Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 November Paper 12 Q2

A Level / CIE / P1

CIE 9709 2023 Nov Paper 12 Paper · Question 2

题目

Problem

Find the exact solution of the equation

16π+tan⁡−1(4x)=−cos⁡−1(123).\frac{1}{6}\pi+\tan^{-1}(4x)=-\cos^{-1}\big(\tfrac{1}{2}\sqrt{3}\big).
(2)
题目中文翻译

求下列方程的精确解:

16π+tan⁡−1(4x)=−cos⁡−1(123).\frac{1}{6}\pi+\tan^{-1}(4x)=-\cos^{-1}\big(\tfrac{1}{2}\sqrt{3}\big).
(2)
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

展开

先用标准角求反余弦,再孤立反正切。取正切时保留负号。

答题过程

展开

The principal value of cos⁡−1(3/2)\cos^{-1}(\sqrt3/2) is π/6\pi/6. Hence

tan⁡−1(4x)=−π6−π6=−π3.\tan^{-1}(4x)=-\frac\pi6-\frac\pi6=-\frac\pi3.

This angle is in the range of tan⁡−1\tan^{-1}, so 4x=tan⁡(−π/3)=−34x=\tan(-\pi/3)=-\sqrt3. The exact solution is x=−3/4x=-\sqrt3/4.