Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 June Paper 12 Q6

A Level / CIE / P1

CIE 9709 2026 June Paper 12 Paper · Question 6

题目

Problem

Solve the equation

cos⁡−1(sin⁡2x+32sin⁡x)=23π\cos^{-1}\big(\sin^2x+\tfrac{3}{2}\sin x\big)=\frac{2}{3}\pi

for 0≤x≤2π0\le x\le2\pi.

(4)
题目中文翻译

在 0≤x≤2π0\le x\le2\pi 内解方程

cos⁡−1(sin⁡2x+32sin⁡x)=23π.\cos^{-1}\big(\sin^2x+\tfrac{3}{2}\sin x\big)=\frac{2}{3}\pi.
(4)
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

两边取余弦,先求正弦的两个可能值。注意 cos⁡−1\cos^{-1} 给出的角 2π/32\pi/3 在主值范围内。

答题过程

展开

Taking cosine gives

sin⁡2x+32sin⁡x=−12⟹(2sin⁡x+1)(sin⁡x+1)=0.\sin^2x+\frac32\sin x=-\frac12 \quad\Longrightarrow\quad (2\sin x+1)(\sin x+1)=0.

If sin⁡x=−1/2\sin x=-1/2, then x=7π/6x=7\pi/6 or 11π/611\pi/6. If sin⁡x=−1\sin x=-1, then x=3π/2x=3\pi/2. For each of these angles, the input to cos⁡−1\cos^{-1} is −1/2-1/2, so they all satisfy the original equation.