(a) 通过画出一对合适的图像的草图，证明方程 $\cot2x=\sec x$ 在区间 $0<x<\frac12\pi$ 内恰有一个根。

(b) 证明：如果由迭代公式

给出的实数数列收敛，那么它收敛于 (a) 问中的根。

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。

余切在整个区间严格递减，正割严格递增，两端顺序交换，因此只有一个交点。

Sketch $y=\cot2x$ and $y=\sec x$ for $0<x<\pi/2$.

The cotangent curve strictly decreases from $+\infty$ to $-\infty$, crossing the horizontal axis at $x=\pi/4$.

The secant curve strictly increases from a limiting value $1$ at $x=0$ to $+\infty$ as $x\to\pi/2$.

Near $x=0$ the cotangent curve lies above the secant curve, while at $x=\pi/4$ it lies below it. Continuity gives an intersection, and opposite strict monotonicity makes it unique.

Mark this single intersection on the sketch. Hence the equation has $\boxed{\text{exactly one root in }(0,\pi/2)}$.

也可取倒数，比较正切与余弦，但要分清正切渐近线及右端点不属于区间。

At any solution in the given interval, both sides are nonzero, so the equation is equivalent to

Sketch these graphs on $0<x<\pi/2$, marking the vertical asymptote $x=\pi/4$ of the tangent curve.

On $(0,\pi/4)$, $\tan2x$ increases from $0$ to $+\infty$, while $\cos x$ decreases and remains positive. Thus there is exactly one intersection.

On $(\pi/4,\pi/2)$, $\tan2x<0$ and $\cos x>0$, so there is no intersection. Although their limits both become zero at $x=\pi/2$, that endpoint is excluded.

Therefore there is $\boxed{\text{exactly one root}}$.

对收敛迭代取极限，再取正切和倒数，恢复原方程。

If the sequence converges to $L$, continuity gives

Hence $\tan2L=\cos L$. The fixed-point formula implies $|L|\leqslant\pi/8$, and then $\cos L>0$, so $L>0$.

Taking reciprocals,

Thus $L$ lies in the interval of (a) and is its unique root.