📚 pdfjoiner_(4)-110: Trigonometric Equations | 三角函数方程求解全攻略
find all angles (in degrees or radians) that satisfy an equation involving sin, cos, or tan within a given interval. The key is to understand the periodic nature of these functions and to apply the CAST diagram method to capture every possible solution. This article, based on the content often found in compiled revision materials such as pdfjoiner_(4)-110, will walk you through the essential techniques, from basic equations to more advanced cases involving identities and multiple angles.
三角函数方程是 Edexcel A-Level 数学的核心内容，经常出现在纯数考卷中。你需要根据给定的区间，找出所有满足包含 sin、cos 或 tan 的方程的角度（以度数或弧度表示）。关键在于理解这些函数的周期性，并运用 CAST 图方法获取每一个可能的解。本文基于 pdfjoiner_(4)-110 这类合辑复习资料中常见的内容，将带你掌握从基础方程到涉及恒等式与倍角的更复杂情形的解题技巧。
1. Understanding Trigonometric Equations | 理解三角函数方程
t involves trigonometric functions of an unknown angle. For example, sin x = 0.5, cos 2θ = √3 any solutions. In A-level exams, you are often asked to find all solutions within a specific range, such as 0° ≤ x ≤ 360° or 0 ≤ θ < 2π.
三角函数方程是指包含未知角的三角函数的方程。例如 sin x = 0.5，cos 2θ = √3/2 或 tan(x-30°) = -1 都是三角函数方程。由于正弦、余弦和正切是周期函数，一个方程通常有无穷多个解。在 A-level 考试中，通常会要求找出特定范围内的所有解，如 0° ≤ x ≤ 360° 或 0 ≤ θ < 2π。
2. The CAST Diagram and Quadrant Rule | CAST 图与象限规则
The CAST diagram (or ASTC rule) tells you which of the three trigonometric functions are positive in each quadrant of the unit circle. Starting from Quadrant I (top right), moving anticlockwise: Quadrant I – All are positive (A); Quadrant II – Sine only (S); Quadrant III – Tangent only (T); Quadrant IV – Cosine only (C). This memory aid is vital for determining the correct sign and the appropriate related acute angle when solving equations.
CAST 图（或 ASTC 规则）告诉你单位圆的每个象限中，哪几个三角函数是正的。从第一象限（右上）开始，沿逆时针方向：第一象限 – 全部为正（A）；第二象限 – 仅正弦为正（S）；第三象限 – 仅正切为正（T）；第四象限 – 仅余弦为正（C）。在解方程时，这个记忆辅助工具对于确定正确的符号以及相关的锐角至关重要。
Quadrant I: All + | Quadrant II: Sin + | Quadrant III: Tan + | Quadrant IV: Cos +
| Quadrant | Functions Positive | Angle Range |
| I | sin, cos, tan | 0° – 90° |
| II | sin | 90° – 180° |
| III | tan | 180° – 270° |
| IV | cos | 270° – 360° |
3. Solving sin x = k | 解 sin x = k 型方程
ositive, by the CAST rule the solutions lie in Quadrants I and II. Thus, the solutions are x = α and x = 180° - α. Therefore, x = 30° and x = 150°.
要解 sin x = 0.5（0° ≤ x ≤ 360°），先通过反正弦求出锐角主值：α = sin⁻¹(0.5) = 30°。由于正弦为正，根据 CAST 规则，解位于第一和第二象限。因此解为 x = α 和 x = 180° - α，即 x = 30° 和 x = 150°。
General solution: x = α, 180° - α for sin positive
n Quadrants III and IV. For example, sin x = -0.5 gives α = 30°; solutions: 180° + 30° = 210° and 360° - 30° = 330°.
若 sin x = -k（负值），主值为负角，通常加 360° 得到正锐角参考值。解则落在第三和第四象限。例如 sin x = -0.5，取 α = 30°；解为 180° + 30° = 210° 和 360° - 30° = 330°。
4. Solving cos x = k | 解 cos x = k 型方程
For cos x = k, the principal value α is found using cos⁻¹(|k| x = -0.5, solutions lie in Quadrants II and III: x = 180° - α and x = 180° + α. With α = 60°, we get x = 120° and x = 240°.
