Can we use taylor series to solve difficult equations (example with cos(x)=x)? Well I saw that the curve of the taylor function series of $\cos(x)$ at $x=0$ marry (it's a french expression to say that is very very near) the curve of $\cos(x)$ between $x=0$ and $x=pi/2$ 
So if I have a lot of terms of this series and I solve the equation, have I the solution ?
Example : $ \cos(x)\underset{x \to 0}=1-\frac{x^2}{2} + \frac{x^4}{24} o(x^2)$

If I solve $x=1-x^2/2+x^4/24$ I find $x=x = 0.73921922...$ (x is positive for \cos(x)) and with my computer I find $x=0.739085... $
Can we do that with all real function? 
Shadock
 A: If, for solving $x=\cos(x)$, you expand $\cos(x)$ as a Taylor series up to order $n$, you will need to solve a polynomial of degree close to $n$ which is not an easy task and the result will depend on the value you select fo $n$. 
Let us call $x_n$ the solution obtained for the Taylor expansion of $\cos(x)$. The results will then be $$x_1=1$$ $$x_2=x_3=0.732051$$ $$x_4=x_5=0.739219$$ $$x_6=x_7=0.739084$$ $$x_8=x_9=0.739085$$ which is the solution for six significant figures. But, to reach this level of accuracy, you had to solve an high degree polynomial for which there is almost no analytical solution (except for very low values of $n$).
As mentioned in the comments, it is much easier to use a simple root-finder as Newton which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let us be very lazy and start at $x_0=1$. The successive iterates will then be : $0.750364$, $0.739113$, $0.739085$. The solution has been obtained with very simple algebra and without the need of solving any polynomial.
If we take advantage of the second solution obtained solving a quadratic $x_0=0.732051$, the successive iterates will then be : $0.739096$, $0.739085$ which show that we only saved one Newton iteration but had to solve first $-\frac{x^2}{2}-x+1=0$.
