Example 6.11.3: Solving a Trigonometric Equation | 例题 6.11.3：求解三角方程

📚 Example 6.11.3: Solving a Trigonometric Equation | 例题 6.11.3：求解三角方程

This worked example is a classic AQA A-Level Mathematics problem. It combines three important skills: using a trigonometric identity, solving a quadratic equation, and finding all solutions within a given interval. By studying Example 6.11.3, you will learn how to move confidently between algebra and trigonometry, and how to avoid the common mistakes that cost marks in the exam.

这一例题是 AQA A-Level 数学中的经典问题。它将三项重要技能结合在一起：运用三角恒等式、求解二次方程，以及在给定范围内找出所有解。通过学习例题 6.11.3，你将学会如何在代数与三角函数之间自如转换，并避免在考试中失分的常见错误。

1. Why This Example Matters | 为什么这个例题重要

Trigonometric equations are always present in AQA Pure Mathematics papers. A typical question will present an equation that looks unfamiliar at first, such as one involving both cos² x and sin x. The key is to reduce the equation to a single trigonometric function before solving. Example 6.11.3 does exactly this: it asks you to solve a quadratic equation in sin x after applying a standard identity.

在 AQA 纯数学试卷中，三角方程几乎从不缺席。典型考题给出的方程乍看起来并不熟悉，例如同时含有 cos² x 和 sin x 的方程。关键在于先把方程化简为只含一个三角函数的方程，然后再求解。例题 6.11.3 做的正是这件事：它要求你先运用标准恒等式，然后求解一个关于 sin x 的二次方程。

In the AQA specification, this topic appears under Pure Mathematics, Topic 5: Trigonometry. The mark schemes reward clear setting out, correct use of identities, and complete answers within the stated interval. Skipping the identity step or failing to discard impossible roots can easily cost you two or three marks.

在 AQA 大纲中，该内容属于纯数学第 5 专题：三角学。评分标准重视清晰的书写、恒等式的正确使用，以及在给定范围内完整的答案。省略恒等式步骤或未能舍去不可能的根，都会轻易让你丢掉 2 至 3 分。

2. The Problem Statement | 题目呈现

Here is the full wording of Example 6.11.3, which you might see in a textbook or an exam:

以下是例题 6.11.3 的完整表述，你在课本或考试中可能会看到类似问题：

Solve, for 0° ≤ x < 360°, the equation 2 cos² x + 3 sin x = 0.

Notice that the range is given in degrees. This means all answers must be given in degrees, not radians. The interval includes 0° but does not include 360°. This is a common condition in AQA questions, and your final answer should contain only values in that half-open interval.

注意题目给出的范围是度数。这意味着所有答案都必须用度数表示，而不是弧度。该区间包含 0°，但不包含 360°。这是 AQA 题目中常见的条件，你的最终答案应只包含该半开区间内的值。

The equation itself has two different trigonometric functions: cos² x and sin x. Because cos² x can be written in terms of sin x using the Pythagorean identity, we can turn this into a familiar quadratic equation.

方程本身包含两个不同的三角函数：cos² x 与 sin x。因为利用勾股恒等式可以将 cos² x 写成 sin x 的表达式，所以我们可以把它转化为熟悉的二次方程。

3. The Key Identity | 关键恒等式

The identity needed for this example is the Pythagorean identity. For any angle θ, the sum of the squares of sine and cosine is always 1:

本例所需的关键恒等式是勾股恒等式。对任意角 θ，正弦的平方与余弦的平方之和恒等于 1：

sin² θ + cos² θ = 1

Therefore, we can rearrange the identity to make cos² θ the subject:

因此，我们可以将恒等式变形，使 cos² θ 成为主项：

cos² θ = 1 − sin² θ

This is the step that unlocks the solution. Without replacing cos² x, the equation would contain two different trig functions and could not be factorised easily. By using the identity, we obtain a single equation in sin x only.

这是解题的关键步骤。如果不替换 cos² x，方程就会同时包含两个不同的三角函数，难以直接因式分解。通过使用恒等式，我们可以得到只含 sin x 的单一方程。

4. Substitution and Rearrangement | 代入与整理

Now we substitute cos² x = 1 − sin² x into the original equation:

现在将 cos² x = 1 − sin² x 代入原方程：

2(1 − sin² x) + 3 sin x = 0

Expanding the bracket gives:

展开括号得到：

2 − 2 sin² x + 3 sin x = 0

It is usually easier to work with a positive leading coefficient, so we rearrange every term to the opposite sign:

为了避免二次项系数为负，通常我们两边都移项并改变正负号：

2 sin² x − 3 sin x − 2 = 0

If you prefer, you could also write the equation as 2 sin² x − 3 sin x − 2 = 0 directly, but always check that the signs are correct before moving on. A single sign error will cause your roots to be wrong.

如果你愿意，也可以直接写出 2 sin² x − 3 sin x − 2 = 0，但在继续之前一定要检查符号是否正确。一个符号错误就会导致根出错。

5. Solving the Quadratic in sin x | 求解关于 sin x 的二次方程

We now treat sin x as a single letter, for example u. The quadratic becomes:

我们把 sin x 看作一个单独的字母，例如 u。二次方程变为：

2u² − 3u − 2 = 0, where u = sin x

Factorising the quadratic gives:

对二次方程因式分解得到：

(2u + 1)(u − 2) = 0

Setting each factor equal to zero gives two possible equations:

令每个因式等于零，得到两个可能的方程：

2u + 1 = 0, so u = −½, meaning sin x = −½

2u + 1 = 0，所以 u = −½，即 sin x = −½
u − 2 = 0, so u = 2, meaning sin x = 2

u − 2 = 0，所以 u = 2，即 sin x = 2

Here is where many students make their first mistake. The sine function has output values only between −1 and 1, inclusive. Since 2 is outside this range, the equation sin x = 2 has no solutions. Therefore we must discard this root and focus entirely on sin x = −½.

