(a) 通过画出合适的两个函数图像的草图，证明方程 $\ln x=\operatorname{cosec}\frac12x$ 在区间 $0<x<\pi$ 内恰有一个根。

(b) 通过计算验证这个根位于 $2.6$ 与 $2.9$ 之间。

(c) 使用迭代公式 $x_{n+1}=\exp\big(\operatorname{cosec}\frac12x_n\big)$ 求这个根，答案精确到小数点后 $3$ 位。每次迭代的结果保留小数点后 $5$ 位。

[$\exp(x)$ 是 $\mathrm{e}^x$ 的另一种记法。]

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。

在同一坐标系画递增的对数曲线和递减的余割曲线。区间两端的相对高低交换，而单调性保证只相交一次。

Sketch $y=\ln x$ and $y=\operatorname{cosec}(x/2)$ for $0<x<\pi$.

The logarithmic curve increases, passes through $(1,0)$ and tends to $-\infty$ as $x\to0^+$.

The cosecant curve decreases from $+\infty$ as $x\to0^+$ towards $(\pi,1)$.

Near $x=0$, the logarithmic curve lies below the cosecant curve. Near $x=\pi$, it lies above it because $\ln\pi>1$. By continuity they intersect, and their opposite strict monotonicity ensures that they intersect exactly once.

Hence the equation has $\boxed{\text{exactly one root in }(0,\pi)}$.

构造连续的两边之差，计算区间端点并检查变号。计算器使用弧度制。

Let

Using radians,

Since $f$ is continuous on $[2.6,2.9]$ and changes sign, the root lies in $\boxed{2.6<x<2.9}$.

用区间内的值开始迭代，每步内部保留完整精度，只把输出写成五位小数，直至三位小数稳定。

Starting with $x_0=2.6$, use

in radians. Retaining full precision during calculation gives

The values settle to $\boxed{x=2.767}$ to 3 decimal places.