Taylor Polynomial Error I was told to find the "bounds on the error" in a problem in Spivak's Calculus, but I have no idea what that means. 

Use the third degree Taylor polynomial of $\cos$ at $0$ to show that the solutions of $x^2=\cos x$ are approximately $\pm\sqrt{2/3}$, and
  find bounds on the error.

I recognize that this reduces to simply $1 - x^2/2! + R(x) = x^2$ and that $R(x)$ is the error. So, um, if we don't have the $R(x)$, since it's probably small, we get the needed approximate solution. Am I on the right track? What do I need to do to make this formal?
 A: You're being told that one solution is $x=\sqrt{2/3}+\epsilon$, and you're being asked to find some positive real number $r$ for which you can guarantee $|\epsilon|\lt r$. You should be able to express $\epsilon$ in terms of $R(x)$ and use that expression to find $r$. 
Oh, then you still have to do the $-\sqrt{2/3}$ thing, but if you can do the one, I'm sure you can do the other.  
A: The error in the Taylor expansion is $\dfrac{x^4}{4!}\cos(\theta_x)$ where $\theta_x$ is between $0$ and $\pi/2$, and therefore
$$1-\frac{x^2}{2}< \cos x<1-\frac{x^2}{2}+\frac{x^4}{24}.$$
The solution near $\sqrt{2/3}$ of the equation 
$$\frac{x^4}{24}-\frac{3x^2}{2}+1=0$$
will then be larger than the true (positive) solution of $\cos x=x^2$.
The above equation is quadratic in $x^2$. The relevant solution turns out to be 
$\sqrt{18-10\sqrt{3}}$.
Calculate. We have $\sqrt{2/3}\approx 0.8164966$, and 
$\sqrt{18-10\sqrt{3}}\approx 0.824313$.  We conclude that the positive solution of $\cos x=x^2$ is larger than $\sqrt{2/3}$, but by less than $0.0078165$.  
By the way, Wolfram Alpha gives the approximation $0.8241323$ for the positive solution of $\cos x=x^2$. It is not a surprise that this is quite a bit closer to our upper bound than it is to $\sqrt{2/3}$. 
By symmetry, $-\sqrt{2/3}$ is larger than the negative solution of $\cos x=x^2$, but by less than $0.0078165$.
