📚 Solving Basic Trigonometric Equations | 基础三角方程的求解策略
Trigonometric equations are a core part of A-Level Mathematics. Mastering their solution requires a clear understanding of the unit circle, reference angles, and the periodic nature of sine, cosine and tangent. This article provides a structured, exam-focused strategy for solving basic trigonometric equations.
三角方程是 A-Level 数学的核心内容。要熟练求解，必须清晰理解单位圆、参考角以及正弦、余弦和正切函数的周期性。本文提供一套结构化、紧扣考点的基础三角方程求解策略。
1. The Unit Circle and Key Angles | 单位圆与特殊角
ometric view helps us visualise why equations such as sin θ = 0.5 have multiple solutions.
单位圆是以原点为圆心、半径为 1 的圆。对于从 x 轴正方向量起的任意角 θ，圆上对应点的坐标为 (cos θ, sin θ)。这种几何视角有助于我们理解为什么像 sin θ = 0.5 这样的方程会有多个解。
You should know the exact values of sine, cosine and tangent for the common angles: 0°, 30°, 45°, 60° and 90° (and their radian equivalents). These values appear repeatedly in exam questions and allow you to recognise solutions without a calculator.
你应该熟记常见角 0°、30°、45°、60° 和 90°（以及对应的弧度制）的正弦、余弦和正切精确值。这些值在考试中反复出现，能让你在不使用计算器的情况下识别出解。
sin 30° = ½, cos 60° = ½, tan 45° = 1
2. Principal Values and the CAST Diagram | 主值与CAST象限图
When solving a trigonometric equation, your calculator gives only one solution, called the principal value. For sin, the principal value lies between −90° and 90°; for cos, between 0° and 180°; for tan, between −90° and 90°. However, the full solution set usually contains more solutions, found using symmetry and periodicity.
求解三角方程时，计算器只给出一个解，称为主值。对于 sin，主值位于 −90° 到 90° 之间；对于 cos，主值位于 0° 到 180° 之间；对于 tan，主值位于 −90° 到 90° 之间。但完整的解集通常包含更多解，这些解需要利用对称性和周期性来寻找。
The CAST diagram tells us which trigonometric functions are positive in each quadrant:
CAST 象限图告诉我们每个象限中哪些三角函数为正：
- Quadrant A (0°–90°): All positive
- 第一象限 (0°–90°)： 全部为正
- Quadrant S (90°–180°): Sin positive
- 第二象限 (90°–180°)： Sin 为正
- Quadrant T (180°–270°): Tan positive
- 第三象限 (180°–270°)： Tan 为正
- Quadrant C (270°–360°): Cos positive
- 第四象限 (270°–360°)： Cos 为正
By drawing the angle in the correct quadrant, you can find all solutions within a given interval.
通过在正确的象限中画出角度，你就能找到给定区间内的所有解。
3. Solving sin x = k | 求解 sin x = k
For the equation sin θ = k (where |k| ≤ 1), the general solutions in degrees are:
对于方程 sin θ = k（其中 |k| ≤ 1），以度为单位的一般解为：
θ = α + 360n° or θ = 180° − α + 360n°, n ∈ ℤ
ue. These two families correspond to the two points on the unit circle with vertical coordinate k.
其中 α = sin⁻¹(k) 是主值。这两族解分别对应单位圆上纵坐标为 k 的两个点。
Example: Solve sin θ = ½ for 0° ≤ θ < 360°.
示例： 求 sin θ = ½ 在 0° ≤ θ < 360° 内的解。
α = 30°, so θ = 30° or θ = 180° − 30° = 150°
Both 30° and 150° lie in the interval, so the solutions are {30°, 150°}.
30° 和 150° 都在区间内，因此解为 {30°, 150°}。
4. Solving cos x = k | 求解 cos x = k
For cos θ = k (where |k| ≤ 1), the general solutions in degrees are:
对于 cos θ = k（其中 |k| ≤ 1），以度为单位的一般解为：
θ = α + 360n° or θ = −α + 360n°, n ∈ ℤ
ue between 0° and 180°. Cosine is positive in the first and fourth quadrants, so the two solutions are symmetric about 0°.
其中 α = cos⁻¹(k) 是位于 0° 到 180° 之间的主值。余弦在第一和第四象限为正，因此两个解关于 0° 对称。
Example: Solve cos θ = √3/2 for 0° ≤ θ < 360°.
