📚 Solving Trigonometric Equations (0° to 180°) | 解三角函数方程（0° 至 180°）
In Edexcel A-Level Mathematics, solving trigonometric equations within a specific interval is a fundamental skill. One of the most commonly tested ranges is from 0° to 180°, which links directly to the unit circle, the CAST diagram, and the graphs of sine, cosine, and tangent. Understanding how each trigonometric function behaves in this interval will help you find all possible solutions efficiently and avoid losing marks.
在 Edexcel A-Level 数学中，在特定区间内求解三角方程是一项基本技能。最常考的区间之一就是从 0° 到 180°，这与单位圆、CAST 图以及正弦、余弦和正切的图像直接相关。理解各个三角函数在此区间内的行为，有助于你高效地找到所有可能的解，避免失分。
1. Introduction to Trigonometric Equations | 三角函数方程简介
ou are often asked to solve such equations for x within a given domain, commonly 0° ≤ x ≤ 180° or 0 ≤ x ≤ π radians. The interval 0° to 180° corresponds to Quadrants I and II on the CAST diagram, which immediately tells you which trigonometric ratios can be positive or negative.
三角方程包含未知角的一个或多个三角函数，例如 sin x = 0.5 或 2 cos² x − 1 = 0。在纯数学中，题目常常要求你在给定定义域内求解 x，通常是 0° ≤ x ≤ 180° 或 0 ≤ x ≤ π 弧度。0° 到 180° 这个区间对应 CAST 图中的第一和第二象限，这直接告诉你哪些三角比可以为正、哪些可以为负。
2. Understanding Degrees and Radians | 理解角度与弧度
nd 30° = π/6 rad. You must be comfortable working in both degrees and radians, as Edexcel exam questions frequently mix the two or specify the mode explicitly.
区间 0° 到 180° 等价于 0 到 π 弧度。转换因子是 180° = π rad。因此，90° = π/2 rad，60° = π/3 rad，45° = π/4 rad，30° = π/6 rad。你必须能熟练地在角度和弧度之间切换，因为 Edexcel 考题经常混合使用或明确指定模式。
Always check whether the question requires the final answer in degrees or radians. Giving the answer in the wrong measure can cause you to lose accuracy marks even if your method is correct.
务必检查题目要求最终答案用角度还是弧度表示。即使方法正确，用错误的度量给出答案也会导致你失去准确度分数。
3. The CAST Diagram for 0° to 180° | 0° 至 180° 区间的 CAST 图
The CAST diagram divides the coordinate plane into four quadrants, starting from the positive x‑axis and moving anticlockwise. For the interval 0° to 180°, we are only concerned with Quadrant I (0° to 90°) and Quadrant II (90° to 180°). In Quadrant I, all three trigonometric ratios (sin, cos, tan) are positive. In Quadrant II, only sin is positive; cos and tan are negative.
CAST 图将坐标平面分为四个象限，从正 x 轴开始逆时针旋转。对于 0° 到 180° 的区间，我们只关心第一象限（0° 到 90°）和第二象限（90° 到 180°）。在第一象限中，三个三角比（sin、cos、tan）均为正。在第二象限中，只有 sin 为正；cos 和 tan 为负。
This simple rule determines how many solutions an equation can have:
这条简单规则决定了一个方程可能有多少个解：
- ns in [0°, 180°]: one acute (Quadrant I) and one obtuse (Quadrant II). / sin x = k (k > 0) 在 [0°, 180°] 内可以有两个解：一个锐角（第一象限）和一个钝角（第二象限）。
- nly one solution in [0°, 180°] because cos is negative in Quadrant II. / cos x = k (k > 0) 在 [0°, 180°] 内只有一个解，因为 cos 在第二象限为负。
- 0°] since tan has a period of 180° and is positive only in Quadrant I. / tan x = k (k > 0) 在 [0°, 180°] 内刚好有一个解，因为 tan 的周期为 180°，且只在第一象限为正。
4. Special Angles and Exact Values | 特殊角与精确值
You must memorise the exact values of sin, cos, and tan for the standard angles in the 0° to 180° range. The table below summarises these values; they form the building blocks for many trigonometric equations.
