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CIE 9709 2022 November Paper 31 Q4

A Level / CIE / P3

CIE 9709 2022 Nov Paper 31 Paper · Question 4

题目

Problem

Solve the equation tan⁡(x+45∘)=2cot⁡x\tan(x+45^\circ)=2\cot x for 0∘<x<180∘0^\circ<x<180^\circ.

(5)
题目中文翻译

解方程 tan⁡(x+45∘)=2cot⁡x\tan(x+45^\circ)=2\cot x,其中 0∘<x<180∘0^\circ<x<180^\circ。

(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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使用正切和角公式,将方程化成关于 tan⁡x\tan x 的二次方程。解出后根据正切正负选象限,并核对原式的无定义角。

答题过程

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Within the stated interval, the original equation is undefined only at x=45∘x=45^\circ. At x=90∘x=90^\circ, it gives −1=0-1=0, so this is not a solution.

For the remaining possible angles, put t=tan⁡xt=\tan x. Using the addition formula,

t+11−t=2t.\frac{t+1}{1-t}=\frac2t.

Here t≠0,1t\ne0,1. Multiplying through,

t(t+1)= 2(1−t),t2+3t−2= 0.\begin{align*} t(t+1)=&\,2(1-t),\\[4mm] t^2+3t-2=&\,0. \end{align*}

Thus

t=−3±172.t=\frac{-3\pm\sqrt{17}}2.

The positive root gives a first-quadrant solution,

x=tan⁡−1(−3+172)=29.3165…∘.x=\tan^{-1}\bigg(\frac{-3+\sqrt{17}}2\bigg) =29.3165\ldots^\circ.

The negative root gives a second-quadrant solution,

x= 180∘+tan⁡−1(−3−172)= 105.6834…∘.\begin{align*} x=&\,180^\circ+\tan^{-1}\bigg(\frac{-3-\sqrt{17}}2\bigg)\\[4mm] =&\,105.6834\ldots^\circ. \end{align*}

Both are in the domain of the original equation. Hence

x=29.3∘or105.7∘.\boxed{x=29.3^\circ\quad\text{or}\quad105.7^\circ}.