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CIE 9709 2025 March Paper 32 Q4

A Level / CIE / P3

CIE 9709 2025 March Paper 32 Paper · Question 4

题目

Problem

By first expressing the equation tan⁡(x−60∘)=2cot⁡x\tan(x-60^\circ)=2\cot x as a quadratic equation in tan⁡x\tan x, solve the equation for 0∘≤x≤180∘0^\circ\le x\le180^\circ.

(6)
题目中文翻译

先将方程 tan⁡(x−60∘)=2cot⁡x\tan(x-60^\circ)=2\cot x 表示为关于 tan⁡x\tan x 的一元二次方程,再解此方程,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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按题意先化成正切的二次方程。解出两个正切值后,要在第二象限补回负正切值所对应的角。

答题过程

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The original equation excludes x=0∘,180∘x=0^\circ,180^\circ and x=150∘x=150^\circ.

Using the tangent subtraction formula and writing T=tan⁡xT=\tan x,

T−31+3T=2T.\frac{T-\sqrt3}{1+\sqrt3T}=\frac2T.

The value x=90∘x=90^\circ, where TT is undefined, is not a solution of the original equation. For the remaining allowed angles, multiplying through gives

T(T−3)=2(1+3T).T(T-\sqrt3)=2(1+\sqrt3T).

Hence

T2−33T−2=0.\boxed{T^2-3\sqrt3T-2=0}.

Solving,

T=33±352.T=\frac{3\sqrt3\pm\sqrt{35}}2.

The positive root gives x=79.8∘x=79.8^\circ. The negative root gives a second-quadrant angle x=160.2∘x=160.2^\circ. Both are defined in the original equation.

Thus

x=79.8∘, 160.2∘.\boxed{x=79.8^\circ,\ 160.2^\circ}.