📚 Example 6.11.4: Solving Trigonometric Equations with Quadratic Forms | 示例6.11.4：解二次型三角方程
ability to recognise the quadratic form, use a substitution, and then find all solutions within the given interval.
在 AQA A Level 数学中，三角方程经常通过 sin²x = 1 - cos²x 这样的恒等式隐藏着一个二次结构。示例 6.11.4 要求我们在 0 ≤ x ≤ 2π 范围内解方程 2 sin²x + 3 cos x = 0。这类题目考查你识别二次形式、使用代换以及在给定区间内找出所有解的能力。
1. Understanding the Problem | 理解问题
The given equation involves both sin²x and cos x. Since the equation contains two different trigonometric functions, we need to rewrite it in terms of a single function. The presence of sin²x suggests using the Pythagorean identity to convert everything into cos x, which will reveal a quadratic equation.
给定方程同时包含 sin²x 和 cos x。由于方程中含有两个不同的三角函数，我们需要将其改写为单一函数的形式。sin²x 的出现提示我们使用勾股恒等式把一切都转化为 cos x，从而揭示出一个二次方程。
2 sin²x + 3 cos x = 0, with 0 ≤ x ≤ 2π
This step is crucial because solving a mixed trigonometric equation directly is usually more difficult than solving a polynomial equation in one variable.
这一步至关重要，因为直接求解混合三角方程通常比求解单变量多项式方程更困难。
2. Key Trigonometric Identities | 关键三角恒等式
The core identity used in this example is the Pythagorean identity. It connects sin²x and cos²x, allowing us to eliminate one of them. For all real x, we have:
本例中使用的核心恒等式是勾股恒等式。它把 sin²x 和 cos²x 联系起来，使我们能够消去其中一个。对于所有实数 x，我们有：
sin²x + cos²x = 1
From this identity, we can express sin²x in terms of cos²x. This is the key algebraic step that will convert the original equation into a quadratic in cos x.
从这个恒等式中，我们可以用 cos²x 来表示 sin²x。这是关键的代数步骤，将把原方程转化为关于 cos x 的二次方程。
sin²x = 1 - cos²x
You should memorise this identity and be able to apply it in both directions without hesitation. It is one of the most frequently used tools in A Level trigonometry.
你应该熟记这个恒等式，并且能够毫不犹豫地双向运用它。它是 A Level 三角学中最常用的工具之一。
3. Converting to a Single Trigonometric Ratio | 转化为单一三角比
e original equation. This removes sin x completely from the problem and gives an equation that only contains cos x.
我们将 sin²x = 1 - cos²x 代入原方程。这样就完全从问题中去掉了 sin x，得到一个只含有 cos x 的方程。
2(1 - cos²x) + 3 cos x = 0
Expanding the bracket gives a simpler expression. The constant term 2 appears from the first product, and the term -2 cos²x is the quadratic part.
展开括号会得到一个更简单的表达式。常数项 2 来自第一个乘积，-2 cos²x 项是二次部分。
2 - 2 cos²x + 3 cos x = 0
We usually prefer to multiply through by -1 so that the coefficient of cos²x is positive. This makes the quadratic easier to factorise.
我们通常更喜欢将方程两边同时乘以 -1，使 cos²x 的系数为正。这样二次方程会更容易因式分解。
2 cos²x - 3 cos x - 2 = 0
Now the equation is clearly a quadratic in cos x. The next step is to solve it just like any other quadratic equation.
现在方程明显是一个关于 cos x 的二次方程。下一步就是像解任何其他二次方程一样求解它。
4. Introducing a Substitution | 引入代换
ion is temporary but very helpful, especially when you are first learning this method. Then the equation becomes:
为了更清楚地显示二次结构，设 y = cos x。这个代换是暂时的，但非常有帮助，尤其是当你刚学习这种方法时。于是方程变为：
2y² - 3y - 2 = 0
This is a standard quadratic equation in y. We can solve it by factorising, completing the square, or using the quadratic formula. Factorising is often the fastest if the factors are simple.
