How to solve $0 = \cos x − \cosh x + 1$?

Question

The question is around solving a dynamical system and finding the equilibrium points, and I'm not sure where to take solving this equation for the 2 values of $x$.

Edit: To clarify, the question asks to analyse the dynamical system graphically.

Answer 1

For $|x|\ge 2\ln 2$ you get $\cosh(x)-1\ge \frac12(4+\frac14)-1=1+\frac18$ so that there can be no solutions of $\cos x=\cosh x-1$ in this region. Inside the interval $|x|<2\ln 2$ the approximations $\cos x\approx 1-\frac12x^2$ and $\cosh x=1+\frac12x^2$ are sufficiently close to predict the roots close to $x^2=1$, $x=\pm 1$. Note that the next terms cancel in $$1=\cosh x-\cos x=x^2+\frac1{12}x^6+\frac2{10!}x^{10}+\dots$$ so that in the next approximation $x^2\approx 1-\frac1{12}$, $x\approx \pm (1-\frac1{24})\approx\pm0.96$ which gives an idea for the range of the exact solution. Newton converges to $x=\pm 0.99862133827...$

Answer 2

the blue coloured graph is of $ \ cosx $ and green coloured graph is of $\cosh x$

zooming inside red rectangular region of $[-2,2] \times [0,3]$

you get $2$ solutions $x=0.999$ and $x=-0.999$ (marked as black points) you can

solve it by numerical technique of finding roots