示例： 求 cos θ = √3/2 在 0° ≤ θ < 360° 内的解。
α = 30°, so θ = 30° or θ = −30° + 360° = 330°
Thus the solutions are {30°, 330°}.
因此解为 {30°, 330°}。
5. Solving tan x = k | 求解 tan x = k
s period 180°, so once you have the principal value α = tan⁻¹(k), the general solution is:
正切的周期为 180°，因此一旦你得到主值 α = tan⁻¹(k)，一般解为：
θ = α + 180n°, n ∈ ℤ
Unlike sine and cosine, tangent gives only one family of solutions, because the two points on the unit circle are diametrically opposite.
与正弦和余弦不同，正切只给出一族解，因为单位圆上的两个对应点正好关于原点对称。
Example: Solve tan θ = 1 for 0° ≤ θ < 360°.
示例： 求 tan θ = 1 在 0° ≤ θ < 360° 内的解。
α = 45°, so θ = 45° or θ = 45° + 180° = 225°
Solutions are {45°, 225°}.
解为 {45°, 225°}。
6. Using Inverse Trigonometric Functions | 使用反三角函数
if sin θ = 0.3, you can press sin⁻¹(0.3) to get a principal value of approximately 17.5°.
当 k 不是标准特殊值时，你必须使用计算器。例如，若 sin θ = 0.3，你可以按 sin⁻¹(0.3) 得到主值约为 17.5°。
Then apply the appropriate general solution formulas. Always give your final answers to the accuracy requested in the question, usually 1 decimal place or 3 significant figures.
然后应用相应的一般解公式。最终答案必须按题目要求的精度给出，通常保留一位小数或三位有效数字。
Example: Solve cos θ = −0.6 for 0° ≤ θ < 360°.
示例： 求 cos θ = −0.6 在 0° ≤ θ < 360° 内的解。
α = cos⁻¹(−0.6) = 126.87°
θ = 126.87° or θ = 360° − 126.87° = 233.13°
Therefore θ ≈ 126.9° or 233.1°.
因此 θ ≈ 126.9° 或 233.1°。
7. Multiple Solutions in a Given Interval | 给定区间内的多解
When a question specifies an interval, you must generate all solutions within that interval by adding or subtracting periods. For sine and cosine, add multiples of 360°; for tangent, add multiples of 180°.
当题目指定了区间时，你必须通过加上或减去周期来生成该区间内的所有解。对于正弦和余弦，加上 360° 的整数倍；对于正切，加上 180° 的整数倍。
Always write down the general solution first, then substitute integer values of n to find the required solutions. This avoids missing answers.
务必先写出一般解，然后代入整数 n 的值来找到所需的解。这样可以避免遗漏答案。
Example: Solve sin θ = ½ for −180° ≤ θ ≤ 180°.
示例： 求 sin θ = ½ 在 −180° ≤ θ ≤ 180° 内的解。
θ = 30° + 360n° or θ = 150° + 360n°
we get −330° (too small) and −210° (too small). No other values fit. So the solutions are {30°, 150°}.
当 n = 0 时，得到 30° 和 150°。当 n = −1 时，得到 −330°（太小）和 −210°（太小）。没有其他值符合。因此解为 {30°, 150°}。
8. Quadratic Trigonometric Equations | 二次型三角方程
θ or tan θ, such as 2 sin²θ − sin θ − 1 = 0. Let x = sin θ and solve the quadratic first.
有些方程可以写成关于 sin θ、cos θ 或 tan θ 的二次形式，例如 2 sin²θ − sin θ − 1 = 0。令 x = sin θ，先解这个二次方程。
Example: Solve 2 sin²θ − sin θ − 1 = 0 for 0° ≤ θ < 360°.
示例： 求 2 sin²θ − sin θ − 1 = 0 在 0° ≤ θ < 360° 内的解。
(2 sin θ + 1)(sin θ − 1) = 0
θ = 1, θ = 90°. From sin θ = −½, the principal value is −30°, but we need solutions in 0°–360°:
因此 sin θ = −½ 或 sin θ = 1。由 sin θ = 1，得 θ = 90°。由 sin θ = −½，主值为 −30°，但我们需要 0°–360° 内的解：
θ = 180° − (−30°) = 210° or θ = 360° + (−30°) = 330°
Hence the full solution set is {90°, 210°, 330°}.
