<s>[INST] show this $x_{n+1}-x_{n}\ge 2\pi$ let $f(x)=e^x\cos{x}-\sin{x}-1$,and $n$ be postive integer,such  $x_{n}$ be a root of $f(x)=0$ ,and
$\dfrac{\pi}{3}+2n\pi<x_{n}<\dfrac{\pi}{2}+2n\pi$,show that
$$x_{n+1}-x_{n}\ge 2\pi,\forall n\in ^{+}\tag{1}$$
My try: since
$$\dfrac{7\pi}{3}+2n\pi<x_{n+1}<\dfrac{5\pi}{2}+2n\pi$$
and
$$-(\dfrac{\pi}{2}+2n\pi)<-x_{n}<-(\dfrac{\pi}{3}+2n\pi)$$
so
$$x_{n+1}-x_{n}\ge \left(\dfrac{7\pi}{3}+2n\pi\right)-\left(-(\dfrac{\pi}{2}+2n\pi)\right)=\dfrac{11\pi}{6}$$
But $2\pi>\dfrac{11\pi}{6}$,so How to prove this inequality $(1)$?
 [/INST] You are looking for the zero's of function$$f(x)=e^x\cos(x)-\sin(x)-1$$ Because of the $\cos(x)$, they are very close to the right bound $x_0^{(n)}=\left(2 \pi  n+\frac{\pi }{2}\right)$.
A first estimate of the solution can be obtained performing one single iteration of Newton method. This would give
$$x_1^{(n)}=2 \pi  n+\frac{\pi }{2}-2 e^{-2 \pi  n-\frac{\pi }{2}}$$
$$x_1^{(n+1)}-x_1^{(n)}=2 \pi+4 \sinh (\pi )\,e^{-\frac{4n+3}{2} \pi  } > 2\pi$$
We could obtain better results using one single iteration of Halley method. This would give
$$x_1^{(n)}=2 \pi  n+\frac{\pi }{2}-\frac{1}{2} \text{csch}^2\left( \left(n+\frac{1}{4}\right)\pi\right)$$
$$x_1^{(n+1)}-x_1^{(n)}=2\pi+\frac{1}{2} \left(\text{csch}^2\left(  \left(n+\frac{1}{4}\right)\pi\right)-\text{csch}^2\left( 
   \left(n+\frac{5}{4}\right)\pi\right)\right) > 2\pi$$ For sure, we could improve the value of the difference using iterative methods of higher order.
For illustration purposes, consider the second and third roots; computed rigorously, they give
$$x_3-x_2=2\pi +\color{red}{1.447189010436905}\times 10^{-6}$$
Using the first formula, we should have
$$x_3-x_2=2\pi+\color{red}{1.44718690}82458479\times 10^{-6}$$
Using the second formula, we should have
$$x_3-x_2=2\pi+\color{red}{1.44718901043}48732\times 10^{-6}$$
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