你必须熟记 0° 到 180° 范围内标准角度的 sin、cos 和 tan 的精确值。下表总结了这些值；它们是许多三角方程的基础构件。
| Angle / 角度 | 0° / 0 | 30° / π/6 | 45° / π/4 | 60° / π/3 | 90° / π/2 | 120° / 2π/3 | 135° / 3π/4 | 150° / 5π/6 | 180° / π |
|---|---|---|---|---|---|---|---|---|---|
| sin | 0 | ½ | √2/2 | √3/2 | 1 | √3/2 | √2/2 | ½ | 0 |
| cos | 1 | √3/2 | √2/2 | ½ | 0 | −½ | −√2/2 | −√3/2 | −1 |
| tan | 0 | √3/3 | 1 | √3 | undefined | −√3 | −1 | −√3/3 | 0 |
5. Solving sin x = k (0° ≤ x ≤ 180°) | 解 sin x = k (0° ≤ x ≤ 180°)
oth Quadrant I and Quadrant II, two solutions exist for positive k within 0° to 180°: the acute angle α and its supplement 180° − α. For example, to solve sin x = ½ for 0° ≤ x ≤ 180°:
当求解 sin x = k 时，首先利用计算器或精确值求出主值（锐角）。由于 sin 在第一象限和第二象限均为正，对于 k > 0，在 0° 到 180° 内存在两个解：锐角 α 及其补角 180° − α。例如，求解 sin x = ½，0° ≤ x ≤ 180°：
x = 30° (acute, Quadrant I) and x = 180° − 30° = 150° (obtuse, Quadrant II)
e means the angle must lie in Quadrant III or IV. Since those quadrants are outside the 0°– the equation has no solution. For example, sin x = −0.5 has no solution for 0° ≤ x ≤ 180°.
如果 k 为负，sin x = 负数意味着角度必定在第三或第四象限。由于这些象限超出了 0°–180° 的范围，方程无解。例如，sin x = −0.5 在 0° ≤ x ≤ 180° 内无解。
6. Solving cos x = k (0° ≤ x ≤ 180°) | 解 cos x = k (0° ≤ x ≤ 180°)
k has exactly one solution: an acute angle in Quadrant I. For a negative k, the solution is an obtuse angle in Quadrant II. For instance, cos x = ½ gives x = 60°, while cos x = −½ gives x = 120°. In general, if cos x = k (where k ≥ 0), x = α; if k < 0, x = 180° − α, where α is the positive reference angle such that cos α = |k|.
余弦在 0° 到 180° 区间内的行为有所不同。该函数从 1 递减到 −1，在 90° 处穿过零。对于正数 k，方程 cos x = k 恰好有一个解：第一象限的锐角。对于负数 k，解是第二象限的钝角。例如，cos x = ½ 给出 x = 60°，而 cos x = −½ 给出 x = 120°。一般来说，若 cos x = k（k ≥ 0），x = α；若 k < 0，x = 180° − α，其中 α 是使得 cos α = |k| 的正参考角。
7. Solving tan x = k (0° ≤ x ≤ 180°) | 解 tan x = k (0° ≤ x ≤ 180°)
nt II (90° < x < 180°). For tan x = k where k > 0, there is a single acute solution. For k < 0, there is a single obtuse solution. For example, tan x = √3 1 gives x = 135°. Always remember that you never include 90° because tan 90° is undefined.
正切函数的周期为 180°，其图像在 90° 处有一条渐近线。在 0° 到 180° 内，tan x 在第一象限（0° < x < 90°）为正，在第二象限（90° < x < 180°）为负。对于 tan x = k（k > 0），只有一个锐角解。对于 k < 0，只有一个钝角解。例如，tan x = √3 给出 x = 60°，tan x = −1 给出 x = 135°。始终记住，不可取 90°，因为 tan 90° 无定义。
8. Using Trigonometric Identities | 使用三角恒等式
r instance, to solve 2 sin² x + cos x = 2 in the interval 0° ≤ x ≤ 180°, replace sin² x with 1 − cos² x:
许多方程需要你使用恒等式 sin² x + cos² x = 1 或其变形，将方程化为单一三角函数。例如，求解方程 2 sin² x + cos x = 2，0° ≤ x ≤ 180°，将 sin² x 替换为 1 − cos² x：
2(1 − cos² x) + cos x = 2 → 2 − 2 cos² x + cos x = 2 → −2 cos² x + cos x = 0
oning, cos x = 0 gives x = 90°, and cos x = ½ gives x = 60°. Both are within the interval.
