📚 Worked Example 4.5.3: Solving Trigonometric Equations in a Given Interval | 示例 4.5.3：在给定区间内解三角方程
This AQA A-Level Mathematics worked example demonstrates how to solve a quadratic trigonometric equation involving sin x. The technique requires recognising a hidden quadratic structure, factorising, rejecting impossible values, and using quadrant symmetry to find every solution in the stated interval. It is a core skill in the Pure Mathematics component of both AS and A-Level AQA courses.
这个 AQA A-Level 数学例题演示如何求解一个含有 sin x 的二次三角方程。解题技巧包括识别隐藏的二次结构、因式分解、舍去不可能的值，以及利用象限对称性求出指定区间内的所有解。这是 AQA AS 和 A-Level 纯数学部分的核心技能。
1. Why Quadratic Trigonometric Equations Matter | 为什么二次三角方程重要
Quadratic trigonometric equations appear regularly in AQA A-Level Mathematics papers because they test two skill sets at the same time: algebraic manipulation and trigonometric reasoning. A student must be fluent in factorising or using the quadratic formula, and must also understand the sine function, its range, and its symmetry.
二次三角方程经常出现在 AQA A-Level 数学试卷中，因为它们同时考查代数运算和三角知识。学生既要熟练因式分解或使用求根公式，又要理解正弦函数、它的取值范围以及它的对称性。
les in the given interval have been found. A correct algebraic answer such as sin x = 1/2 is therefore only a halfway result.
这类方程不只是贴上三角标签的代数题。变量 x 是一个角，只有当给定区间内所有角都被找到时，方程才算解完。因此，像 sin x = 1/2 这样的代数中间结果只是完成了一半。
Because the equation is quadratic in form, it can be solved by substitution. Because it is trigonometric, the final answer must be expressed as angles, usually in degrees or radians depending on the interval stated in the question.
由于方程在形式上呈二次，可以通过换元求解。又由于它是三角方程，最终答案必须以角的形式表示，通常根据题目给出的区间使用角度或弧度。
2. The Problem Statement and the Given Interval | 题目陈述与给定区间
tion is not solved by simply rearranging it, because it contains both sin²x and sin x. It must first be treated as a quadratic in sin x.
在本例中，要求解的方程是 2sin²x + 3sin x − 2 = 0。给定区间为 0° ≤ x ≤ 360°。这个方程不能仅靠移项求解，因为它同时含有 sin²x 和 sin x。必须先把它看作关于 sin x 的二次方程。
2sin²x + 3sin x − 2 = 0, 0° ≤ x ≤ 360°
lving method is identical, but the final answers would be π/6 and 5π/6 instead of 30° and 150°. The interval must always be read carefully because it controls how many solutions are valid.
如果区间用弧度给出，可能写成 0 ≤ x < 2π。解法完全相同，但最终答案会是 π/6 和 5π/6，而不是 30° 和 150°。必须仔细阅读区间，因为它决定有多少个解是有效的。
Examiners expect the interval to be copied or clearly referenced in the solution. A final answer without an interval check can lose marks, even if the algebraic work is correct.
阅卷人希望考生在解答中抄写或明确引用区间。如果最终答案没有检查区间，即使代数过程正确，也可能失分。
3. Recognising the Hidden Quadratic Structure | 识别隐藏的二次结构
powers of sin x are 2, 1 and 0, which matches the pattern of a quadratic equation in the variable sin x.
方程 2sin²x + 3sin x − 2 = 0 与 2u² + 3u − 2 = 0 结构相同。sin x 的幂次分别为 2、1 和 0，正好符合关于变量 sin x 的二次方程模式。
den quadratic is recognised by asking: can this be written as a(sin x)² + b(sin x) + c = 0? a = 2, b = 3 and c = −2. Once this is seen, the trigonometry temporarily becomes algebra.
识别隐藏二次结构的方法是问：这是否可以写成 a(sin x)² + b(sin x) + c = 0？这里 a = 2，b = 3，c = −2。一旦看出这一点，三角问题就暂时转化为代数问题。
Students should not confuse this with a trigonometric identity. It is an equation to be solved for x, not an identity that is true for all x. The goal is to find specific angles in the stated interval.
学生不应把它与三角恒等式混淆。这是一个要求解 x 的方程，不是对所有 x 都成立的恒等式。目标是求出给定区间内的特定角。
4. Substitution: Turning Trigonometry into Algebra | 换元：把三角问题化为代数问题
thematically compulsory, but it makes the quadratic structure visible and is a presentation feature that examiners look for.
令 u = sin x。方程变为 2u² + 3u − 2 = 0。这个换元在数学上并非必须，但它使二次结构清晰可见，也是阅卷人希望看到的书写方式。
≤ 1. Therefore any solution for u outside the interval −1 ≤ u ≤ 1 must be rejected later.
由于 u = sin x，u 的可能值受到正弦函数值域的限制。对所有实数 x，都有 −1 ≤ sin x ≤ 1。因此，任何落在区间 −1 ≤ u ≤ 1 之外的 u 解，之后都必须舍去。
e final answer must be in terms of x, not u. Writing the back-substitution explicitly avoids confusion and secures method marks.
解出关于 u 的二次方程后，必须用 u = sin x 回代。最终答案必须用 x 表示，而不是 u。明确写出回代过程可以避免混淆，并获得方法分。
5. Factorising the Quadratic | 因式分解
ackets gives 2u² + 4u − u − 2 = 2u² + 3u − 2, which confirms the factorisation is correct.
