📚 Solving Quadratic Trigonometric Equations Example 6.11.2 | 解二次三角方程例 6.11.2
In AQA A-Level Mathematics, Example 6.11.2 is a standard problem that asks you to solve a trigonometric equation which can be rewritten as a quadratic in sin θ or cos θ. This type of question appears regularly in Paper 1 and Paper 2, and it tests your ability to combine algebraic factorising with trigonometric angle finding.
在 AQA A-Level 数学中，例 6.11.2 是一道标准题，要求解一个可以改写成关于 sin θ 或 cos θ 的二次型三角方程。这类题目经常出现在 Paper 1 和 Paper 2 中，考查你是否能把代数因式分解与三角求角结合起来。
1. What This Example Covers | 本例涵盖内容
u need to use both algebraic techniques and the CAST diagram or graph knowledge for trigonometric functions.
本例涵盖求解形如 a sin² θ + b sin θ + c = 0 的方程，并且通常在 AQA AS 和 A-Level 单元中给出区间 0° ≤ θ ≤ 360°。你需要同时使用代数技巧以及 CAST 图或三角函数图像知识。
It is important to read the interval carefully. Some questions use radians, but Example 6.11.2 in many AQA-style textbooks uses degrees to build confidence before moving on to radian measure.
仔细阅读给定区间非常重要。有些题目使用弧度制，但许多 AQA 风格教材中的例 6.11.2 使用角度制，先建立信心，再过渡到弧度制。
2. The Given Equation | 题目给出的方程
For this worked example, the equation is:
本例给出的方程是：
2 sin² θ − 3 sin θ + 1 = 0, 0° ≤ θ ≤ 360°
This equation is not linear in sin θ, because sin θ appears squared. However, it has the structure of a quadratic expression if we treat sin θ as the variable.
这个方程并不是关于 sin θ 的线性方程，因为 sin θ 出现了平方。不过，如果我们把 sin θ 看作变量，它就具有二次表达式的结构。
3. Recognising the Quadratic Structure | 识别二次结构
uld notice that 2 sin² θ − 3 sin θ + 1 has the same form as 2x² − 3x + 1, where x = sin θ.
本题首先考查的是模式识别能力。你应当注意到 2 sin² θ − 3 sin θ + 1 与 2x² − 3x + 1 具有相同形式，其中 x = sin θ。
In AQA mark schemes, identifying this hidden quadratic is often awarded the first method mark. If you do not spot it, you may try to rearrange incorrectly or use identities that are not needed.
在 AQA 评分标准中，识别出这个隐藏的二次型通常能得到第一个方法分。如果没有发现这一点，你可能会错误地移项或使用不必要的恒等式。
4. Using Substitution to Simplify | 用换元简化
y to proceed is to let y = sin θ. The original equation then becomes a familiar quadratic:
一种稳妥的方法是设 y = sin θ。原方程就变为一个熟悉的二次方程：
2y² − 3y + 1 = 0
This substitution step is not always written in your final answer, but it is very useful for avoiding mistakes when factorising or solving.
这个换元步骤并不一定要写在最终答案中，但它在因式分解或求解时非常有用，可以避免出错。
5. Factorising the Quadratic | 因式分解二次式
ultiply to 2 × 1 = 2 and add to −3. Those numbers are −2 and −1, so split the middle term:
现在对 2y² − 3y + 1 进行因式分解。寻找两个数，乘积为 2 × 1 = 2，且相加为 −3。这两个数是 −2 和 −1，因此拆分中间项：
2y² − 2y − y + 1 = 0
Factor by grouping:
分组分解：
2y(y − 1) − 1(y − 1) = 0
This gives:
得到：
(2y − 1)(y − 1) = 0
Alternatively, you can factorise directly if you are confident. Both methods are acceptable in AQA exams.
如果你有把握，也可以直接因式分解。在 AQA 考试中两种方法都可以接受。
6. Solving the Two Basic Trig Equations | 解两个基本三角方程
Since the product is zero, set each factor equal to zero:
由于乘积为零，令每个因式等于零：
2y − 1 = 0 or y − 1 = 0
Solving these gives:
解这两个方程得到：
y = 1/2 or y = 1
Now replace y with sin θ:
现在将 y 替换回 sin θ：
sin θ = 1/2 or sin θ = 1
You must now solve each simple trigonometric equation within the interval 0° ≤ θ ≤ 360°.
接下来你必须在区间 0° ≤ θ ≤ 360° 内分别求解每个简单的三角方程。
7. Using the CAST Diagram and Interval | 使用 CAST 图与区间
ve in the first and second quadrants. The principal value is θ = 30°, since sin 30° = 1/2.
