Example 6.10.1: Solving Trigonometric Equations | 示例6.10.1：求解三角方程

📚 Example 6.10.1: Solving Trigonometric Equations | 示例6.10.1：求解三角方程

Welcome to the TutorHao A-Level Mathematics Revision Series for AQA. Example 6.10.1 is a classic examination-style question that tests your ability to solve a trigonometric equation using algebraic factorisation, the unit circle and the specified domain. This worked example is deliberately selected because it combines three essential skills: substitution, factorisation, and the systematic identification of all solutions within a restricted interval.

欢迎阅读 TutorHao 为 AQA 编写的 A-Level 数学复习系列。示例 6.10.1 是一道典型的考试风格题目，考查你运用代数因式分解、单位圆以及指定区间来求解三角方程的能力。这道例题经过精心挑选，因为它综合了三个关键技能：换元、因式分解，以及在限定区间内系统地找出所有解。

1. The Problem | 问题陈述

Example 6.10.1 asks us to find every value of x in the interval 0° ≤ x ≤ 360° that satisfies the equation below.

示例 6.10.1 要求我们求出区间 0° ≤ x ≤ 360° 内满足下列方程的所有 x 值。

2 sin² x − sin x − 1 = 0

At first glance the equation looks unusual because it contains both sin² x and sin x, but no plain x term. Recognising that the expression is quadratic in the variable sin x is the most important first step, and it immediately suggests an algebraic method of solution.

乍一看，这个方程有些不同寻常，因为它同时含有 sin² x 和 sin x，却没有单独的 x 项。认识到这个表达式是关于变量 sin x 的二次式，是最关键的第一步，它立刻提示我们可以用代数方法求解。

2. Why This Equation Matters | 为什么这个方程很重要

Equations of the form a sin² x + b sin x + c = 0 appear regularly in AQA A-Level Mathematics papers, both in Core Trigonometry and in later applications such as differentiation and integration. Mastering this example gives you a reusable model for solving any quadratic trigonometric equation, including those involving cos² x or tan² x.

形如 a sin² x + b sin x + c = 0 的方程在 AQA A-Level 数学试卷中经常出现，既出现在核心三角学中，也出现在后续的微分与积分应用中。掌握这道例题，你就拥有了一套可复用的模型，用来求解任何二次型三角方程，包括含有 cos² x 或 tan² x 的方程。

The AQA specification requires you to solve trigonometric equations using exact values and graphs, and to give all solutions within a stated interval. This example directly addresses those assessment objectives.

AQA 教学大纲要求你运用精确值和图形求解三角方程，并在给定的区间内写出全部解。这道例题直接对应这些考核目标。

3. Making the Substitution | 换元法

The cleanest approach is to replace sin x with a single letter. Let y = sin x. The original equation then becomes an ordinary quadratic equation in y.

最简洁的做法是用一个字母替换 sin x。令 y = sin x，原方程就变成关于 y 的普通二次方程。

2y² − y − 1 = 0

This substitution is not a new mathematical idea; it simply makes the structure visible. Many students find it easier to factorise 2y² − y − 1 than to stare at the trigonometric form. You may perform this step mentally or on paper, but writing it down reduces the chance of error.

这个换元并不是新的数学思想，它只是让结构变得清晰。许多学生发现对 2y² − y − 1 因式分解比盯着三角形式更容易。你可以心算或在草稿纸上完成这一步，但写下来能减少出错的可能。

4. Factorising the Quadratic | 因式分解二次式

We now factorise the quadratic expression. We are looking for two brackets that multiply to give 2y² − y − 1.

现在我们对这个二次式进行因式分解。我们要找两个括号，使它们相乘后得到 2y² − y − 1。

2y² − y − 1 = (2y + 1)(y − 1)

It is always worth checking the factorisation by expanding: (2y + 1)(y − 1) = 2y² − 2y + y − 1 = 2y² − y − 1, which matches the original expression. Since the product of the two brackets equals zero, the zero product property tells us that at least one bracket must be zero.

值得通过展开来检验因式分解是否正确：(2y + 1)(y − 1) = 2y² − 2y + y − 1 = 2y² − y − 1，与原式一致。由于两个括号的乘积为零，零乘积性质告诉我们至少有一个括号必须为零。

5. Two Simple Equations | 两个简单方程

Setting each bracket to zero gives two separate linear equations.

令每个括号分别等于零，就得到两个独立的线性方程。

2y + 1 = 0 or y − 1 = 0

Substituting back y = sin x, the first equation gives 2 sin x + 1 = 0, so sin x = −½. The second equation gives sin x − 1 = 0, so sin x = 1. The problem is now reduced to solving two basic trigonometric equations.

代回 y = sin x，第一个方程给出 2 sin x + 1 = 0，因此 sin x = −½。第二个方程给出 sin x − 1 = 0，因此 sin x = 1。现在问题被简化成求解两个基本三角方程。

6. Using the Unit Circle | 借助单位圆

We now solve each equation separately within the domain 0° ≤ x ≤ 360°. Start with sin x = 1. On the unit circle, the sine value equals the y-coordinate of the point on the circle. The y-coordinate reaches 1 only at the top of the circle, which corresponds to x = 90°.

现在我们在区间 0° ≤ x ≤ 360° 内分别求解每个方程。先看 sin x = 1。在单位圆上，正弦值等于圆周上点的纵坐标。纵坐标只有在圆周顶端才等于 1，对应 x = 90°。

Next consider sin x = −½. The reference angle is the acute angle whose sine is ½, which is 30°. Since sin x is negative, the solutions must lie in the third and fourth quadrants. In the third quadrant, x = 180° + 30° = 210°. In the fourth quadrant, x = 360° − 30° = 330°.

