(a) 通过画出合适的两个函数图像的草图，证明方程

在区间 $0<x<\pi$ 内只有一个根。

(b) 通过计算证明这个根位于 $1$ 与 $1.5$ 之间。

(c) 使用迭代公式 $x_{n+1}=2-2\sin\frac12x_n$，以 $1.03$ 为初值，计算这个根，答案精确到小数点后 $2$ 位。每次迭代的结果保留小数点后 $4$ 位。

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。

左半段绝对值曲线递减而正弦曲线递增，因而只有一个交点。右半段要另作比较，排除第二个交点。

Sketch $y=|x-2|$ and $y=2\sin(x/2)$ for $0<x<\pi$.

The modulus graph has vertex $(2,0)$ and tends to $(0,2)$ at the left endpoint. The sine curve increases from $(0,0)$ towards $(\pi,2)$, with downward curvature.

On $(0,2)$, $2-x$ strictly decreases and $2\sin(x/2)$ strictly increases. Their ordering reverses, so they intersect exactly once.

For $2\leqslant x<\pi$, let

Since $h'(x)=\cos(x/2)-1<0$, $h$ decreases towards

Thus the sine curve stays above $x-2$ throughout this second interval. There is no further intersection, so the equation has $\boxed{\text{exactly one root}}$.

对两边之差计算端点值。绝对值函数在给定区间连续，可直接使用变号结论。

Let $f(x)=|x-2|-2\sin(x/2)$. Using radians,

and

By continuity and the sign change, the root lies in $\boxed{1<x<1.5}$.

必须使用题目指定的初值。交替迭代收敛较慢，继续到连续结果的两位小数稳定。

Starting with the required $x_0=1.03$, use

in radians. Keeping full precision during calculation gives

Thus the root is $\boxed{x=1.02}$ to 2 decimal places.