📚 Solving Trigonometric Equations: Worked Example 5.3.1 | 解三角方程：例题 5.3.1
In this AQA A-Level Mathematics revision article, we work carefully through Example 5.3.1 on solving a trigonometric equation in a specified interval. The technique shown here - finding a principal value, then using the symmetry of the sine curve and the CAST diagram to locate every other solution - applies directly to Paper 1 and Paper 2 pure mathematics questions involving sine, cosine and tangent.
在这篇 AQA A-Level 数学复习文章中，我们将完整、细致地讲解例题 5.3.1——在指定区间内求解三角方程。这里展示的方法（先求主值，再利用正弦曲线的对称性与 CAST 图找出其余所有解）直接适用于纯数学试卷 1 和试卷 2 中涉及正弦、余弦和正切的考题。
1. The Worked Problem | 例题呈现
Example 5.3.1 asks us to solve sin θ = 1/2 for 0° ≤ θ ≤ 360°, giving all the solutions that lie within this interval. The instruction "all solutions" is the crucial part: a single calculator answer will not earn full marks.
例题 5.3.1 要求解方程 sin θ = 1/2，其中 0° ≤ θ ≤ 360°，并给出该区间内的所有解。"所有解"这四个字是关键：只写一个计算器给出的答案无法拿到满分。
sin θ = 1/2, 0° ≤ θ ≤ 360°
The interval 0° ≤ θ the unit circle, and the inequality signs are inclusive, so angles equal to 0° or 360° would count as valid answers if they happened to satisfy the equation. In this case they do not, but examiners do test that boundary condition in other questions.
区间 0° ≤ θ ≤ 360° 恰好覆盖单位圆的一整圈，而且不等号均取等号，因此若 0° 或 360° 恰好满足方程，它们也算作有效答案。本题中这两个边界值并不满足方程，但在其他题目中主考方确实会考查这一边界条件。
This is a core trigonometry skill: finding every angle in a full turn that has the same sine value, and doing it without relying on guesswork.
这是一项核心三角学技能：找出一个完整圆周内具有相同正弦值的所有角，而且不能靠试数猜测。
2. Understanding the Sine Function | 理解正弦函数
The graph of y = sin θ has a period of 360° and is symmetric about the vertical line θ epetition, for most sine values between −1 and 1 there are two solutions in one full turn.
y = sin θ 的图像周期为 360°，并关于直线 θ = 90° 对称。由于这种重复性，对于 −1 到 1 之间的大多数正弦值，在一个完整圆周内都会有两个解。
The sign of sine across the four quadrants follows a fixed pattern that you should be able to recall instantly:
正弦在四个象限中的正负号遵循固定规律，应当能够立刻回忆出来：
- sin θ is positive in the 1st and 2nd quadrants.
- sin θ is negative in the 3rd and 4th quadrants.
- sin θ 在第 1 和第 2 象限为正。
- sin θ 在第 3 和第 4 象限为负。
Since 1/2 is positive, both of our expected solutions must lie in the first two quadrants, that is between 0° and 180°. That single observation already rules out three-quarters of the circle and is worth stating in your working.
由于 1/2 是正数，我们预期的两个解必然落在前两个象限，即 0° 到 180° 之间。这一观察本身就排除了圆周的四分之三，值得写在解题过程中。
The key point is that one sine value can correspond to more than one angle, so a calculator answer alone is never the complete solution.
关键在于，一个正弦值可以对应多个角，因此仅凭计算器给出的答案永远不是完整解。
3. Finding the Principal Value | 求主值
n to find the principal value. Since sin 30° = 1/2, we obtain the first solution directly:
我们首先使用反正弦函数求主值。因为 sin 30° = 1/2，所以可以直接得到第一个解：
θ = sin⁻¹(1/2) = 30°
The principal value of arcsin always lies between −90° and 90°, so a calculator in degree mode will return 30° and nothing else. This restricted range is a deliberate design choice: without it, the inverse function would not be well defined, because infinitely many angles share the same sine.
反正弦的主值始终位于 −90° 到 90° 之间，因此处于角度模式的计算器只会返回 30°。这一受限的值域是刻意设定的：若不限制，反函数就无法定义，因为有无穷多个角具有相同的正弦值。
However, the principal value is only the first solution. We must use the symmetry of the sine function to find all the other solutions inside the requested interval, and then confirm that each one genuinely lies in the interval.
然而，主值只是第一个解。我们必须利用正弦函数的对称性，求出所给区间内的所有其他解，然后逐一确认它们确实落在该区间内。
Make it a habit to write the principal value on its own line before doing anything else. It keeps your working readable and protects you from mixing up the two solutions later.
要养成先把主值单独写一行的习惯。这样既能让解题过程清晰易读，也能避免后面把两个解弄混。
4. Using the CAST Diagram | 使用 CAST 图
The CAST diagram tells us which trigonometric ratios are positive in each quadrant:
CAST 图告诉我们每个象限中哪些三角比为正：
| Quadrant | Positive ratios | 中文 |
|---|---|---|
| I | All | 全部为正 |
| II | Sine | 正弦为正 |
| III | Tangent | 正切为正 |
| IV | Cosine | 余弦为正 |
Since sin θ y. The principal value 30° already sits in quadrant I, so the missing solution must be the corresponding angle in quadrant II, measured from the negative θ-axis back towards the positive y-axis.
