I am trying to approximate the solutions for $x^2=\cos x$. Using Taylor quadratic approximation, we get $$\cos x \approx 1-\frac{x^2}2.$$ Solving yields $x=\pm\sqrt{\frac32}$.
However, I am having trouble calculating the error for this approximation, as Lagrange's Error Bound formula cannot be directly used. Is there another way to calculate the maximum error?
Sorry, but that's not very likely: since $\cos x\le1$, a reasonable solution of your equation should satisfy $|x|\le1$, too. In reality, your approximation yields $x=\pm\sqrt{\frac23}$, and that's not bad, as $\cos\sqrt{\frac23}=0.68477852552355028302824662222974449974$ is not so far from $\frac23$.
There is a nice, simple, $\color{red}{\large 1,400}$ years old approximation $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$ which gives as an estimate $$x^2=\frac{1}{2} \left(-4-\pi ^2+\sqrt{16+12 \pi ^2+\pi ^4}\right)$$ that is to say $$x=\pm 0.823660$$ while the exact solution, given by Newton method, is $x=\pm 0.824132$.
You can polish the root using the fact that the estimate is close to $\frac \pi 4$ and then use a series expansion $$x^2-\cos(x)=\left(\frac{\pi ^2}{16}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}+\frac{\pi }{2}\right) \left(x-\frac{\pi }{4}\right)+\left(1+\frac{1}{2 \sqrt{2}}\right) \left(x-\frac{\pi }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ Solving the quadratic gives $$x=\pm\frac{-4 \sqrt{2}+\sqrt{2} \pi +2 \sqrt{24+\frac{8192+2048 \pi -256 \pi ^2}{128\sqrt{2}}}}{4 \left(4+\sqrt{2}\right)}=\pm 0.824129$$
Edit
Even if it does not make any sense, notice that $$\cos \left(\frac{35 \pi }{229}\right)-\sin \left(\frac{\pi }{50}\right)$$ approximates the solution within an error of $4.53 \times 10^{-9}$ (this was found by a trigonometric inverse calculator of mine).