对于 cos x = k，利用 cos⁻¹(|k|) 求出主值 α。若 k 为正，解在第一和第四象限：x = α 和 x = 360° - α。例如 cos x = 0.5，α = 60°，解为 60° 和 300°。若 k 为负，例如 cos x = -0.5，解位于第二和第三象限：x = 180° - α 和 x = 180° + α。取 α = 60°，得到 x = 120° 和 x = 240°。
cos x = k → x = α, 360°-α (positive); x = 180°∓α (negative)
5. Solving tan x = k | 解 tan x = k 型方程
. For tan x = 1, the principal value α = tan⁻¹(1) = 45°. Solutions in 0° ≤ x ≤ 360° are x = 45° and x = 45° + 180° = 225°. There is no quadrant restriction based on sign; merely add 180° to get the next solution.
正切方程的周期为 180°，因此找到一个解之后，加或减 180° 的倍数即可得到区间内的其他解。对于 tan x = 1，主值 α = tan⁻¹(1) = 45°。在 0° ≤ x ≤ 360° 内的解为 x = 45° 和 x = 45° + 180° = 225°。符号不影响象限查找，只需每次加 180° 即可得到下一个解。
tan x = k → x = α + n·180°
6. Equations with Multiple Angles | 含倍角的方程
in u = 0.5 for 0° ≤ u ≤ 720°. Solve for u: u = 30°, 150°, 390°, 510°. Then divide by 2 to find θ: θ = 15°, 75°, 195°, 255°. Always check that the final θ values lie within the original interval.
当角度前面有系数时，例如 sin 2θ = 0.5（0° ≤ θ ≤ 360°），必须先转换定义域。令 u = 2θ，则方程变为 sin u = 0.5，且 0° ≤ u ≤ 720°。解出 u = 30°, 150°, 390°, 510°。然后除以 2 得到 θ = 15°, 75°, 195°, 255°。始终要检查最终的 θ 值是否落在原区间内。
Example: sin 2θ = 0.5 → change domain, solve, then halve
7. Quadratic Trigonometric Equations | 二次三角函数方程
. So cos x = -½ or cos x = 1. Solve each: for cos x = 1, x = 0°, 360°; for cos x = -½, principal α = 60°, solutions 120°, 240°. Collect all valid solutions in the range.
像 2 cos² x - cos x - 1 = 0 这样的方程可以看作关于 cos x 的二次方程。设 y = cos x，则 2y² - y - 1 = 0，分解因式得 (2y + 1)(y - 1) = 0，故 y = -½ 或 y = 1。即 cos x = -½ 或 cos x = 1。分别求解：cos x = 1 ⇒ x = 0°, 360°；cos x = -½ ⇒ 主值 α = 60°，解为 120°, 240°。收集区间内所有有效解。
2 cos² x - cos x - 1 = 0 → (2 cos x + 1)(cos x - 1) = 0
ilarly, equations with tan² x can be reduced. For instance, tan² x - 3 = 0 gives tan x = ±√3 h by finding α = 60° and using the period of 180°, you will get x = 60°, 120°, 240°, 300°.
类似地，含有 tan² x 的方程也可化简。例如 tan² x - 3 = 0 得 tan x = ±√3。求出 α = 60°，利用周期 180° 可得 x = 60°, 120°, 240°, 300°。
8. Using Trigonometric Identities | 运用三角恒等式
= 1 for 0° ≤ x ≤ 360°. Replace cos² x with 1 - sin² x: 3(1 - sin² x) - sin x = 1 → 3 - 3 sin² x - sin x = 1 → 3 sin² x + sin x - 2 = 0. This is a quadratic in sin x: (3 sin x - 2)(sin x + 1) = 0 → sin x = 2/3 or sin x = -1. sin x = 2/3 gives two solutions (Quadrant I and II); sin x = -1 gives x = 270°.
