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CIE 9709 2025 June Paper 35 Q2

A Level / CIE / P3

CIE 9709 2025 June Paper 35 Paper · Question 2

题目

Problem

Solve the equation 3cot⁡θ−4cosec⁡2θ+5=03\cot\theta-4\operatorname{cosec}^2\theta+5=0 for −π≤θ≤π-\pi\le\theta\le\pi.

(5)
题目中文翻译

解方程 3cot⁡θ−4cosec⁡2θ+5=03\cot\theta-4\operatorname{cosec}^2\theta+5=0,其中 −π≤θ≤π-\pi\le\theta\le\pi。

(5)
原卷通用作答要求

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。

解答

解法一

思路

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用余割与余切的平方关系化为二次方程,逐个列出指定区间内的角。

答题过程

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Using cosec⁡2θ=1+cot⁡2θ\operatorname{cosec}^2\theta=1+\cot^2\theta,

3cot⁡θ−4(1+cot⁡2θ)+5=0,3\cot\theta-4(1+\cot^2\theta)+5=0,

so

4cot⁡2θ−3cot⁡θ−1=0.4\cot^2\theta-3\cot\theta-1=0.

Factoring,

(4cot⁡θ+1)(cot⁡θ−1)=0.(4\cot\theta+1)(\cot\theta-1)=0.

For cot⁡θ=1\cot\theta=1, the solutions in [−π,π][-\pi,\pi] are θ=π/4\theta=\pi/4 and θ=−3π/4\theta=-3\pi/4.

For cot⁡θ=−1/4\cot\theta=-1/4, tan⁡θ=−4\tan\theta=-4, giving

θ=−tan⁡−14=−1.325817…\theta=-\tan^{-1}4=-1.325817\ldots

and

θ=π−tan⁡−14=1.815774….\theta=\pi-\tan^{-1}4=1.815774\ldots.

Therefore

θ=−3π4, −1.33, π4, 1.82.\boxed{\theta=-\frac{3\pi}{4},\ -1.33,\ \frac\pi4,\ 1.82}.

The values −π,0,π-\pi,0,\pi are excluded because the original cotangent and cosecant are undefined there.