(a) 通过画出一组合适的两个函数图像的草图，证明方程

在区间 $0<x\le\frac12\pi$ 内有一个根。

(b) 通过计算证明此根位于 $0.6$ 与 $0.8$ 之间。

(c) 使用迭代公式 $x_{n+1}=\tan^{-1}\bigg(\dfrac1{2-\cos x_n}\bigg)$ 求此根，结果保留 $2$ 位小数。每次迭代的结果保留 $4$ 位小数。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

在给定区间内，余切曲线严格递减，而 $2-\cos x$ 严格递增。比较两端的位置，即可说明它们恰交一次。

Sketch $y=\cot x$ for $0<x\le\pi/2$. It has vertical asymptote $x=0$, decreases from $+\infty$, and meets the $x$-axis at $(\pi/2,0)$.

On the same axes sketch $y=2-\cos x$, which increases from the limiting value $1$ at $x=0$ to $2$ at $x=\pi/2$. Mark the intersection.

Near $x=0$, $\cot x>2-\cos x$, whereas at $x=\pi/2$, $\cot x<2-\cos x$. Since one curve decreases strictly while the other increases strictly, they cross exactly once. Thus the equation has $\boxed{\text{one root in the given interval}}$.

用两边之差构造连续函数，注意计算器使用弧度。

Let $f(x)=\cot x-2+\cos x$. Working in radians,

The function is continuous on $[0.6,0.8]$, so there is a root in this interval. By part (a), this is the required unique root.

按给出的迭代式计算，内部保留精度，表中列出四位小数，直到结果稳定。

Use

in radians, starting with $x_0=0.7$:

Therefore the root is $\boxed{0.68}$ to 2 decimal places.