因式分解得 cos x (1 − 2 cos x) = 0，故 cos x = 0 或 cos x = ½。根据前面的推理，cos x = 0 给出 x = 90°，cos x = ½ 给出 x = 60°。两者均在区间内。
9. Quadratic Trigonometric Equations | 二次三角方程
sider 2 sin² x − 3 sin x + 1 = 0 for 0° ≤ x ≤ 180°. Factorising yields (2 sin x − 1)(sin x − 1) = 0, so sin x = ½ or sin x = 1. sin x = ½ produces x = 30°, 150°; sin x = 1 gives x = 90°. Therefore the solution set is {30°, 90°, 150°}.
形如 a sin² x + b sin x + c = 0 或 a cos² x + b cos x + c = 0 的方程，其解法与普通二次方程相同，但变量是三角函数。考虑方程 2 sin² x − 3 sin x + 1 = 0，0° ≤ x ≤ 180°。因式分解得 (2 sin x − 1)(sin x − 1) = 0，因此 sin x = ½ 或 sin x = 1。sin x = ½ 产生 x = 30°、150°；sin x = 1 给出 x = 90°。因此解集为 {30°, 90°, 150°}。
Always check that each factor' 1 and 1 before solving. If a factor gives sin x = 2, for instance, discard it immediately.
在求解之前，务必检查每个因子的取值是否在 −1 和 1 之间。例如，若某个因子给出 sin x = 2，应立即舍弃。
10. Checking Solutions within the Interval | 在区间内检验解
des the endpoints 0° and 180° if the function is defined there. Substitute each value back into the original equation to ensure it satisfies the equality, not just an intermediate step where squaring or factoring might have introduced extraneous roots.
在求出候选解之后，必须验证每个解是否落在指定的定义域内。记住，区间 0° ≤ x ≤ 180° 包含端点 0° 和 180°，前提是函数在该处有定义。将每个值代回原方程，以确保其满足等式，而不仅仅是代入某个中间步骤，因为平方或因式分解可能引入了增根。
alid because tan 0° = 0 and tan 180° = 0, and neither coincides with the asymptote at 90°.
例如，在 0° ≤ x ≤ 180° 范围内求解 tan x = 0，得到 x = 0°、180°。两者均有效，因为 tan 0° = 0 且 tan 180° = 0，且它们均不触及 90° 处的渐近线。
11. Common Mistakes to Avoid | 常见错误避免
- Forgetting the second sin solution: When sin x = positive, students often give only the acute angle and forget 180° − α. / 忘记 sin 的第二个解：当 sin x = 正数时，学生常常只给出锐角，而忽略 180° − α。
- Confusing quadrants for negative values: Assuming a negative sin means no solution without checking the interval; in [0°, 180°], negative sin indeed has no solution, but negative cos and tan do have solutions. / 混淆负值所在的象限：未检查区间就假定负的 sin 无解；在 [0°, 180°] 内，负 sin 确实无解，但负 cos 和负 tan 有解。
- Using the wrong period for tan: Treating tan x like sin (180° symmetry) or cos (360° period) leads to missed or extra solutions. / 对 tan 使用错误的周期：把 tan x 当作 sin（180° 对称）或 cos（360° 周期）处理，会导致漏解或增解。
- Radians/Degrees mix-up: Setting your calculator to the wrong mode and obtaining a radian value when the question expects degrees. / 弧度/角度混淆：计算器模式设置错误，题目要求角度却得到弧度值。
- Discarding the cos x = negative case: Overlooking that cos x = −k still yields a solution in the second quadrant. / 舍弃 cos x = 负值的情况：忽视 cos x = −k 在第二象限仍有解。
12. Exam Tips and Summary | 考试技巧与总结
To tackle any trigonometric equation in the 0° to 180° range with confidence, follow this structured approach: identify the function, determine the sign, find the reference angle using exact values or a calculator, apply the CAST quadrant rules, and list all solutions within the interval. Always write your final answer clearly, using either degrees or radians as specified.
为了自信地处理 0° 到 180° 范围内的任何三角方程，请遵循以下结构化方法：识别函数，判断符号，利用精确值或计算器求出参考角，应用 CAST 象限规则，列出区间内的所有解。始终按照题目规定，清晰地写出最终答案，使用角度或弧度。
Practise with a wide variety of equations, including those that require identities and factorisation, and double-check that your solutions satisfy the original equation. Mastering this topic not only secures marks on the Pure 2 paper but also lays the groundwork for calculus problems involving trigonometric functions.
多练习各种类型的方程，包括需要使用恒等式和因式分解的题目，并反复检查你的解是否满足原方程。掌握这一专题不仅能确保你在 Pure 2 试卷上得分，还能为涉及三角函数的微积分问题奠定基础。
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