因此完整解集为 {90°, 210°, 330°}。
9. Equations Involving Compound Angles | 复合角方程
If the angle is of the form 2θ, θ + 30°, or 3θ − 15°, you must adjust the interval accordingly. Replace the variable and solve for the compound expression first, then divide or shift to find θ.
如果角的形式为 2θ、θ + 30° 或 3θ − 15°，你必须相应地调整区间。将复合表达式视为整体先求解，然后再除以系数或平移来求 θ。
Example: Solve sin(2θ) = ½ for 0° ≤ θ < 360°.
示例： 求 sin(2θ) = ½ 在 0° ≤ θ < 360° 内的解。
Since 0° ≤ θ < 360°, we have 0° ≤ 2θ < 720°. Solve for 2θ first:
由于 0° ≤ θ < 360°，所以 0° ≤ 2θ < 720°。先解 2θ：
2θ = 30°, 150°, 390°, 510°
Dividing by 2 gives θ = 15°, 75°, 195°, 255°.
除以 2 得到 θ = 15°, 75°, 195°, 255°。
This technique prevents missing solutions caused by the compressed interval.
这种技巧可以防止因角度范围扩大而遗漏解。
10. Using Identities to Simplify | 用恒等式化简
Sometimes you need to rewrite an equation using fundamental identities before solving. The most common ones are:
有时你需要先用基本恒等式改写方程再求解。最常见的有：
sin²θ + cos²θ = 1, tan θ = sin θ / cos θ
1 − sin²θ) can be simplified using the Pythagorean identity to obtain a quadratic in sin θ.
例如，方程 3 sin²θ = 2(1 − sin²θ) 可以使用毕达哥拉斯恒等式化简为关于 sin θ 的二次方程。
Always check whether an identity can reduce the equation to a single trigonometric function. This is often the key to solving more complex equations.
始终检查恒等式是否可以将方程化简为单一三角函数。这往往是解决更复杂方程的关键。
11. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Many students lose marks on trigonometric equations due to avoidable errors. The table below lists typical pitfalls and solutions.
许多学生因为可避免的错误在三角方程上失分。下表列出了典型陷阱和应对方法。
| Mistake | 错误 | Why it happens | 发生原因 | Correction | 纠正 |
| Forgetting to adjust interval for compound angles | 忘记调整复合角的区间 | Only solving for the inner angle | 只解内角 | Write the interval for 2θ or 3θ first | 先写出 2θ 或 3θ 的区间 |
| Losing solutions when dividing by a trig function | 除以三角函数时丢失解 | Dividing by sin θ without considering sin θ = 0 | 未考虑 sin θ = 0 就除以 sin θ | Factorise instead of dividing | 使用因式分解而非除法 |
| Using degrees instead of radians (or vice versa) | 混淆角度制与弧度制 | Mixing systems in the same solution | 在同一解法中混用两种制度 | Check the question and keep one consistent system | 检查题目并保持一致 |
Always substitute your found solutions back into the original equation to verify them.
始终将找到的解代入原方程验证。
12. Practice Questions | 练习题
Test your understanding with these exam-style questions. Answers are given at the end.
用以下考试风格题目测试你的理解。答案见末尾。
Q1. Solve cos θ = ½ for 0° ≤ θ < 360°.
题1. 求 cos θ = ½ 在 0° ≤ θ < 360° 内的解。
Q2. Solve tan θ = −1 for 0° ≤ θ ≤ 360°.
题2. 求 tan θ = −1 在 0° ≤ θ ≤ 360° 内的解。
Q3. Solve 2 cos²θ − cos θ = 0 for 0° ≤ θ < 360°.
题3. 求 2 cos²θ − cos θ = 0 在 0° ≤ θ < 360° 内的解。
Answers: Q1: {60°, 300°} Q2: {135°, 315°} Q3: {60°, 90°, 270°, 300°}
If you got all three correct, you are well prepared for basic trigonometric equations. If not, revisit the relevant section and practise again.
如果你三道题全部正确，说明你对基础三角方程已经掌握得很好。如果没有，请重新阅读相关章节并再练习。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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