Find an integer $k$ such that $\cos(x)=x$ in the interval $\left(\frac{k}{10^6}, \frac{k+1}{10^6}\right)$

Question

I'm currently struggling on how I should get started with this question. So far I have sketched a graph, but still cannot see what I can do. Can anyone provide some help for me please?

Answer 1

Since there is just one point of the plane in which $\cos(x)=x$, take first the solution of this trascendental equation. We have $x_0\approx0.739085$.

Now we need $0.739085\in[\frac{k}{10^6},\frac{k+1}{10^6}]$ so $k\le739085\le k+1$.

Because of the approximation of $x_0$ we can just get an approximated solution. Taking, for example $k=739085$ we get the two following points: $$(0,739085,0.739085229),\quad (0.739086,0.739084549)$$ These two points are very close in the plane.If you have chosen a finer approximation for $x_0$, the answer given would have varied slightly.By continuity of the function cosinus this approximation is valid since we have $$\cos(0.739085)\gt0.739085 \text{ and } \cos(0.739086)\lt0.739085$$ in which the solution of the equation above has been assumed to be $x_0=0.739085$