题目
Problem
Solve the equation 2 cot 2 x + 3 cot x = 5 2\cot 2x+3\cot x=5 2 cot 2 x + 3 cot x = 5 0 ∘ < x < 180 ∘ 0^\circ<x<180^\circ 0 ∘ < x < 18 0 ∘
(6)
题目中文翻译
解方程 2 cot 2 x + 3 cot x = 5 2\cot 2x+3\cot x=5 2 cot 2 x + 3 cot x = 5 0 ∘ < x < 180 ∘ 0^\circ<x<180^\circ 0 ∘ < x < 18 0 ∘
(6)
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
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用余切二倍角公式把方程化为关于 tan x \tan x tan x
答题过程
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The original equation is undefined at x = 90 ∘ x=90^\circ x = 9 0 ∘ t = tan x t=\tan x t = tan x
cot 2 x = 1 − tan 2 x 2 tan x . \cot2x=\frac{1-\tan^2x}{2\tan x}. cot 2 x = 2 tan x 1 − tan 2 x . Then
2 1 − t 2 2 t + 3 t = 5 , 2\frac{1-t^2}{2t}+\frac3t=5, 2 2 t 1 − t 2 + t 3 = 5 , so
1 − t 2 + 3 = 5 t , t 2 + 5 t − 4 = 0. \begin{align*}
1-t^2+3=&\,5t,\\[4mm]
t^2+5t-4=&\,0.
\end{align*} 1 − t 2 + 3 = t 2 + 5 t − 4 = 5 t , 0. Thus
t = − 5 ± 41 2 . t=\frac{-5\pm\sqrt{41}}2. t = 2 − 5 ± 41 . The positive root gives
x = tan − 1 ( − 5 + 41 2 ) = 35.0520 … ∘ . x=\tan^{-1}\bigg(\frac{-5+\sqrt{41}}2\bigg)
=35.0520\ldots^\circ. x = tan − 1 ( 2 − 5 + 41 ) = 35.0520 … ∘ . The negative root gives
x = 180 ∘ + tan − 1 ( − 5 − 41 2 ) = 99.9479 … ∘ . \begin{align*}
x=&\,180^\circ+\tan^{-1}\bigg(\frac{-5-\sqrt{41}}2\bigg)\\[4mm]
=&\,99.9479\ldots^\circ.
\end{align*} x = = 18 0 ∘ + tan − 1 ( 2 − 5 − 41 ) 99.9479 … ∘ . Both are in the domain of the original equation. Therefore
x = 35.1 ∘ or 99.9 ∘ . \boxed{x=35.1^\circ\quad\text{or}\quad99.9^\circ}. x = 35. 1 ∘ or 99. 9 ∘ . 