Solving Trigonometric Equations | 三角方程的求解技巧

📚 Solving Trigonometric Equations | 三角方程的求解技巧

Trigonometric equations are a core topic in A-Level Mathematics. Mastering them requires a clear understanding of periodic functions, reference angles, and domain restrictions. This article breaks down the essential strategies for solving trig equations accurately and efficiently.

三角方程是A-Level数学的核心内容。掌握它需要对周期函数、参考角和定义域限制有清晰的理解。本文将系统讲解求解三角方程的关键策略，帮助你在考试中准确高效地得分。

1. Basic Concepts and Periodicity | 基本概念与周期性

The three main trigonometric functions are sine, cosine, and tangent. Each function is periodic: sine and cosine have a period of 360° (2π radians), while tangent has a period of 180° (π radians). This means that without domain restrictions, a trigonometric equation generally has infinitely many solutions.

三大主要三角函数是正弦、余弦和正切。每个函数都是周期性的：正弦和余弦的周期为360°（2π弧度），而正切的周期为180°（π弧度）。这意味着在没有定义域限制时，三角方程通常有无限多个解。

For example, the equation sin θ = 0.5 has solutions θ = 30° + 360°n and θ = 150° + 360°n for any integer n. When a question specifies an interval such as 0° ≤ θ ≤ 360°, you must list only the solutions inside that interval.

例如，方程 sin θ = 0.5 的解为 θ = 30° + 360°n 和 θ = 150° + 360°n，其中n为任意整数。当题目指定区间如 0° ≤ θ ≤ 360° 时，你只需列出该区间内的解。

2. Special Angle Values | 特殊角的值

Knowing the exact values of trigonometric functions for common angles is essential. The table below summarises the values for angles in degrees.

熟记常见角的三角函数精确值是解题的基础。下表总结了角度制下一些常见角的值。

Angle θ	sin θ	cos θ	tan θ
0°	0	1	0
30°	½	√3/2	√3/3
45°	√2/2	√2/2	1
60°	√3/2	½	√3
90°	1	0	undefined

You should also memorise the radian equivalents: 30° = π/6, 45° = π/4, 60° = π/3, and 90° = π/2. These appear in nearly every exam.

同时需要记住对应的弧度制：30° = π/6，45° = π/4，60° = π/3，90° = π/2。这些在几乎所有考试中都会出现。

3. Using the Unit Circle and Graphs | 使用单位圆与图象

The unit circle helps visualise the signs of sin, cos, and tan in each quadrant. The CAST rule summarises this:

单位圆可以帮助直观理解sin、cos、tan在各个象限的正负。CAST法则总结了这一点：

Quadrant A (0°–90°): All functions are positive.

第一象限 (0°–90°)： 所有函数均为正。
Quadrant S (90°–180°): Only sine is positive.

第二象限 (90°–180°)： 只有正弦为正。
Quadrant T (180°–270°): Only tangent is positive.

第三象限 (180°–270°)： 只有正切为正。
Quadrant C (270°–360°): Only cosine is positive.

第四象限 (270°–360°)： 只有余弦为正。

For example, if sin θ = 0.5, the reference angle is 30°. In the first quadrant, θ = 30°; in the second quadrant, θ = 180° − 30° = 150°. Drawing a quick sketch of the sine curve can also confirm these solutions.

例如，如果 sin θ = 0.5，参考角为30°。在第一象限，θ = 30°；在第二象限，θ = 180° − 30° = 150°。快速画出正弦曲线草图也可以验证这些解。

4. Principal Values and General Solutions | 主值与一般解

A calculator gives the principal value, which is the solution in a restricted domain: for sin and tan, usually −90° to 90°; for cos, 0° to 180°. However, you must adjust this to find all solutions in the required interval.

计算器给出的是主值，即在受限定义域内的解：对sin和tan通常是 −90° 到 90°，对cos是 0° 到 180°。但你必须调整主值以得到给定区间内的所有解。

For an angle α, the general solutions are:

For sin θ = sin α: θ = α + 360°n or θ = 180° − α + 360°n

For cos θ = cos α: θ = ±α + 360°n

For tan θ = tan α: θ = α + 180°n

In radians, replace 360° with 2π and 180° with π. For example, tan θ = √3 gives θ = 60° + 180°n, so in [0°, 360°] the solutions are 60° and 240°.

