Solving cos(x)=x Algebraically I think a graphical approach is the best way, but can algebra be used? With IVT, from x:[0,1] a solution must exist.

As said in comments and answers, you need a numerical method such as Newton.
Consider the function $$f(x)=x-\cos(x)\qquad f'x)=1+\sin(x)$$ The iterates will be given by
$$x_{n+1}=\frac{x_n \sin (x_n)+\cos (x_n)}{1+\sin (x_n)}$$ You can have a very good starting point building the $[2,2]$ Padé approximant of the function at $x=0$; this would give
$$f(x)\approx \frac{-1+\frac{7 }{6}x+\frac{1}{4}x^2 } { 1-\frac{1}{6}x+\frac{1}{12}x^2}$$ and then $x_0=\frac{1}{3} \left(\sqrt{85}-7\right)$.
Now, the iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.73984815243096243667 \\
 1 & 0.73908526166480706885 \\
 2 & 0.73908513321516428479 \\
 3 & 0.73908513321516064166
\end{array}
\right)$$ which is the solution for twenty significant figures.
Edit
For the fun of it, let us build the sama Pade approximant around $x=\frac \pi 4$. This will give as solution
$$4 \left(48+34 \sqrt{2}+7 \pi \right)x_0=$$ $$-720-408 \sqrt{2}+\pi  \left(36+46 \sqrt{2}+7 \pi \right)+4 \sqrt{65976+45744
   \sqrt{2}-3 \pi  \left(1328+940 \sqrt{2}+\left(131+62 \sqrt{2}\right) \pi
   \right)}$$ which is $\approx 0.73908513347523013264$. Quite close, isn't it ?