这是一个关于 y 的标准二次方程。我们可以通过因式分解、配方法或使用二次公式来求解。如果因式简单，因式分解通常是最快的。
- Factorise the quadratic: 2y² - 3y - 2 = (2y + 1)(y - 2)
- Set each factor equal to zero to find possible values of y.
将二次式因式分解：2y² - 3y - 2 = (2y + 1)(y - 2)；令每个因子等于零，求出 y 的可能值。
5. Solving the Quadratic Equation | 解二次方程
From the factorised form, we obtain two possible values for y. These correspond to two possible values for cos x, but one of them may not be valid because of the range of the cosine function.
从因式分解后的形式中，我们得到 y 的两个可能值。它们对应于 cos x 的两个可能值，但由于余弦函数的值域限制，其中一个可能无效。
(2y + 1)(y - 2) = 0
Setting each factor equal to zero gives:
令每个因子等于零，得到：
2y + 1 = 0 ⇒ y = -½
y - 2 = 0 ⇒ y = 2
The cosine function always lies between -1 and 1 inclusive, so cos x = 2 has no solution.
由于 y = cos x，我们现在考虑每个值是否可能。余弦函数的值始终在 -1 到 1 之间（含端点），因此 cos x = 2 无解。
cos x = -½ is the only valid equation.
all. Always check that your solutions for cos x, sin x, or tan x fall within their natural ranges.
无效值 y = 2 是一个常见的陷阱。务必检查你解出的 cos x、sin x 或 tan x 是否落在它们的自然值域内。
6. Finding Principal Solutions | 求主解
tells us that the solutions are in the second and third quadrants, where cosine is negative.
现在我们需要在区间 0 ≤ x ≤ 2π 内解 cos x = -½。负号告诉我们解在第二和第三象限，因为余弦在这些象限为负。
es α = π/3, the angle with the same cosine value but negative must be π - π/3 and π + π/3.
我们首先求出参考角。因为 cos α = ½ 给出 α = π/3，所以具有相同余弦绝对值但符号为负的角度必须是 π - π/3 和 π + π/3。
x = π - π/3 = 2π/3
x = π + π/3 = 4π/3
olutions in the given domain. Always check that they lie inside 0 ≤ x ≤ 2π, which they do.
这些就是给定定义域内的主解。务必检查它们是否位于 0 ≤ x ≤ 2π 内，确实如此。
- Solution 1: 2π/3
- Solution 2: 4π/3
Using the unit circle or a CAST diagram can help you visualise why the cosine is negative in these two quadrants.
使用单位圆或 CAST 图可以帮助你直观理解为什么余弦在这两个象限为负。
7. Casting the General Solutions | 求通解
write the general solution is useful. For cos x = -½, the general solution in radians is:
在 AQA 考试中，你经常被要求给出特定区间内的所有解，但能够写出通解也很有用。对于 cos x = -½，以弧度表示的通解为：
x = 2π/3 + 2nπ or x = 4π/3 + 2nπ, n ∈ ℤ
This form uses the periodicity of cosine, which repeats every 2π. Adding multiples of 2π generates all possible solutions beyond the given interval.
这种形式利用了余弦的周期性，即每 2π 重复一次。加上 2π 的整数倍可以生成给定区间之外的所有可能解。
e. Checking the general form is a good way to verify that you have not missed any solutions.
对于特定区间 0 ≤ x ≤ 2π，本例中只有 n = 0 能给出有效答案。检查通解形式是验证你没有遗漏任何解的好方法。
8. Checking the Domain | 检查定义域
≤ x ≤ 2π. We must ensure that every solution we report lies strictly within this interval.
定义域检查在三角方程中至关重要。在本例中，定义域给定为 0 ≤ x ≤ 2π。我们必须确保所报告的每个解都严格落在该区间内。
| Candidate solution | Check | Valid? |
|---|---|---|
| 2π/3 | 0 ≤ 2π/3 ≤ 2π | Yes |
| 4π/3 | 0 ≤ 4π/3 ≤ 2π | Yes |
| -2π/3 | Outside domain | No |
| 8π/3 | Outside domain | No |
This table highlights common extra solutions that appear if you forget the interval or mix radians with degrees. Always match the domain unit to the angle measure used in the question.