在弧度制中，将360°替换为2π，180°替换为π。例如，tan θ = √3 给出 θ = 60° + 180°n，因此在 [0°, 360°] 内的解为60°和240°。

5. Transformations and Substitutions | 变换与替换

When the angle is not simply θ, such as sin(2x + 30°) = 0.5, treat the whole expression as a new variable. Let u = 2x + 30°, solve for u first, then solve for x.

当角度不是简单的θ，例如 sin(2x + 30°) = 0.5，可以把整个式子视为一个新变量。令 u = 2x + 30°，先解出u，再解出x。

For example, solve sin u = 0.5 for 0° ≤ x ≤ 360°. Then u = 30° + 360°n and u = 150° + 360°n. Since u = 2x + 30°, we write 2x + 30° = 30° + 360°n or 2x + 30° = 150° + 360°n. Solving gives x = 360°n or x = 60° + 180°n. Now list the values in the given domain.

例如，在 0° ≤ x ≤ 360° 内解 sin u = 0.5。则 u = 30° + 360°n 以及 u = 150° + 360°n。由于 u = 2x + 30°，所以 2x + 30° = 30° + 360°n 或 2x + 30° = 150° + 360°n。解得 x = 360°n 或 x = 60° + 180°n。然后列出给定定义域内的值。

Be careful: a smaller interval for x may correspond to a larger interval for u. Always extend the u interval before solving.

注意：x的区间较小时，u的区间可能更大。务必先扩展u的区间再求解。

6. Quadratic Trigonometric Equations | 二次型三角方程

Equations like 2sin²θ − sinθ − 1 = 0 are quadratic in sinθ. Let y = sinθ, then solve 2y² − y − 1 = 0 by factorisation:

像 2sin²θ − sinθ − 1 = 0 这样的方程是关于sinθ的二次方程。令 y = sinθ，然后通过因式分解解 2y² − y − 1 = 0：

(2y + 1)(y − 1) = 0 ⇒ y = −½ or y = 1

Thus sin θ = −½ or sin θ = 1. The second equation gives θ = 90°; the first gives θ = 210° and θ = 330° within 0° ≤ θ < 360°.

因此 sin θ = −½ 或 sin θ = 1。第二个方程给出 θ = 90°；第一个在 0° ≤ θ < 360° 内给出 θ = 210° 和 θ = 330°。

Always remember that sinθ and cosθ are bounded between −1 and 1. If a quadratic root gives a value outside this range, discard it immediately.

永远记住 sinθ 和 cosθ 的取值范围在 −1 到 1 之间。如果二次方程的根超出此范围，应立即舍去。

7. Using Identities to Simplify | 恒等式化简

Some equations mix different trigonometric functions. Use fundamental identities to reduce them to a single function.

有些方程混合了不同的三角函数。使用基本恒等式将它们化为单一函数。

sin²θ + cos²θ = 1, tanθ = sinθ / cosθ

For example, solve 3cos²θ + 2sinθ = 1 for 0° ≤ θ ≤ 360°. Replace cos²θ with 1 − sin²θ:

例如，在 0° ≤ θ ≤ 360° 内解 3cos²θ + 2sinθ = 1。将 cos²θ 替换为 1 − sin²θ：

3(1 − sin²θ) + 2sinθ = 1 ⇒ 3sin²θ − 2sinθ − 2 = 0

This is now a quadratic in sinθ. Solve it carefully, then find the corresponding angles within the domain.

这变成了关于sinθ的二次方程。谨慎求解，然后在定义域内找出相应的角度。

8. Multiple Solutions and Domain Restrictions | 多解范围与取舍

When a domain is given, you must include the endpoints correctly. For example, “0° ≤ θ ≤ 360°” includes both 0° and 360°, while “0° ≤ θ < 360°" excludes 360°. Always check the inequality signs.

当给定定义域时，必须正确包含端点。例如，”0° ≤ θ ≤ 360°” 同时包含0°和360°，而 “0° ≤ θ < 360°" 不包含360°。始终检查不等号。

To list all solutions, write the general solution first, then substitute integer values of n and select those within the interval. It may require trying n = 0, ±1, ±2, etc.