许多方程需要你先用恒等式替换某项再进行求解。最常用的就是 sin² x + cos² x = 1。例如解 3 cos² x - sin x = 1（0° ≤ x ≤ 360°）。将 cos² x 换成 1 - sin² x：3(1 - sin² x) - sin x = 1 → 3 - 3 sin² x - sin x = 1 → 3 sin² x + sin x - 2 = 0。这是关于 sin x 的二次方程：(3 sin x - 2)(sin x + 1) = 0 → sin x = 2/3 或 sin x = -1。sin x = 2/3 给出两个解（第一、二象限）；sin x = -1 得 x = 270°。
sin² x + cos² x ≡ 1
Double-angle formulas are also frequently used. For example, cos 2x can be expressed as 2 cos² x - 1 or 1 - 2 sin² x, which helps reduce the equation to a single trig function.
倍角公式也经常用到。例如 cos 2x 可写成 2 cos² x - 1 或 1 - 2 sin² x，这有助于将方程化为单一三角函数。
9. Equations with Phase Shifts | 含相位移动的方程
0°. Solve sin u = 0.8: u ≈ 53.13°, 126.87°, 413.13° (by adding 360° to the first solution). Then subtract 30° from each to recover x: x ≈ 23.13°, 96.87°, 383.13°. Discard the last if it exceeds 360°. Always be mindful of the adjusted interval.
解 sin(x + 30°) = 0.8 这类方程时，采用与倍角类似的方法。令 u = x + 30°，则定义域发生平移。若 0° ≤ x ≤ 360°，则 30° ≤ u ≤ 390°。解 sin u = 0.8，得 u ≈ 53.13°, 126.87°, 413.13°（将第一个解加 360° 得到）。然后每个解减去 30° 还原 x：x ≈ 23.13°, 96.87°, 383.13°。若最后一个超过 360° 则舍去。始终注意调整后的区间。
10. Radian Measure Solutions | 弧度制下的解
o give solutions in radians between 0 and 2π. The same CAST principle applies. For sin x = √3 yields x = 3π/4 and 7π/4. Remember the period of tan is π, so you can add multiples of π.
在 A-level 考试中，经常要求用 0 至 2π 之间的弧度给出解。同样应用 CAST 原理。对于 sin x = √3/2，在 0 ≤ x ≤ 2π 内，锐角为 π/3，解为 x = π/3 和 x = 2π/3。对于 cos x = -½，α = π/3，解为 2π/3 和 4π/3。tan x = -1 的解为 x = 3π/4 和 7π/4。记住 tan 的周期是 π，因此可以加上 π 的倍数。
π rad = 180°; key angles: π/6, π/4, π/3, π/2
11. Common Pitfalls and How to Avoid Them | 常见错误及规避方法
One frequent mistake is forgetting to check that all solutions fall within the specified interval after solving for a transformed angle. Students often stop after finding one or two solutions, especially when the domain changes. Another pitfall is misapplying the CAST quadrants for negative values. Always draw a quick sketch of the waveform or the CAST diagram to visualise where the solutions should be. Additionally, when squaring both sides of an equation, extraneous solutions may be introduced, so always verify your final answers.
一个常见错误是，在求出变换后的角之后，忘记检查所有解是否落在指定的区间内。学生经常在找出一两个解后就停下，尤其是在定义域改变时。另一个易错点是对负值误用 CAST 象限。始终快速画一张波形草图或 CAST 图，以便直观看出解该往哪里取。此外，当对方程两边平方时，可能会引入增根，因此最后一定要验证所得的解。
12. Exam Tips and Summary | 考试技巧与总结
s, find the principal acute angle, and generate all solutions within the range. When the equation is quadratic or contains multiple functions, simplify using identities before solving. Manage transformed domains carefully by writing them out explicitly. With consistent practice, the method becomes routine and your confidence will grow. Remember, the pdfoiner_(4)-110 material encapsulates many of these strategies into a single revision resource.
要在 Edexcel A-Level 考试中掌握三角函数方程，需要系统练习：先识别函数，使用 CAST 确定象限，求出主锐角，然后生成范围内的所有解。当方程为二次式或包含多种函数时，解题前先用恒等式化简。处理变换后的定义域时要明确写出新的范围。持续练习将使这套方法成为本能，你的信心也会随之增强。请记住，pdfoiner_(4)-110 这份材料将诸多策略凝聚成了一本高效的复习资源。
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