这个表格突出显示了如果你忘记区间或混淆弧度与角度时可能出现的常见多余解。务必使定义域的单位与题目中使用的角度度量一致。
9. Common Mistakes to Avoid | 常见错误
Many students lose marks on this type of question not because they cannot solve quadratics, but because they make predictable trigonometric errors. Knowing these pitfalls will help you avoid them in the exam.
许多学生在这类题目上失分，不是因为他们不会解二次方程，而是因为他们犯了可预见的三角学错误。了解这些陷阱将帮助你在考试中避免它们。
- Forgetting to check if the quadratic roots are within the range of cosine or sine.
- Using the wrong quadrant for a negative trigonometric ratio.
- Leaving the equation in terms of two different functions and trying to factorise directly.
- Writing degree symbols when the domain is given in radians.
- Stopping after finding only one solution when two exist in the interval.
忘记检查二次方程的根是否在余弦或正弦的值域内；对负的三角比使用错误的象限；将方程保留为两个不同函数的形式并试图直接因式分解；在定义域以弧度给出时写上度数符号；在区间内存在两个解时只找到一个就停止。
By being systematic and using a CAST diagram or unit circle, you can avoid these errors and present a complete, accurate solution.
通过系统地解题并使用 CAST 图或单位圆，你可以避免这些错误，并给出完整、准确的解答。
10. Exam-Style Tips | 考试技巧
AQA examiners often look for clear method marks. Even if you make a small arithmetic slip later, showing the correct substitution and quadratic factorisation can earn you most of the marks.
AQA 考官通常会寻找清晰的方法分。即使你后来出现小的算术失误，展示正确的代换和二次因式分解也能为你赢得大部分分数。
u reject cos x = 2, because examiners want to see that you understand the range of cosine.
明确写出每一步：恒等式、代换、因式分解和参考角。不要跳过拒绝 cos x = 2 的步骤，因为考官希望看到你理解余弦的值域。
- Always state the identity you are using.
- Label your substitution clearly.
- Show the factorisation step fully.
- State the final solutions in the required form, usually as exact multiples of π.
始终说明你使用的恒等式；清楚地标注你的代换；完整展示因式分解步骤；以所需形式陈述最终解，通常是 π 的精确倍数。
11. Worked Alternative Method | 另一种解法示例
d you are careful with signs. Starting from 2y² - 3y - 2 = 0, the quadratic formula gives:
有些学生更喜欢使用二次公式而不是因式分解。这完全可以接受，只要你注意符号。从 2y² - 3y - 2 = 0 开始，二次公式给出：
y = [-(-3) ± √((-3)² - 4(2)(-2))] / (2 × 2)
½ and y = 2. This alternative method is useful if you do not immediately spot the factors.
化简判别式和分母会得到同样的两个根，y = -½ 和 y = 2。如果你不能立即看出因式，这种替代方法会很有用。
y = (3 ± √9 + 16) / 4 = (3 ± 5) / 4
g the previous factorisation. Choose whichever method is faster and more reliable for you.
因此 y = 2 或 y = -½，证实了之前的因式分解。选择对你来说更快、更可靠的方法。
12. Summary | 小结
Example 6.11.4 demonstrates a standard AQA A Level method for solving quadratic trigonometric equations. The steps are: use a Pythagorean identity to express everything in one trigonometric ratio, introduce a substitution, solve the quadratic, reject invalid roots, and find all angles in the given domain.
示例 6.11.4 展示了 AQA A Level 解二次型三角方程的标准方法。步骤如下：使用勾股恒等式将所有项转化为同一个三角比，引入代换，解二次方程，拒绝无效根，并在给定定义域内求出所有角度。
Final answer: x = 2π/3 and x = 4π/3 for 0 ≤ x ≤ 2π
Practising this type of question will build your confidence in combining algebraic skills with trigonometric knowledge, which is exactly what AQA examiners expect.
练习这类题目将帮助你建立将代数技能与三角学知识结合起来的信心，这正是 AQA 考官所期望的。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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