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CIE 9709 2025 June Paper 32 Q4

A Level / CIE / P3

CIE 9709 2025 June Paper 32 Paper · Question 4

题目

Problem

Solve the equation 3cot⁡x−4cot⁡2x=33\cot x-4\cot2x=3 for 0∘≤x≤180∘0^\circ\le x\le180^\circ.

(6)
题目中文翻译

解方程 3cot⁡x−4cot⁡2x=33\cot x-4\cot2x=3,其中 0∘≤x≤180∘0^\circ\le x\le180^\circ。

(6)
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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用余切二倍角公式变成关于 cot⁡x\cot x 的二次方程。原式不允许使 sin⁡x\sin x 或 sin⁡2x\sin2x 为零的角。

答题过程

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The original equation excludes x=0∘,90∘,180∘x=0^\circ,90^\circ,180^\circ.

Let c=cot⁡xc=\cot x. Using

cot⁡2x=cot⁡2x−12cot⁡x,\cot2x=\frac{\cot^2x-1}{2\cot x},

the equation becomes

3c−4c2−12c=3.3c-4\frac{c^2-1}{2c}=3.

Multiplying by c≠0c\ne0,

3c2−2(c2−1)=3c,3c^2-2(c^2-1)=3c,

so

c2−3c+2= 0,(c−1)(c−2)= 0.\begin{align*} c^2-3c+2=&\,0,\\[4mm] (c-1)(c-2)=&\,0. \end{align*}

Thus cot⁡x=1\cot x=1 or cot⁡x=2\cot x=2. On the stated interval, the valid angles are

x=45.0∘, 26.6∘.\boxed{x=45.0^\circ,\ 26.6^\circ}.

Neither angle is excluded by the original equation.