - 9x - 1 = 0$, correct to 3 decimal places.

Assuming that $x = \cos 3\theta$, you can substitute it into the equation so that you get:
$4(\cos\theta)^3 - 9(\cos\theta) = 1$
The problem with the above is that there is a $9$ in front of the $\cos\theta$. If there was a $3$ instead, I could have replaced the LHS of the equation with $\cos 3\theta$, and then solved for $\theta$. How can I solve this equation? Any insights are appreciated.

Render $x=a\cos\theta$.  Thereby
$4x^3-9x=4a^3\cos^3\theta-9a\cos\theta$
$=a^3(4\cos^3\theta-(9/a^2)\cos\theta)$
If we put $a=\sqrt{3}$ then $9/a^2=3$ and then
$4x^3-9x=3\sqrt{3}\cos(3\theta)=1$
where the "=1" cones from the constant term of the original equation.  So $\cos(3\theta)=\sqrt{3}/9$, from which we can get values of $\theta$ to four decimal places.  That should be enough accuracy as $x=a\cos\theta=\sqrt{3}\cos\theta$ has a derivative within $\pm 2$ for all $\theta$.