因为 sin θ = 1/2 > 0，我们只需要第 I 和第 II 象限。主值 30° 已经位于第 I 象限，所以缺失的解必定是第 II 象限中对应的那个角，即从负 θ 轴反向量向正 y 轴得到的角。
A useful shorthand in exams is to write "S" and "A" next to your sketch and tick the two quadrants you need. It takes three seconds and prevents the most common error in the whole topic.
考试中一个实用的速记方法是在草图上标出 "S" 和 "A"，并勾出需要的两个象限。这只需三秒，却能避免整个专题中最常见的错误。
5. Finding the Second Solution by Symmetry | 利用对称性求第二个解
The sine curve satisfies the identity sin(180° − θ) = sin θ for every angle θ. Applying it to our principal value gives the second solution:
正弦曲线满足恒等式 sin(180° − θ) = sin θ，对任意角 θ 都成立。把它用于主值即可得到第二个解：
θ₂ = 180° − 30° = 150°
1/2, so 150° genuinely satisfies the original equation. Verification is not optional in a "show that" or "hence solve" question, and it also catches arithmetic slips.
我们可以立即验证：sin 150° = sin(180° − 30°) = sin 30° = 1/2，所以 150° 确实满足原方程。在"证明"或"由此求解"类题目中，验证并非可选步骤，它还能帮你发现计算失误。
Both 30° and 150° lie inside 0° ≤ θ sine function repeats every 360°, adding or subtracting 360° from either answer would take us outside the interval, so no further solutions exist here.
30° 和 150° 都落在 0° ≤ θ ≤ 360° 之内，因此都有效。由于正弦函数每 360° 重复一次，对任一答案加减 360° 都会超出该区间，所以本题不存在其他解。
6. The Unit Circle Check | 单位圆检验
A second, more visual way to confirm the solution set is to use the unit circle. For a point on the unit circle at angle θ measured anticlockwise from the positive x-axis, the y-coordinate equals sin θ.
另一种更直观的验证方法是使用单位圆。对于单位圆上从正 x 轴逆时针量得角度 θ 的点，其 y 坐标就等于 sin θ。
Solving sin θ = 1/2 therefore means drawing the horizontal line y d asking where it cuts the circle during one full turn. A horizontal line strictly between y = −1 and y ircle at exactly two points, giving two angles: 30° in quadrant I and 150° in quadrant II.
因此，解 sin θ = 1/2 就等于画出水平直线 y = 1/2，并问它在一整圈中与圆相交于何处。一条严格位于 y = −1 与 y = 1 之间的水平直线恰好与圆交于两点，得到两个角：第 I 象限的 30° 和第 II 象限的 150°。
This picture explains why the two solutions always add up to 180° when the sine value is positive: they are mirror images in the vertical axis. If the sine value were negative, the same logic would place the two solutions in quadrants III and IV, adding up to 360° instead.
这幅图解释了为什么当正弦值为正时，两个解之和总是 180°：它们关于竖直轴互为镜像。如果正弦值为负，同样的逻辑会把两个解放到第 III 和第 IV 象限，此时两者之和为 360°。
7. Writing the Full Solution Set | 写出完整解集
Collecting the results, the complete solution set in the interval 0° ≤ θ ≤ 360° is:
综合以上结果，在区间 0° ≤ θ ≤ 360° 内的完整解集为：
θ = 30°, 150°
If the question had asked for the general solution instead of solutions in an interval, we would attach the period to each family of answers:
如果题目要求的是通解而非区间内的解，我们就需要给每一族答案加上周期：
θ = 30° + 360°n or θ = 150° + 360°n, n ∈ ℤ
The symbol ℤ reminds the examiner that n is any integer, positive, negative or zero. Writing "n is an integer" in words is equally acceptable and sometimes safer if you are unsure of the symbol.
符号 ℤ 提醒阅卷人 n 可以是任意整数，包括正整数、负整数和零。用文字写出"n 为整数"同样可以接受，在不确定符号时甚至更稳妥。
Notice how the two families never overlap: 30° + 360°n and 150° + 360°n differ by 120°, which is not a multiple of 360°, so no angle is counted twice. This is exactly the kind of neatness examiners expect in a fully correct answer.
注意这两族解永不会重叠：30° + 360°n 与 150° + 360°n 相差 120°，而 120° 不是 360° 的整数倍，所以没有角被重复计算。这正是阅卷人期望在完整正确答案中看到的严谨性。
8. Common Errors and Exam Tips | 常见错误与考试技巧
The single most frequent mistake is to stop after the principal value and answer 30° only. In a two-mark question that earns one mark; in a three-mark question it typically earns one. Always ask yourself how many solutions the interval and the sign pattern imply before you write your final line.
