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.
 A: 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 $\theta$.
A: Let $x=a\cos\theta$. We want $4x^3-9x$ as a multiple of $\cos3\theta$; that is to say $\frac{4a^3}{9a}=\frac43$, so $a=\sqrt3$. Thus, we find $3\sqrt3\cos3\theta=1$ and then $x=\sqrt3\cos\left(\frac13\arccos\frac1{3\sqrt3}\right)$.
A: As you received as answers shows that
$$x=\sqrt{3} \cos \left(\frac{1}{3} \cos ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$$ what you want to evaluate to a given accuracy assuming that your calculator does not have trigonoetric functions at all.
Let us write
$$x=\sqrt{3} \cos \left(\frac{1}{3} \cos ^{-1}(\epsilon )\right)\qquad \text{where} \qquad \epsilon=\frac{1}{3 \sqrt{3}}$$ and use composition of Taylor series
$$\cos ^{-1}(\epsilon )=\frac{\pi }{2}-\epsilon -\frac{\epsilon ^3}{6}-\frac{3 \epsilon
   ^5}{40}+O\left(\epsilon ^7\right)$$
$$\frac{1}{3} \cos ^{-1}(\epsilon )=\frac{\pi }{6}-\frac{\epsilon }{3}-\frac{\epsilon ^3}{18}-\frac{\epsilon
   ^5}{40}+O\left(\epsilon ^7\right)$$
$$\cos \left(\frac{1}{3} \cos ^{-1}(\epsilon )\right)=\frac{\sqrt{3}}{2}+\frac{\epsilon }{6}-\frac{\epsilon ^2}{12 \sqrt{3}}+\frac{2
   \epsilon ^3}{81}-\frac{35 \epsilon ^4}{1296 \sqrt{3}}+O\left(\epsilon ^5\right)$$
$$x=\frac{3}{2}+\frac{\epsilon }{2 \sqrt{3}}-\frac{\epsilon ^2}{12}+\frac{2 \epsilon
   ^3}{27 \sqrt{3}}-\frac{35 \epsilon ^4}{1296}+O\left(\epsilon ^5\right)$$
Now, making $\epsilon=\frac{1}{3 \sqrt{3}}$, computing each term separately, we have
$$x=\frac{3}{2}+\frac{1}{18}-\frac{1}{324}+\frac{2}{6561}$$ and you see that the last term is already smaller than what is required (remember that we face an alternating series). So, with thiese numbers
$$x=\frac{1467001}{944784}\approx 1.55274$$ while the exact value would be ... the same.
If you forget the last term, $x=\frac{503}{324}\approx 1.55247$ is more than correct.
