📚 Example 2.8.1: Solving Quadratic Trigonometric Equations | 例题 2.8.1：解二次三角方程
In AQA A-Level Mathematics, trigonometric equations often appear in forms that require algebraic manipulation before standard angle solutions can be applied. Worked Example 2.8.1 demonstrates a quadratic trigonometric equation, where substituting a trigonometric function with a variable turns the problem into a familiar quadratic factorisation task.
在 AQA A-Level 数学中，三角方程经常以需要先进行代数处理的形式出现，然后才能应用标准的角度解法。例题 2.8.1 展示了一个二次三角方程，其中将三角函数替换为变量，使问题转化为熟悉的二次因式分解任务。
1. Problem Statement | 题目陈述
s in degrees. This is a classic AQA-style question because it combines trigonometric functions with quadratic structure.
在 0° ≤ θ ≤ 360° 的范围内解方程 2 sin²θ − 3 sinθ + 1 = 0，并给出所有以度为单位的解。这是一道经典的 AQA 风格题目，因为它将三角函数与二次结构结合在一起。
2. Recognising the Quadratic Form | 识别二次形式
s substitution is a powerful strategy because it allows us to use standard algebraic methods.
虽然方程包含 sinθ，但 sin²θ 和 sinθ 的幂次提示了二次模式。如果令 x = sinθ，方程变为 2x² − 3x + 1 = 0。这种代换是一种有效的策略，因为它使我们能够使用标准的代数方法。
It is important to state the substitution clearly in your working, as AQA examiners look for evidence that you have identified the quadratic structure rather than guessing solutions.
在解题过程中清晰地写出代换非常重要，因为 AQA 考官会寻找你已识别出二次结构的证据，而不是猜测答案。
3. Substitution and Factorisation | 代换与因式分解
= 1. The middle term split is: 2x² − 2x − x + 1 = 2x(x − 1) − 1(x − 1) = (2x − 1)(x − 1).
使用 x = sinθ，我们对 2x² − 3x + 1 进行因式分解。因式分解后的形式为 (2x − 1)(x − 1) = 0。令每个括号为零，得到 x = 1/2 或 x = 1。中间项拆分如下：2x² − 2x − x + 1 = 2x(x − 1) − 1(x − 1) = (2x − 1)(x − 1)。
2x² − 3x + 1 = (2x − 1)(x − 1)
4. Returning to Trigonometric Values | 还原为三角函数值
We must now solve each equation separately within the given interval. At this stage, it is essential to remember that sinθ has two principal solutions in one full turn, except at its maximum and minimum.
由于 x = sinθ，这两个代数解变成了两个更简单的三角方程：sinθ = 1/2 和 sinθ = 1。现在我们必须分别在给定区间内求解每个方程。在这个阶段，必须记住 sinθ 在一个完整周期内通常有两个基本解，除非处于最大值或最小值。
The substitution step is reversible because sine is a well-defined function over the interval, so every solution for x gives a valid equation for θ.
代换步骤是可逆的，因为 sin 在该区间上是一个定义明确的函数，因此每个 x 的解都会给出一个有效的 θ 方程。
5. Solving sinθ = 1/2 | 解 sinθ = 1/2
erefore θ = 30° and θ = 180° − 30° = 150°. These are the only two solutions for this part.
方程 sinθ = 1/2 具有与 30° 参考角相关的精确解。在 0° ≤ θ ≤ 360° 的区间内，sin 在第一象限和第二象限为正。因此 θ = 30° 和 θ = 180° − 30° = 150°。这是该部分的仅有两个解。
sinθ = 1/2 → θ = 30° or θ = 150°
aph of y = sinθ can help visualise why the second solution appears in the second quadrant.
使用 CAST 图或 y = sinθ 的图像可以帮助直观理解为什么第二个解出现在第二象限。
6. Solving sinθ = 1 | 解 sinθ = 1
° ≤ θ ≤ 360°, as the next occurrence would be at 450°, which lies outside the given range.
方程 sinθ = 1 在一个完整周期内只有一个解，因为 sin 仅在一个点达到最大值 1，即 θ = 90°。在 0° ≤ θ ≤ 360° 的区间内没有第二个解，因为下一次出现将在 450°，这超出了给定范围。
2: maximum and minimum values occur only once per cycle, whereas intermediate values occur twice.
这是与 sinθ = 1/2 的一个关键区别：最大值和最小值每个周期只出现一次，而中间值会出现两次。
7. Full Solution Set | 完整解集
150°. It is useful to present these in ascending order and to state clearly that they satisfy the original quadratic equation. In radian measure, these correspond to π/6, π/2, and 5π/6.
将两个较简单方程的所有解合并，得到 θ = 30°、90°、150°。按升序列出这些解并明确指出它们满足原二次方程是有用的。用弧度表示，它们分别对应 π/6、π/2 和 5π/6。
| Trigonometric equation | Solutions in 0° ≤ θ ≤ 360° |
|---|---|
| sinθ = 1/2 | 30°, 150° |
| sinθ = 1 | 90° |
8. Verification by Substitution | 代入验证
² − 3(1) + 1 = 0. For θ = 150°, the same calculation as 30° applies because sin150° = 1/2.
将每个解代入原方程可以确认其正确性。对于 θ = 30°，2(1/2)² − 3(1/2) + 1 = 1/2 − 3/2 + 1 = 0。对于 θ = 90°，2(1)² − 3(1) + 1 = 0。对于 θ = 150°，由于 sin150° = 1/2，计算结果与 30° 相同。
Verification is not always required in AQA mark schemes, but it is a good habit that can catch arithmetic errors during practice.
验证并不总是 AQA 评分方案的要求，但它是一个好习惯，可以在练习中帮助发现算术错误。
9. Common Mistakes | 常见错误
ake is forgetting that the interval is in degrees and mixing degree and radian answers. Always check the range and whether exact values are required.
一个常见错误是在解题过程中将两边同时除以 sinθ，这会丢失 sinθ = 1 这个解。另一个错误是忘记区间以度为单位，并将角度制和弧度制答案混用。务必检查范围以及是否需要精确值。
- Do not cancel sinθ unless you are certain it is not zero — 除非确定 sinθ 不为零，否则不要约去 sinθ
- Sketch or recall the CAST diagram to find all quadrant solutions — 绘制或回忆 CAST 图以找到所有象限解
- State the final answer in the requested unit — 按照要求的单位写出最终答案
10. AQA Exam Tips | AQA 考试提示
n make your method clear to the examiner. Always give final answers in the form requested.
在 AQA 考试中，代换步骤、因式分解以及列出所需区间内的所有解都能获得分数。写出中间方程 2x² − 3x + 1 = 0 并注明 x = sinθ，可以使你的方法对考官更清晰。始终按照题目要求的形式给出最终答案。
If the question uses radians, convert your reference angles at the end rather than mixing systems throughout the working.
如果题目使用弧度制，在最后转换你的参考角，而不是在整个解题过程中混用两种单位制。
11. Extension to Radian Measure | 弧度制扩展
ns θ = π/6, π/2, and 5π/6. Understanding both systems is essential for AQA A-Level papers.
如果区间给定为 0 ≤ θ ≤ 2π，则应用相同方法，但解用弧度表示。方程 2sin²θ − 3sinθ + 1 = 0 的解将是 θ = π/6、π/2 和 5π/6。理解两种单位制对于 AQA A-Level 试卷至关重要。
30° = π/6, 90° = π/2, 150° = 5π/6
12. Summary | 总结
ns. The final solution set is θ = 30°, 90°, 150° in degrees, or π/6, π/2, 5π/6 in radians.
例题 2.8.1 展示了如何将
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