很多学生会在这里犯第一个错误。正弦函数的值域是 [−1, 1]，即其输出值只在 −1 到 1 之间。因为 2 超出该范围，方程 sin x = 2 无解。因此我们必须舍去这个根，把注意力全部集中在 sin x = −½ 上。

6. Finding All Solutions in the Given Interval | 在给定区间内求所有解

We now need to solve:

现在我们需要求解：

sin x = −½, for 0° ≤ x < 360°

First, find the acute angle whose sine is ½. From the standard exact values, we know that sin 30° = ½. However, sin x is negative, so the solutions lie in the third and fourth quadrants.

首先找出正弦值为 ½ 的锐角。根据标准特殊角值，我们知道 sin 30° = ½。但由于 sin x 是负数，解位于第三象限和第四象限。

Using the CAST diagram, or the symmetry of the sine graph, we can find both solutions:

利用 CAST 象限图，或正弦图像的对称性，我们可以找到两个解：

In the third quadrant, x = 180° + 30° = 210°

在第三象限，x = 180° + 30° = 210°
In the fourth quadrant, x = 360° − 30° = 330°

在第四象限，x = 360° − 30° = 330°

Therefore, the full set of solutions is x = 210° and x = 330°. It is important to include both values. In the range 0° ≤ x < 360°, no other angle has a sine of −½.

因此，完整解集为 x = 210° 和 x = 330°。必须把两个值都写出来。在 0° ≤ x < 360° 的范围内，没有其他角的正弦值为 −½。

7. Verification | 验证

Checking answers is a good habit in any A-Level exam. Let us verify each solution by substituting back into the original equation.

在做任何 A-Level 考试题时，检验答案都是好习惯。我们通过回代来验证每个解是否满足原方程。

For x = 210°:

当 x = 210° 时：

cos 210° = −√3/2, so cos² 210° = 3/4

sin 210° = −1/2

Substitute into 2 cos² x + 3 sin x:

代入 2 cos² x + 3 sin x：

2 × (3/4) + 3 × (−1/2) = 3/2 − 3/2 = 0

So x = 210° is correct. For x = 330°, the values are cos 330° = √3/2, so cos² 330° = 3/4, and sin 330° = −1/2. The substitution works in exactly the same way and gives 0.

所以 x = 210° 是正确的。对于 x = 330°，有 cos 330° = √3/2，因此 cos² 330° = 3/4，sin 330° = −1/2。代入计算过程完全相同，结果也是 0。

Verification is not a waste of time. If you make a sign error when factorising, substitution will reveal it immediately, allowing you to correct your work before the final answer.

验证并不是浪费时间。如果你在因式分解时出现符号错误，代入很快就能暴露问题，让你在写出最终答案前及时纠正。

8. Common Pitfalls | 常见错误

Let us look at the most common mistakes made by AQA students when solving this type of example.

让我们来看看 AQA 学生在求解这类例题时最常见的错误。

Forgetting to discard sin x = 2. Since sine values are between −1 and 1, this equation has no solution. Keep this range in mind at all times.

忘记舍去 sin x = 2。由于正弦值在 −1 与 1 之间，该方程无解。请时刻牢记这个值域。
Finding only 210° and missing 330°. Many students stop after finding the first angle in the third quadrant. Always use the graph or CAST diagram to find all angles in the given interval.

只找到 210° 而漏掉 330°。许多学生在找到第三象限的一个角之后就停止了。务必利用图像或 CAST 象限图找出给定区间内的所有角。
Making a sign error when rearranging the quadratic. Writing −2 sin² x + 3 sin x + 2 = 0 can lead to incorrect factorisation. Rearrange so that the coefficient of sin² x is positive.

整理二次方程时出现符号错误。写成 −2 sin² x + 3 sin x + 2 = 0 可能导致错误的因式分解。最好整理为使 sin² x 系数为正的形式。
Giving answers in radians by mistake. An AQA question that states 0° ≤ x < 360° expects degrees. Read the range condition carefully.

误将答案写成弧度。AQA 题目若注明 0° ≤ x < 360°，则期望答案使用度数。请仔细阅读范围条件。
Not showing the factorisation step. If you write the quadratic and jump straight to the roots, a marker may not be able to award method marks for factorising.

不展示因式分解步骤。如果你写出二次方程后直接跳到最后根，阅卷老师可能无法给出因式分解的方法分。

9. AQA Exam Tips | AQA 考试提示

In the real AQA examination, this type of question is usually worth 4 or 5 marks. Here is how the marks are typically distributed:

在真正的 AQA 考试中，这类问题通常值 4 至 5 分。以下是一般分布情况：

1 mark for substituting the Pythagorean identity correctly, replacing cos² x with 1 − sin² x.

1 分：正确代入勾股恒等式，用 1 − sin² x 替换 cos² x。
1 mark for rearranging into a quadratic equation in sin x and factorising.

1 分：整理为关于 sin x 的二次方程并因式分解。
1 mark for correctly obtaining sin x = −½ and identifying that sin x = 2 to be discarded.

1 分：正确得到 sin x = −½，并判断 sin x = 2 需要舍去。
2 marks for finding both solutions, 210° and 330°, within the interval.

2 分：在区间内找到两个解，210° 与 330°。

You should always write down the identity before substituting it. This tells the examiner exactly what

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Example 6.11.3: Solving a Trigonometric Equation | 例题 6.11.3：求解三角方程