题目
Problem
Solve the equation 3 cos 2 θ = 3 cos θ + 2 3\cos2\theta=3\cos\theta+2 3 cos 2 θ = 3 cos θ + 2 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ\leq\theta\leq360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
(5)
题目中文翻译
解方程 3 cos 2 θ = 3 cos θ + 2 3\cos2\theta=3\cos\theta+2 3 cos 2 θ = 3 cos θ + 2 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ\leq\theta\leq360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
(5)
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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用余弦二倍角公式将方程化为关于 cos θ \cos\theta cos θ
答题过程
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Using cos 2 θ = 2 cos 2 θ − 1 \cos2\theta=2\cos^2\theta-1 cos 2 θ = 2 cos 2 θ − 1
3 ( 2 cos 2 θ − 1 ) = 3 cos θ + 2. 3(2\cos^2\theta-1)=3\cos\theta+2. 3 ( 2 cos 2 θ − 1 ) = 3 cos θ + 2. Hence
6 cos 2 θ − 3 cos θ − 5 = 0 , 6\cos^2\theta-3\cos\theta-5=0, 6 cos 2 θ − 3 cos θ − 5 = 0 , so
cos θ = 3 ± 9 + 120 12 = 3 ± 129 12 . \cos\theta=\frac{3\pm\sqrt{9+120}}{12}
=\frac{3\pm\sqrt{129}}{12}. cos θ = 12 3 ± 9 + 120 = 12 3 ± 129 . The plus sign gives a value greater than 1 1 1
θ = cos − 1 ( 3 − 129 12 ) = 134.1456 … ∘ . \theta=\cos^{-1}\bigg(\frac{3-\sqrt{129}}{12}\bigg)
=134.1456\ldots^\circ. θ = cos − 1 ( 12 3 − 129 ) = 134.1456 … ∘ . The second solution is 360 ∘ − θ = 225.8543 … ∘ 360^\circ-\theta=225.8543\ldots^\circ 36 0 ∘ − θ = 225.8543 … ∘
θ = 134.1 ∘ or 225.9 ∘ . \boxed{\theta=134.1^\circ\quad\text{or}\quad225.9^\circ}. θ = 134. 1 ∘ or 225. 9 ∘ . 