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CIE 9709 2023 Nov Paper 33 Q6

A Level / CIE / P3

CIE 9709 2023 Nov Paper 33 Paper · Question 6

题目

Problem

(a) Show that the equation cot⁡2θ+2cos⁡2θ=4\cot^2\theta+2\cos2\theta=4 can be written in the form

4sin⁡4θ+3sin⁡2θ−1=0.4\sin^4\theta+3\sin^2\theta-1=0.
(3)

(b) Hence solve the equation cot⁡2θ+2cos⁡2θ=4\cot^2\theta+2\cos2\theta=4, for 0∘<θ<360∘0^\circ<\theta<360^\circ.

(3)
题目中文翻译

(a) 证明方程 cot⁡2θ+2cos⁡2θ=4\cot^2\theta+2\cos2\theta=4 可以写成

4sin⁡4θ+3sin⁡2θ−1=04\sin^4\theta+3\sin^2\theta-1=0

的形式。

(3)

(b) 由此解方程 cot⁡2θ+2cos⁡2θ=4\cot^2\theta+2\cos2\theta=4，其中 0∘<θ<360∘0^\circ<\theta<360^\circ。

(3)
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

(a)

解法一

思路

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用 cot⁡2θ=1/sin⁡2θ−1\cot^2\theta=1/\sin^2\theta-1 与二倍角公式，把方程统一写成 sin⁡2θ\sin^2\theta 的式子，再清分母。

答题过程

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Where the original equation is defined, sin⁡θ≠0\sin\theta\ne0. Using

cot⁡2θ= 1sin⁡2θ−1,cos⁡2θ= 1−2sin⁡2θ,\begin{align*} \cot^2\theta=&\,\frac1{\sin^2\theta}-1,\\[4mm] \cos2\theta=&\,1-2\sin^2\theta, \end{align*}

the equation becomes

1sin⁡2θ−1+2(1−2sin⁡2θ)=4.\frac1{\sin^2\theta}-1+2(1-2\sin^2\theta)=4.

Thus

1sin⁡2θ−3−4sin⁡2θ=0.\frac1{\sin^2\theta}-3-4\sin^2\theta=0.

Multiplying by sin⁡2θ\sin^2\theta and rearranging gives

4sin⁡4θ+3sin⁡2θ−1=0.\boxed{4\sin^4\theta+3\sin^2\theta-1=0}.

(b)

解法一

思路

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把(a)看成关于 sin⁡2θ\sin^2\theta 的二次方程，排除不可能的负值，再找全四个象限中的角。

答题过程

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From part (a),

(4sin⁡2θ−1)(sin⁡2θ+1)=0.(4\sin^2\theta-1)(\sin^2\theta+1)=0.

The second factor is positive for every real θ\theta, so

sin⁡2θ=14,sin⁡θ=±12.\sin^2\theta=\frac14,\qquad \sin\theta=\pm\frac12.

In 0∘<θ<360∘0^\circ<\theta<360^\circ, the solutions are

θ=30∘, 150∘, 210∘, 330∘.\boxed{\theta=30^\circ,\ 150^\circ,\ 210^\circ,\ 330^\circ}.

All have sin⁡θ≠0\sin\theta\ne0, as required by the original equation.