对于 sin θ = 1/2，你知道 sin θ 为正。正弦在第一象限和第二象限为正。主值是 θ = 30°，因为 sin 30° = 1/2。
To find the second-quadrant solution, use the symmetry property of sine:
要求第二象限的解，可以利用正弦的对称性：
θ = 180° − 30° = 150°
90°. This is because sin 90° = 1, and no other angle in the interval has sine equal to 1.
对于 sin θ = 1，在 0° ≤ θ ≤ 360° 内唯一解是 θ = 90°。这是因为 sin 90° = 1，并且该区间内没有其他角的正弦值等于 1。
8. Listing All Solutions in 0° to 360° | 列出 0° 到 360° 的所有解
Combine the three angles found:
把找到的三个角合并起来：
θ = 30°, 90°, 150°
It is good practice to state your final answer in increasing order. In AQA mark schemes, the final mark is often for all three solutions, with no extra values in the interval.
建议按从小到大顺序写出最终答案。在 AQA 评分标准中，最后一个分数通常要求三个解都要出现，并且区间内不能有多余的值。
9. Checking by Substitution | 代入检验
nfident, substitute each value back into the original equation 2 sin² θ − 3 sin θ + 1 = 0.
为了更有把握，可以把每个值代回原方程 2 sin² θ − 3 sin θ + 1 = 0 进行检验。
90° = 1, so 2(1)² − 3(1) + 1 = 0. For θ = 150°: sin 150° = 1/2, so the result is again 0.
当 θ = 30° 时：sin 30° = 1/2，所以 2(1/2)² − 3(1/2) + 1 = 0。当 θ = 90° 时：sin 90° = 1，所以 2(1)² − 3(1) + 1 = 0。当 θ = 150° 时：sin 150° = 1/2，结果仍然为 0。
This check is not always required, but it can help you catch sign errors or missing solutions before moving on.
这种检验并不总是必须的，但它能帮助你在继续之前发现符号错误或漏解。
10. AQA Mark Scheme Focus | AQA 评分标准要点
In a typical AQA mark scheme for a question like Example 6.11.2, the marks are awarded as follows:
在典型的 AQA 评分标准中，像例 6.11.2 这类题目的分数通常按以下方式分配：
| Mark | What it is for | 分数 | 给分原因 |
|---|---|---|---|
| M1 | Recognising or creating the quadratic in sin θ | M1 | 识别或构造关于 sin θ 的二次式 |
| M1 | Factorising correctly | M1 | 正确因式分解 |
| A1 | Obtaining sin θ = 1/2 and sin θ = 1 | A1 | 得到 sin θ = 1/2 和 sin θ = 1 |
| A1 | All three angles in the interval | A1 | 区间内三个角全部正确 |
Although mark schemes vary, the same principles apply: show clear substitution, factorisation, and angle finding.
尽管评分标准可能略有不同，但原则相同：要展示清晰的换元、因式分解和求角过程。
11. Common Mistakes to Avoid | 常见错误规避
solution for sin θ = 1/2. You must check the number of solutions expected in the interval.
许多学生在这类题目上丢分，是因为求出 sin θ = 1/2 的一个解后就停止了。你必须检查区间内预期的解的数量。
210° = −1/2, not +1/2. Sine is positive in the first and second quadrants, not the third.
另一个常见错误是把 sin θ = 1/2 的解写成 θ = 30° 和 θ = 210°。这是错误的，因为 sin 210° = −1/2，而不是 +1/2。正弦在第一和第二象限为正，第三象限为负。
Also, avoid dividing both sides of the original equation by sin θ. Dividing by a variable expression can remove valid solutions and is generally not allowed unless you know it cannot be zero.
此外，不要在原方程两边同时除以 sin θ。除以含变量的表达式可能会消去有效解，除非你知道它不可能为零，否则通常不允许这样做。
12. Practice and Extension | 练习与拓展
Try this similar question to consolidate your understanding:
尝试下面的同类问题来巩固理解：
Solve 3 cos² θ − 5 cos θ + 2 = 0 for 0° ≤ θ ≤ 360°.
actorise, solve the two basic cosine equations, and then use the symmetry of the cosine graph or CAST diagram to find all solutions.
解题过程相同：设 y = cos θ，因式分解，解两个基本余弦方程，然后利用余弦图像的对称性或 CAST 图求出所有解。
= 3. Here you need the identity sin² θ + cos² θ = 1 before forming the quadratic in cos θ.
作为拓展，你还可以考虑诸如 2 sin² θ + 3 cos θ = 3 这样的方程。这里需要先使用恒等式 sin² θ + cos² θ = 1，再构造关于 cos θ 的二次方程。
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