二次方程 2u² + 3u − 2 = 0 可因式分解为 (2u − 1)(u + 2) = 0。展开括号得到 2u² + 4u − u − 2 = 2u² + 3u − 2，确认因式分解正确。
(2u − 1)(u + 2) = 0
r u + 2 = 0. Hence u = 1/2 or u = −2. Alternatively, the quadratic formula gives u = (−3 ± √9 + 16)/4 = (−3 ± 5)/4, which also produces u = 1/2 and u = −2.
令每个因式等于零，得到 2u − 1 = 0 或 u + 2 = 0。因此 u = 1/2 或 u = −2。或者，用求根公式得到 u = (−3 ± √9 + 16)/4 = (−3 ± 5)/4，同样得到 u = 1/2 和 u = −2。
Either factorisation or the quadratic formula is acceptable in an AQA examination, provided the working is clear. Factorisation is usually faster, but the formula is a reliable fallback when factors are not obvious.
在 AQA 考试中，因式分解或求根公式都可以接受，只要过程清晰。因式分解通常更快，但当因式不明显时，求根公式是可靠的备选方法。
6. Rejecting Impossible Values of sin x | 舍去 sin x 的不可能值
x = −2, is impossible because the sine function only takes values from −1 to 1 inclusive.
由于 u = sin x，两个代数解变为 sin x = 1/2 和 sin x = −2。第二个方程 sin x = −2 不可能成立，因为正弦函数的值只在 −1 到 1 之间，包括端点。
on because −1 ≤ sin x ≤ 1. Simply ignoring the −2 without comment may not earn full marks.
阅卷人希望看到明确舍去无效解。学生应该写出类似：sin x = −2 无解，因为 −1 ≤ sin x ≤ 1。如果只是默默忽略 −2 而不加说明，可能拿不到满分。
his is the trigonometric equation that must now be solved over the interval 0° ≤ x ≤ 360°.
舍去之后，只剩下 sin x = 1/2。这个三角方程就是接下来要在区间 0° ≤ x ≤ 360° 内求解的对象。
7. The Principal Solution and the CAST Diagram | 主解与 CAST 图
e knowledge, sin⁻¹(1/2) = 30°. In radians, this is π/6. This is the acute reference angle.
要求解 sin x = 1/2，先求主值。使用计算器或精确值知识，sin⁻¹(1/2) = 30°。用弧度表示就是 π/6。这是锐角参考角。
Because sin x is positive, the solutions must lie in the first and second quadrants. The CAST diagram reminds us that sine is positive in the first quadrant (A) and in the second quadrant (S).
因为 sin x 为正，所以解必须位于第一和第二象限。CAST 图提醒我们，正弦在第一象限 (A) 和第二象限 (S) 为正。
sin x = 1/2 → reference angle = 30°
The CAST diagram is not a substitute for understanding the unit circle, but it is a compact way to remember which trigonometric ratios are positive in each quadrant. For sine, the positive quadrants are the first and second.
CAST 图不能替代对单位圆的理解，但它是记住每个象限中哪些三角比为正的简洁方法。对于正弦，正的象限是第一和第二象限。
8. Using Quadrant Symmetry to Find All Solutions | 利用象限对称求出所有解
− 30° = 150°. Both angles satisfy sin x = 1/2 and both lie in the interval 0° ≤ x ≤ 360°.
在第一象限，解为 x = 30°。在第二象限，利用对称性得到 x = 180° − 30° = 150°。这两个角都满足 sin x = 1/2，且都位于区间 0° ≤ x ≤ 360° 内。
x = 30°, x = 150°
= 390°, which is outside the interval. The angle 360° itself gives sin 360° = 0, not 1/2.
在这个区间内没有其他解，因为正弦函数每 360° 重复一次。下一个正解会是 30° + 360° = 390°，超出了区间。而 360° 本身的正弦值为 0，不是 1/2。
3/2 − 2 = 1/2 + 3/2 − 2 = 0. For x = 150°, sin 150° = 1/2, so the same calculation holds.
把每个解代回原方程检验是良好的习惯。对于 x = 30°：2(1/2)² + 3(1/2) − 2 = 2(1/4) + 3/2 − 2 = 1/2 + 3/2 − 2 = 0。对于 x = 150°，sin 150° = 1/2，所以同样的计算成立。
9. Presenting the Full Solution Set | 写出完整解集
. In radians, over the interval 0 ≤ x < 2π, the same solution set is x = π/6 and x = 5π/6.
用角度表示的完整解集是 x = 30° 和 x = 150°。用弧度表示，在区间 0 ≤ x < 2π 内，同样的解集是 x = π/6 和 x = 5π/6。
quadrant reasoning. The final answer should be clearly separated and given in the units required by the question.
一份书写良好的解答应展示换元 u = sin x、因式分解后的二次方程、舍去 sin x = −2、主值，以及象限推理。最终答案应清楚分开，并按照题目要求使用单位。
| Degrees | Radians | Quadrant |
| 30° | π/6 | First |
| 150° | 5π/6 | Second |
0° and x = 150° still, but if negative angles were included, x = −330° would be equivalent to 30° and would not be added unless the interval demanded it.
如果区间不同，例如 −180° ≤ x ≤ 180°，解集会改变。对于 sin x = 1/2，在该区间内解仍然是 x = 30° 和 x = 150°，但如果包含负角，x = −330° 与 30° 等价，除非区间要求，否则不会额外加入。
10. Radians Version and Calculator Work | 弧度制版本与计算器操作
. The principal value is sin⁻¹(1/2) = π/6. The second quadrant solution is π − π/6 = 5π/6.
如果区间给出为 0 ≤ x < 2π，方法相同，但最终答案必须用弧度。主值是 sin⁻¹(1/2) = π/6。第二象限的解是 π − π/6 = 5π/6。
x = π/6, x = 5π/6
Calculator work must be done in the correct mode. If the calculator is in degree mode, sin⁻¹(1/2) returns 30
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