Exercise 2D: Solving Trigonometric Equations | 练习2D：解三角方程

📚 Exercise 2D: Solving Trigonometric Equations | 练习2D：解三角方程

Trigonometric equations form a fundamental part of the IB Mathematics syllabus, testing your ability to combine algebraic techniques with a deep understanding of periodic functions. In Exercise 2D, we will explore methods for solving a variety of trigonometric equations, from basic linear forms to more complex quadratic and multiple-angle cases. This article provides a comprehensive revision guide, complete with step-by-step examples, to help you master this topic and excel in your exams.

三角方程是 IB 数学大纲的基本组成部分，考察你将代数技巧与对周期函数的深刻理解相结合的能力。在练习 2D 中，我们将探索解各种三角方程的方法，从基本的线性形式到更复杂的二次方程和多倍角情形。本文提供一份全面的复习指南，包含分步示例，帮助你掌握该主题并在考试中取得优异成绩。

---

1. Introduction to Trigonometric Equations | 三角方程简介

ly required to find all solutions within a specified interval, such as 0 ≤ x ≤ 2π or 0° ≤ θ ≤ 360°.

包含未知角的一个或多个三角函数的方程称为三角方程。与多项式方程不同，由于正弦、余弦和正切的周期性，三角方程通常有无穷多个解。在 IB 考试中，通常要求你在指定区间内找出所有解，例如 0 ≤ x ≤ 2π 或 0° ≤ θ ≤ 360°。

Solving these equations effectively requires a solid grasp of the unit circle, trigonometric identities, and algebraic techniques like factoring and substitution. Exercise 2D is designed to build these skills through progressive problem-solving.

有效解这些方程需要扎实掌握单位圆、三角恒等式以及因式分解和代换等代数技巧。练习 2D 旨在通过渐进式问题解决来培养这些技能。

---

2. Solving Basic Equations: sin x = k, cos x = k, tan x = k | 解基本方程：sin x = k, cos x = k, tan x = k

π/6. Therefore, the solutions in the interval [0, 2π] are x = π/6 and x = π − π/6 = 5π/6.

对于像 sin x = 0.5 这样的基本方程，首先认识到正弦函数在第一和第二象限为正。参考角为 α = arcsin(0.5) = π/6。因此，在区间 [0, 2π] 内的解为 x = π/6 和 x = π − π/6 = 5π/6。

Similarly, for cos x = −√2/2, cosine is negative in the second and third quadrants. The reference angle is arccos(√2/2) = π/4. Hence, x = π − π/4 = 3π/4 and x = π + π/4 = 5π/4.

类似地，对于 cos x = −√2/2，余弦在第二和第三象限为负。参考角为 arccos(√2/2) = π/4。因此，x = π − π/4 = 3π/4 和 x = π + π/4 = 5π/4。

π/4 + kπ, where k is any integer. In the interval [0, 2π], the solutions are π/4 and 5π/4.

正切函数的周期为 π，因此方程 tan x = 1 的解为 x = π/4 + kπ，其中 k 为整数。在区间 [0, 2π] 内，解为 π/4 和 5π/4。

General solution: sin x = k ⇒ x = arcsin k + 2πn or x = π − arcsin k + 2πn, n ∈ ℤ.

通解: sin x = k ⇒ x = arcsin k + 2πn 或 x = π − arcsin k + 2πn, n ∈ ℤ.

---

广告 / Sponsored

3. Leveraging the Unit Circle | 利用单位圆

ine x = k for cosine, you can immediately see the symmetries that give multiple solutions.

单位圆是可视化解不可或缺的工具。圆上的每个点 (cos θ, sin θ) 对应一个角度。通过为 sine 画水平线 y = k 或为 cosine 画垂直线 x = k，你可以立即看到产生多个解的对称性。

at two points in the first and second quadrants, giving two solutions per period. Always sketch the circle to avoid missing solutions due to sign errors.

例如，解 sin x = 0.8 时，画出 y = 0.8；它与单位圆在第一和第二象限的两个点相交，每周期给出两个解。始终画出草图以避免因符号错误而遗漏解。

---

4. General Solutions vs. Specific Intervals | 通解与特定区间

f cos 2x = ½ is 2x = π/3 + 2πn or 2x = 5π/3 + 2πn, giving x = π/6 + πn or x = 5π/6 + πn, n ∈ ℤ bstitute integer values for n to find solutions within the required range, say 0 ≤ x ≤ 2π.

IB 题目通常要求找出给定区间内的所有解，但你可能会先推导出通解。例如，cos 2x = ½ 的通解为 2x = π/3 + 2πn 或 2x = 5π/3 + 2πn，得到 x = π/6 + πn 或 x = 5π/6 + πn, n ∈ ℤ。然后代入 n 的整数值，找出在规定区间内（如 0 ≤ x ≤ 2π）的解。

Be meticulous when listing solutions; check boundaries and remember that n can be negative if the interval starts below zero. Organize your work clearly to prevent missing any valid n.

列举解时要一丝不苟；检查边界，并记住如果区间从零以下开始，n 可以为负数。清晰组织你的解答过程，以防遗漏任何有效的 n。

---

5. Transforming to Quadratic Form | 转化为二次形式

− 1)(u − 1) = 0. Thus, sin x = ½ or sin x = 1. Then solve each basic equation separately.

像 2sin²x − 3sin x + 1 = 0 这样的方程可按二次方程处理。令 u = sin x，得到 2u² − 3u + 1 = 0，因式分解得 (2u − 1)(u − 1) = 0。因此，sin x = ½ 或 sin x = 1。然后分别解每个基本方程。

that solution as impossible. This technique also applies to equations with tan x or cos x.

始终检查三角函数的值是否在其范围内。例如，若二次方程得出 sin x = 2，则因不可能而舍去。此技巧也适用于带有 tan x 或 cos x 的方程。

---

6. Using Trigonometric Identities to Simplify | 使用三角恒等式化简

os²x − sin²x = 1 can be rewritten using sin²x = 1 − cos²x, giving 3cos²x − (1 − cos²x) = 1 ⇒ 4cos²x − 1 = 1 ⇒ cos²x = ½ ⇒ cos x = ±√2/2.

许多方程在应用恒等式之前不能直接求解。例如，3cos²x − sin²x = 1 可利用 sin²x = 1 − cos²x 重写为 3cos²x − (1 − cos²x) = 1 ⇒ 4cos²x − 1 = 1 ⇒ cos²x = ½ ⇒ cos x = ±√2/2。

agorean identities. Substituting these can turn a complex expression into a manageable quadratic or linear form.

其他有用的恒等式包括 tan x = sin x / cos x，倍角公式如 sin 2x = 2sin x cos x，以及勾股恒等式。代入这些可将复杂表达式转化为易于处理的二次或线性形式。

---

7. Handling Multiple Angles | 处理多倍角

c function is a multiple, like 2x or 3θ, first solve for that multiple angle. For sin 3x =

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）