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IAL 2023 Oct Q3

A Level / Edexcel / P2

IAL 2023 Oct Paper · Question 3

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(a) Solve, for 0<θ360°0 < \theta \leqslant 360°, the equation

2tanθ+3sinθ=02\tan\theta + 3\sin\theta = 0

giving your answers, as appropriate, to one decimal place.

(b) Hence, or otherwise, find the smallest positive solution of

2tan(2x+40°)+3sin(2x+40°)=02\tan(2x + 40°) + 3\sin(2x + 40°) = 0

giving your answer to one decimal place.

(7)
题目中文翻译

本题必须展示所有解题步骤,不能完全依赖计算器技术。

(a) 解方程 2tanθ+3sinθ=02\tan\theta + 3\sin\theta = 0,其中 0<θ360°0 < \theta \leqslant 360°,答案保留一位小数。

(b) 由此或 otherwise,求 2tan(2x+40°)+3sin(2x+40°)=02\tan(2x + 40°) + 3\sin(2x + 40°) = 0 的最小正解,答案保留一位小数。

(7分)

解答

解法一

思路

tanθ=sinθcosθ\tan\theta = \dfrac{\sin\theta}{\cos\theta} 代入,提取公因子 sinθ\sin\theta,分别解 sinθ=0\sin\theta = 0cosθ=23\cos\theta = -\dfrac{2}{3}

答题过程

(a) Solve 2tanθ+3sinθ=02\tan\theta + 3\sin\theta = 0 for 0<θ360°0 < \theta \leqslant 360°:

2 \cdot \frac{\sin\theta}{\cos\theta} + 3\sin\theta =&\, 0 \\[2mm] \sin\theta \left(\frac{2}{\cos\theta} + 3\right) =&\, 0 \\[2mm] \sin\theta \left(\frac{2 + 3\cos\theta}{\cos\theta}\right) =&\, 0 \end{align*}$$ So $\sin\theta = 0$ or $2 + 3\cos\theta = 0$. **Case 1:** $\sin\theta = 0$ $$\theta = 180°, \quad 360° \qquad (\theta = 0° \text{ excluded since } 0 < \theta)$$ **Case 2:** $\cos\theta = -\dfrac{2}{3}$ $$\theta = 131.8°, \quad 228.2°$$ $$\boxed{\theta = 131.8°, \quad 180°, \quad 228.2°, \quad 360°}$$ **(b)** Let $\phi = 2x + 40°$. The equation becomes $2\tan\phi + 3\sin\phi = 0$. From part (a), the smallest positive solution is $\phi = 131.8°$: $$\begin{align*} 2x + 40° =&\, 131.8° \\ 2x =&\, 91.8° \\ x =&\, 45.9° \end{align*}$$ $$\boxed{x = 45.9°}$$