题目
Problem
(a) Show that the equation tan 3 x + 2 tan 2 x − tan x = 0 \tan^3x+2\tan2x-\tan x=0 tan 3 x + 2 tan 2 x − tan x = 0 may be expressed as
tan 4 x − 2 tan 2 x − 3 = 0 \tan^4x-2\tan^2x-3=0 tan 4 x − 2 tan 2 x − 3 = 0
for tan x ≠ 0 \tan x\ne0 tan x = 0 .
(3)
(b) Hence solve the equation tan 3 2 θ + 2 tan 4 θ − tan 2 θ = 0 \tan^32\theta+2\tan4\theta-\tan2\theta=0 tan 3 2 θ + 2 tan 4 θ − tan 2 θ = 0 for 0 < θ < π 0<\theta<\pi 0 < θ < π . Give your answers in exact form.
(3)
题目中文翻译
(a) 证明当 tan x ≠ 0 \tan x\ne0 tan x = 0 时,方程 tan 3 x + 2 tan 2 x − tan x = 0 \tan^3x+2\tan2x-\tan x=0 tan 3 x + 2 tan 2 x − tan x = 0 可以表示为
tan 4 x − 2 tan 2 x − 3 = 0. \tan^4x-2\tan^2x-3=0. tan 4 x − 2 tan 2 x − 3 = 0.
(3)
(b) 由此解方程 tan 3 2 θ + 2 tan 4 θ − tan 2 θ = 0 \tan^32\theta+2\tan4\theta-\tan2\theta=0 tan 3 2 θ + 2 tan 4 θ − tan 2 θ = 0 ,其中 0 < θ < π 0<\theta<\pi 0 < θ < π 。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
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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 x T=\tan x T = tan x . The original equation requires T T T to be finite and 1 − T 2 ≠ 0 1-T^2\ne0 1 − T 2 = 0 .
Using tan 2 x = 2 T / ( 1 − T 2 ) \tan2x=2T/(1-T^2) tan 2 x = 2 T / ( 1 − T 2 ) ,
T 3 + 4 T 1 − T 2 − T = 0. T^3+\frac{4T}{1-T^2}-T=0. T 3 + 1 − T 2 4 T − T = 0.
Multiplying through,
T 3 ( 1 − T 2 ) + 4 T − T ( 1 − T 2 ) = 0. T^3(1-T^2)+4T-T(1-T^2)=0. T 3 ( 1 − T 2 ) + 4 T − T ( 1 − T 2 ) = 0.
Expanding,
− T 5 + 2 T 3 + 3 T = 0. -T^5+2T^3+3T=0. − T 5 + 2 T 3 + 3 T = 0.
For T ≠ 0 T\ne0 T = 0 , dividing by − T -T − T gives
tan 4 x − 2 tan 2 x − 3 = 0 . \boxed{\tan^4x-2\tan^2x-3=0}. tan 4 x − 2 tan 2 x − 3 = 0 .
(b)
解法一
思路
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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 2 θ T=\tan2\theta T = tan 2 θ .
First, T = 0 T=0 T = 0 gives 2 θ = π 2\theta=\pi 2 θ = π in 0 < 2 θ < 2 π 0<2\theta<2\pi 0 < 2 θ < 2 π , so
θ = π 2 . \theta=\frac\pi2. θ = 2 π .
Direct substitution gives tan 3 π + 2 tan 2 π − tan π = 0 \tan^3\pi+2\tan2\pi-\tan\pi=0 tan 3 π + 2 tan 2 π − tan π = 0 , so this is valid.
For T ≠ 0 T\ne0 T = 0 , use (a):
T 4 − 2 T 2 − 3 = 0 , ( T 2 − 3 ) ( T 2 + 1 ) = 0. \begin{align*}
T^4-2T^2-3=&\,0,\\[4mm]
(T^2-3)(T^2+1)=&\,0.
\end{align*} T 4 − 2 T 2 − 3 = ( T 2 − 3 ) ( T 2 + 1 ) = 0 , 0.
Thus T = ± 3 T=\pm\sqrt3 T = ± 3 , giving
2 θ = π 3 , 2 π 3 , 4 π 3 , 5 π 3 . 2\theta=\frac\pi3,\ \frac{2\pi}{3},\
\frac{4\pi}{3},\ \frac{5\pi}{3}. 2 θ = 3 π , 3 2 π , 3 4 π , 3 5 π .
All corresponding angles have both tangents defined. Therefore the complete solution is
θ = π 6 , π 3 , π 2 , 2 π 3 , 5 π 6 . \boxed{\theta=\frac\pi6,\ \frac\pi3,\ \frac\pi2,\
\frac{2\pi}{3},\ \frac{5\pi}{6}}. θ = 6 π , 3 π , 2 π , 3 2 π , 6 5 π .