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CIE 9709 2024 November Paper 32 Q7

A Level / CIE / P3

CIE 9709 2024 Nov Paper 32 Paper · Question 7

题目

Problem

(a) Show that the equation tan⁡3x+2tan⁡2x−tan⁡x=0\tan^3x+2\tan2x-\tan x=0 may be expressed as

tan⁡4x−2tan⁡2x−3=0\tan^4x-2\tan^2x-3=0

for tan⁡x≠0\tan x\ne0.

(3)

(b) Hence solve the equation tan⁡32θ+2tan⁡4θ−tan⁡2θ=0\tan^32\theta+2\tan4\theta-\tan2\theta=0 for 0<θ<π0<\theta<\pi. Give your answers in exact form.

(3)
题目中文翻译

(a) 证明当 tan⁡x≠0\tan x\ne0 时,方程 tan⁡3x+2tan⁡2x−tan⁡x=0\tan^3x+2\tan2x-\tan x=0 可以表示为

tan⁡4x−2tan⁡2x−3=0.\tan^4x-2\tan^2x-3=0.
(3)

(b) 由此解方程 tan⁡32θ+2tan⁡4θ−tan⁡2θ=0\tan^32\theta+2\tan4\theta-\tan2\theta=0,其中 0<θ<π0<\theta<\pi。答案以精确形式表示。

(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

展开

用正切二倍角代入,先去分母,再按题设非零条件除去一次因子。

答题过程

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Let T=tan⁡xT=\tan x. The original equation requires TT to be finite and 1−T2≠01-T^2\ne0.

Using tan⁡2x=2T/(1−T2)\tan2x=2T/(1-T^2),

T3+4T1−T2−T=0.T^3+\frac{4T}{1-T^2}-T=0.

Multiplying through,

T3(1−T2)+4T−T(1−T2)=0.T^3(1-T^2)+4T-T(1-T^2)=0.

Expanding,

−T5+2T3+3T=0.-T^5+2T^3+3T=0.

For T≠0T\ne0, dividing by −T-T gives

tan⁡4x−2tan⁡2x−3=0.\boxed{\tan^4x-2\tan^2x-3=0}.

(b)

解法一

思路

展开

先保留(a)除掉的零正切情形,再解非零部分的四次方程,避免漏解。

答题过程

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Put T=tan⁡2θT=\tan2\theta.

First, T=0T=0 gives 2θ=π2\theta=\pi in 0<2θ<2π0<2\theta<2\pi, so

θ=π2.\theta=\frac\pi2.

Direct substitution gives tan⁡3π+2tan⁡2π−tan⁡π=0\tan^3\pi+2\tan2\pi-\tan\pi=0, so this is valid.

For T≠0T\ne0, use (a):

T4−2T2−3= 0,(T2−3)(T2+1)= 0.\begin{align*} T^4-2T^2-3=&\,0,\\[4mm] (T^2-3)(T^2+1)=&\,0. \end{align*}

Thus T=±3T=\pm\sqrt3, giving

2θ=π3, 2π3, 4π3, 5π3.2\theta=\frac\pi3,\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{5\pi}{3}.

All corresponding angles have both tangents defined. Therefore the complete solution is

θ=π6, π3, π2, 2π3, 5π6.\boxed{\theta=\frac\pi6,\ \frac\pi3,\ \frac\pi2,\ \frac{2\pi}{3},\ \frac{5\pi}{6}}.