接着考虑 sin x = −½。参考角是正弦值为 ½ 的锐角，即 30°。由于 sin x 为负，解必须位于第三象限和第四象限。在第三象限，x = 180° + 30° = 210°。在第四象限，x = 360° − 30° = 330°。

7. Listing All Solutions | 列出所有解

Combining the results from both equations, the complete solution set for Example 6.10.1 is:

综合两个方程的求解结果，示例 6.10.1 的完整解集为：

x = 90°, 210°, 330°

Notice that x = 90° arises from sin x = 1, while x = 210° and x = 330° both arise from sin x = −½. There are exactly three distinct solutions in the given interval, and every solution has been verified geometrically using the unit circle.

注意 x = 90° 来自 sin x = 1，而 x = 210° 和 x = 330° 都来自 sin x = −½。在给定区间内恰好有三个不同的解，每个解都已通过单位圆从几何上加以验证。

8. Verification Using the CAST Diagram | 用CAST图验证

The CAST diagram, also known as the ASTC rule, provides a quick algebraic check for the quadrant locations. The letters C, A, S and T indicate which trigonometric functions are positive in each quadrant: All positive in the first quadrant, Sin positive in the second, Tan positive in the third, and Cos positive in the fourth.

CAST 图，也称为 ASTC 法则，可以快速检验解的象限位置。字母 C、A、S、T 表示各象限中哪些三角函数为正：第一象限全部为正，第二象限正弦为正，第三象限正切为正，第四象限余弦为正。

Quadrant	Sine sign	Solutions for sin x = −½
Third (180° to 270°)	Negative	180° + 30° = 210°
Fourth (270° to 360°)	Negative	360° − 30° = 330°

For sin x = 1, the sine function is positive, and the only position where sine reaches its maximum value of 1 is x = 90°. The CAST diagram confirms that no other solution exists for this equation within the domain.

对于 sin x = 1，正弦函数为正，而正弦达到最大值 1 的唯一位置是 x = 90°。CAST 图确认该方程在区间内没有其他解。

9. Common Pitfalls | 常见错误

Several mistakes are frequently made by students attempting this type of question. Being aware of them will help you avoid losing marks in the examination.

学生在做这类题目时经常犯几种错误。了解这些错误有助于你在考试中避免失分。

Forgetting that a negative sine value has two solutions in 0° to 360°, not just one. Students often write only x = 210° and miss x = 330°.
忘记负的正弦值在 0° 到 360° 内有两个解，而不仅仅是一个。学生常常只写出 x = 210°，漏掉了 x = 330°。

Incorrectly factorising the quadratic. Always check by expanding your brackets before proceeding.
因式分解出错。在继续求解之前，务必通过展开括号来检验。

Using radians and degrees interchangeably. The domain here is stated in degrees, so the final answers must also be given in degrees.
混用弧度和角度。本题的区间以度为单位，因此最终答案也必须以度为单位。

Dropping the solution x = 90° because sin x = 1 is considered too obvious to write down. Every solution must be recorded explicitly.
因为 sin x = 1 被认为太明显而漏写 x = 90°。每一个解都必须明确写出。

10. Practice Question | 练习题目

To consolidate your understanding, attempt the following similar question without looking at the solution below first.

为了巩固理解，请先尝试下面这道类似的题目，再查看下面的解答。

Solve 2 cos² x − cos x − 1 = 0 for 0° ≤ x ≤ 360°

Following the same method, substitute y = cos x to obtain 2y² − y − 1 = 0, which factorises as (2y + 1)(y − 1) = 0. Hence cos x = −½ or cos x = 1. The solutions are x = 0°, x = 120°, x = 240° and x = 360°.

按照同样的方法，设 y = cos x，得到 2y² − y − 1 = 0，因式分解为 (2y + 1)(y − 1) = 0。因此 cos x = −½ 或 cos x = 1。解为 x = 0°、x = 120°、x = 240° 和 x = 360°。

11. Summary and Key Takeaways | 总结与要点

Example 6.10.1 demonstrates a systematic four-step strategy for quadratic trigonometric equations: substitute to reveal the quadratic structure, factorise, solve the resulting simple trigonometric equations, and collect all valid solutions within the specified domain.

示例 6.10.1 展示了求解二次三角方程的系统性四步策略：换元以揭示二次结构，因式分解，求解得到的简单三角方程，然后在指定区间内汇总所有有效解。

For AQA examinations, always show the factorisation step explicitly and indicate how you located each solution, whether by unit circle, CAST diagram or graph sketching. Correct method marks are awarded generously, but final answers must match the stated domain exactly.

对于 AQA 考试，务必明确写出因式分解的步骤，并说明你是如何通过单位圆、CAST 图或草图找到每个解的。方法分给得很宽松，但最终答案必须与题目给定的区间完全一致。

Mastering this technique will prepare you for more advanced topics, including solving equations with multiple angles such as sin 2x or cos (x − 30°), which appear frequently in later sections of the course.

掌握这一技巧将为你学习更高级的内容做好准备，包括求解含倍角或平移角的方程，例如 sin 2x 或 cos (x − 30°)，这些内容在课程后续部分经常出现。

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Example 6.10.1: Solving Trigonometric Equations | 示例6.10.1：求解三角方程