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 $(1)$?
 A: We have $f'(x) = \mathrm{e}^x(\cos x - \sin x) - \cos x  < 0$ on $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$. Thus, $f(x)$ is strictly decreasing on $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$. Also, $f(\pi/3 + 2n\pi) > 0$ and $f(\pi/2 + 2n\pi) < 0$. So, $f(x)=0$ has exactly one real solution (namely, $x_n$)
on  $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$.
Let
\begin{align*}
y_n &= 2n\pi + \frac{\pi}{2} - \frac{1}{3}\mathrm{e}^{-2n\pi},\\
z_n &= 2n\pi + \frac{\pi}{2} - \frac{1}{3}\mathrm{e}^{-2(n- 1)\pi}.
\end{align*}
Clearly, $y_n, z_n \in (\pi/3 + 2n\pi, \pi/2 + 2n\pi)$.
We have
\begin{align*}
f(y_n) &\le \mathrm{e}^{2n\pi + \pi/2}\sin\frac{\mathrm{e}^{-2n\pi}}{3} 
- \cos \frac{\mathrm{e}^{-2n\pi}}{3}  - 1 \\ 
&\le \mathrm{e}^{2n\pi + \pi/2} \frac{\mathrm{e}^{-2n\pi}}{3} - \cos \frac{\pi}{6} - 1\\
&= \frac{\mathrm{e}^{\pi/2}}{3} - \frac{\sqrt{3}}{2} - 1 \\
&< 0
\end{align*}
where we have used $\sin u \le u$ for all $u \ge 0$,
and $\frac{\mathrm{e}^{-2n\pi}}{3} < \frac{1}{3} < \frac{\pi}{6}$.
Also, we have
\begin{align*}
f(z_n) &\ge \mathrm{e}^{2n\pi}\sin\frac{\mathrm{e}^{-2(n - 1)\pi}}{3}
- 1  - 1\\
&\ge \mathrm{e}^{2n\pi}\cdot \frac{2}{\pi}\frac{\mathrm{e}^{-2(n - 1)\pi}}{3} - 2\\
&= \frac{2}{3\pi}\mathrm{e}^{2\pi} - 2\\
&> 0
\end{align*}
where we have used $\sin u \ge \frac{2}{\pi}{u} $ for all $u \in [0, \pi/2]$.
Thus, we have $z_n < x_n < y_n$ that is
$$2n\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2(n- 1)\pi} < x_n < 2n\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2n\pi}.$$
So,
$$2(n + 1)\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2n\pi} < x_{n + 1} < 2(n + 1)\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2(n + 1)\pi}$$
and
$$ - 2n\pi - \frac{\pi}{2} + \frac{1}{\pi}\mathrm{e}^{-2n\pi}
< - x_n < - 2n\pi - \frac{\pi}{2} + \frac{1}{\pi}\mathrm{e}^{-2(n- 1)\pi}.$$
Thus, $x_{n + 1} - x_n > 2\pi$.
We are done.
A: 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
$$x_3-x_2=2\pi+\color{red}{1.44718901043}48732\times 10^{-6}$$