要列出所有解，先写出通解，然后代入整数n并选择区间内的解。可能需要尝试 n = 0, ±1, ±2 等。

For example, for tan(3θ) = 1, the general solution is 3θ = 45° + 180°n, so θ = 15° + 60°n. In 0° ≤ θ < 180°, n = 0,1,2 give θ = 15°, 75°, 135°.

例如，对于 tan(3θ) = 1，通解为 3θ = 45° + 180°n，因此 θ = 15° + 60°n。在 0° ≤ θ < 180° 内，n = 0,1,2 给出 θ = 15°, 75°, 135°。

9. Common Pitfalls and Tips | 常见陷阱与注意事项

Students often lose marks due to avoidable mistakes. The checklist below will help you stay accurate.

学生常常因为可避免的错误而失分。以下清单帮助你保持准确。

Missing the second solution: When solving sin θ = c in 0°–360°, always find both quadrants.

漏掉第二个解： 在 0°–360° 内解 sin θ = c 时，务必找到两个象限的解。
Forgetting the period: tan repeats every 180°, not 360°.

忘记周期： tan每180°重复一次，不是360°。
Applying inverse functions incorrectly: The calculator’s principal value often needs adjusting.

错误使用反函数： 计算器的主值通常需要调整。
Discarding valid roots: For a quadratic in sinθ, retain every root in [−1, 1].

丢弃有效根： 对于关于sinθ的二次方程，保留所有在 [−1, 1] 内的根。
Not using radian mode when needed: If the domain is in radians, switch your calculator to radian mode.

需要时未使用弧度模式： 如果定义域是弧度，请将计算器切换到弧度模式。

10. Worked Examples | 实际例题演练

Let us apply these techniques to two typical exam-style problems.

我们将这些技巧应用到两道典型考试风格的题目中。

Example 1: Solve 2cos²θ − 3sinθ = 0 for 0° ≤ θ ≤ 360°.

例1： 解方程 2cos²θ − 3sinθ = 0，其中 0° ≤ θ ≤ 360°。

Use the identity cos²θ = 1 − sin²θ:

使用恒等式 cos²θ = 1 − sin²θ：

2(1 − sin²θ) − 3sinθ = 0 ⇒ 2sin²θ + 3sinθ − 2 = 0

(2sinθ − 1)(sinθ + 2) = 0 ⇒ sinθ = ½ or sinθ = −2

Since sinθ cannot be −2, only sinθ = ½. This gives θ = 30° and θ = 150° in the given domain.

因为 sinθ 不可能等于 −2，所以只有 sinθ = ½。在给定定义域内得到 θ = 30° 和 θ = 150°。

Example 2: Solve tan(2x − π/4) = √3 for 0 ≤ x ≤ 2π (radians).

例2： 解方程 tan(2x − π/4) = √3，其中 0 ≤ x ≤ 2π（弧度）。

Let u = 2x − π/4. Then tan u = √3, so u = π/3 + πn. But x must lie in [0, 2π], so u lies in [−π/4, 15π/4] roughly. We need integers n such that u is in this extended interval.

令 u = 2x − π/4。则 tan u = √3，因此 u = π/3 + πn。但x必须在 [0, 2π] 内，所以u大约在 [−π/4, 15π/4] 内。我们需要满足条件的整数n。

List u values: −2π/3, π/3, 4π/3, 7π/3, 10π/3. Only those in [−π/4, 15π/4] are π/3, 4π/3, 7π/3, 10π/3. Now solve 2x − π/4 = each u:

列出u值：−2π/3, π/3, 4π/3, 7π/3, 10π/3。在 [−π/4, 15π/4] 内的为 π/3, 4π/3, 7π/3, 10π/3。然后解 2x − π/4 = 每个u：

x = (u + π/4)/2 ⇒ x = 7π/24, 19π/24, 31π/24, 43π/24

These four solutions are all within 0 ≤ x ≤ 2π.

这四个解都在 0 ≤ x ≤ 2π 内。

By mastering these methods — understanding periodicity, using exact values, applying CAST, handling transformations, and avoiding common pitfalls — you can confidently solve any trigonometric equation in your exam.

通过掌握这些方法——理解周期性、使用精确值、应用CAST法则、处理变换以及避免常见陷阱——你可以自信地解决考试中任何三角方程。

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Solving Trigonometric Equations | 三角方程的求解技巧