(a) 通过画出合适的两个函数图像的草图，证明方程 $\cot2x=2\sin2x-1$ 在区间 $0<x<\frac12\pi$ 内恰有一个根。

(b) 通过计算证明这个根位于区间 $0.4<x<0.6$ 内。

(c) 使用迭代公式 $x_{n+1}=\frac12\tan^{-1}\bigg(\frac1{2\sin2x_n-1}\bigg)$ 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。

余切递减，另一条曲线先增后减。左半区间由单调性得唯一交点，右半区间要说明没有第二次相交。

Sketch $y=\cot2x$ and $y=2\sin2x-1$ for $0<x<\pi/2$.

The cotangent curve decreases from $+\infty$ to $-\infty$, crosses the axis at $x=\pi/4$, and has vertical asymptotes at the interval endpoints.

The other curve starts towards $-1$, rises to its maximum $1$ at $x=\pi/4$, then falls towards $-1$.

On $(0,\pi/4)$ the first curve strictly decreases and the second strictly increases. Their ordering reverses, so they intersect exactly once.

On $[\pi/4,5\pi/12]$, $\cot2x\leqslant0$ while $2\sin2x-1\geqslant0$, with no equality between the two curves. For $5\pi/12<x<\pi/2$,

Thus there is no further intersection. Therefore the equation has $\boxed{\text{exactly one root}}$ in the given interval.

对两边之差计算端点值，利用没有渐近线穿过该小区间的连续性。

Let $f(x)=\cot2x-2\sin2x+1$. Using radians,

The function is continuous on $[0.4,0.6]$, and the values have opposite signs. Hence the root lies in $\boxed{0.4<x<0.6}$.

从区间中间的值开始，内部保留完整精度，输出四位小数。迭代交替靠近根，继续到两位小数一致。

Using radians, start with $x_0=0.5$ and apply

This gives

These values settle to $\boxed{x=0.49}$ to 2 decimal places.