最常见的错误是只求出主值便写下答案 30°。在 2 分题中这只能拿 1 分，在 3 分题中通常也只能拿 1 分。在写下最终答案前，先问问自己：区间和正负号规律暗示应该有几个解。
A second common error is to use the wrong symmetry rule. The rule 360° − θ belongs to cosine, not sine. For sine with a positive value use 180° − θ; for sine with a negative value find the negative principal value first and then add 360°.
第二个常见错误是用错对称规则。规则 360° − θ 属于余弦，不属于正弦。正弦值为正时用 180° − θ；正弦值为负时先求出负的主值，再加上 360°。
- Check that your calculator is in degree mode unless the question uses radians.
- Sketch the curve or the unit circle before writing the final answer.
- Test each answer in the original equation, especially if it came from a rearrangement.
- Check the interval: a solution equal to 360° may be rejected if the inequality is strict.
- 除非题目使用弧度，否则确认计算器处于角度模式。
- 在写出最终答案前，先画出曲线草图或单位圆。
- 把每个答案代回原方程检验，尤其是经过变形得到的答案。
- 检查区间：若不等号为严格不等号，等于 360° 的解可能被舍去。
Finally, remember that a small, accurate sketch is worth more than a long paragraph of explanation. Examiners award method marks for a correctly marked quadrant or a correctly drawn horizontal line, so make your reasoning visible on the page.
最后请记住，一幅准确的小草图比一大段文字解释更有价值。阅卷人会为正确标出的象限或正确画出的水平线给出方法分，所以要让你的推理过程在卷面上清晰可见。
9. Generalising: Solving sin θ = k | 推广：解方程 sin θ = k
Once you understand Example 5.3.1, you can solve any equation of the form sin θ = k for 0° ≤ θ ≤ 360° using a fixed routine:
掌握了例题 5.3.1 之后，你就可以用一套固定流程求解任意形如 sin θ = k、0° ≤ θ ≤ 360° 的方程：
θ₁ = sin⁻¹k θ₂ = 180° − θ₁
The formula for θ₁ is used exactly as the calculator returns it, which may be negative. When that happens, θ₂ is also negative, and you must add 360° to both values to bring them into the required interval.
θ₁ 的公式按计算器返回的结果直接使用，它可能是负数。此时 θ₂ 也是负数，你需要给两者都加上 360°，把它们移入所要求的区间。
| Equation | θ₁ | θ₂ | Solutions in 0° to 360° |
|---|---|---|---|
| sin θ = 1/2 | 30° | 150° | 30°, 150° |
| sin θ = 0.8 | 53.1° | 126.9° | 53.1°, 126.9° |
| sin θ = −1/2 | −30° | −150° | 210°, 330° |
| sin θ = 1 | 90° | 90° | 90° only |
| sin θ = 0 | 0° | 180° | 0°, 180°, 360° |
The last two rows show why you must never apply a formula blindly: when k = 1 or k = −1 the two solutions coincide, and when k the value 360° becomes an extra solution because it lies on the boundary of the interval.
最后两行说明为什么绝不能盲目套用公式：当 k = 1 或 k = −1 时两个解重合；当 k = 0 时，由于 360° 位于区间边界上，它会成为额外的第三个解。
10. Extension: Cosine and Tangent Equations | 拓展：余弦与正切方程
The same three-step method - principal value, quadrant analysis, symmetry - handles cosine and tangent, but the symmetry rules differ because the underlying graphs differ.
同样的三步法（求主值、判断象限、利用对称性）也适用于余弦和正切，但由于对应图像不同，对称规则也不同。
cos θ = k: θ₁ = cos⁻¹k, θ₂ = 360° − θ₁
tan θ = k: θ₁ = tan⁻¹k, θ₂ = θ₁ + 180°
n 0° and add up to 360°. Tangent has period 180° rather than 360°, so in the interval 0° ≤ θ ≤ 360° a tangent equation normally has exactly two solutions separated by 180°.
余弦关于水平轴对称，因此它在一个完整圆周内的两个解关于 0° 反射，二者之和为 360°。正切的周期是 180° 而非 360°，所以在区间 0° ≤ θ ≤ 360° 内，正切方程通常恰好有两个相差 180° 的解。
- cos θ = 1/2 gives θ = 60° and θ = 300°.
- cos θ = −1/2 gives θ = 120° and θ = 240°.
- tan θ = 1 gives θ = 45° and θ = 225°.
- tan θ = −1 gives θ = 135° and θ = 315°.
- cos θ = 1/2 给出 θ = 60° 和 θ = 300°。
- cos θ = −1/2 给出 θ = 120° 和 θ = 240°。
- tan θ = 1 给出 θ = 45° 和 θ = 225°。
- tan θ = −1 给出 θ = 135° 和 θ = 315°。
If the question is set in radians, replace 180° by π, 360° by 2π and 90° by π/2. For example, sin θ = 1/2 for 0 ≤ θ ≤ 2π has the solutions θ = π/6 and θ
