📚 Example 3.5.2: Solving Trigonometric Equations | 例题3.5.2：解三角方程
Welcome to TutorHao's worked example series. In this article, we will walk through Example 3.5.2, a typical trigonometric equation problem from the AQA A-Level Mathematics specification. We will break down the solution step by step, explain the underlying concepts, and provide useful tips for your exam.
欢迎阅读 TutorHao 的例题精讲系列。本文我们将详细解析例题 3.5.2——一道来自 AQA A-Level 数学大纲的典型三角方程问题。我们会逐步拆解解题过程，解释背后的概念，并提供实用的考试技巧。
1. Problem Statement | 题目重述
own below for 0° ≤ x ≤ 360°. Give your answers correct to 1 decimal place where necessary.
求解下方方程在 0° ≤ x ≤ 360° 范围内的解。如有必要，答案保留一位小数。
2cos² x + 3sin x = 0
This equation mixes both sine and cosine squared. Our first instinct should be to use a trigonometric identity to reduce it to a single trigonometric function. This is a classic AQA-style question because it tests your ability to select and apply the correct identity from the formula booklet.
该方程同时包含正弦和余弦平方。我们的第一反应应该是利用三角恒等式将其化为单一三角函数。这是一道典型的 AQA 风格题目，因为它考查你从公式册中选择并正确应用恒等式的能力。
2. Key Identity Review | 核心恒等式回顾
n identity states that sin² x + cos² x = 1 for all real x. Therefore, cos² x = 1 − sin² x.
基本毕达哥拉斯恒等式指出：对于任意实数 x，sin² x + cos² x = 1。因此，cos² x = 1 − sin² x。
erstanding this derivation helps you remember the identity and also helps you see why it is always true, regardless of the quadrant.
这个恒等式可以从单位圆推导出来：单位圆上一点的坐标是 (cos θ, sin θ)，而单位圆的方程是 x² + y² = 1。因此 sin² θ + cos² θ = 1。理解这个推导过程有助于你记住恒等式，也有助于你明白为什么它总是成立，无论角在第几象限。
In this example, we need to replace cos² x with 1 − sin² x. This is a substitution that converts a product or sum of different trigonometric functions into a single function, making the equation suitable for solving as an algebraic quadratic.
在这个例子中，我们需要用 1 − sin² x 替换 cos² x。这种代换将不同三角函数的乘积或和化为单一函数，使方程适合作为代数二次方程来求解。
3. Transforming the Equation | 方程变形
Substitute cos² x = 1 − sin² x into the original equation:
将 cos² x = 1 − sin² x 代入原方程：
2(1 − sin² x) + 3sin x = 0
Expand the brackets. Be careful with the negative sign: multiplying 2 by (1 − sin² x) gives 2 − 2sin² x. The expression then becomes:
展开括号。注意负号：2 乘以 (1 − sin² x) 得到 2 − 2sin² x。原式变为：
2 − 2sin² x + 3sin x = 0
Now rearrange the terms to write the quadratic in standard form. Standard form means the term with the highest degree comes first. We multiply the whole equation by −1 to make the coefficient of sin² x positive, which makes factoring easier:
现在重新整理各项，将二次方程写成标准形式。标准形式意味着最高次项在前。我们将方程两边乘以 −1，使 sin² x 的系数为正，这样更便于因式分解：
2sin² x − 3sin x − 2 = 0
3sin x − 2 = 0. Always check that the constant term has the correct sign after this step.
乘以 −1 会改变所有符号：2 − 2sin² x + 3sin x = 0 变为 −2 + 2sin² x − 3sin x = 0，这等价于 2sin² x − 3sin x − 2 = 0。此步之后务必检查常数项的符号是否正确。
4. Factoring the Quadratic | 因式分解二次方程
Let u = sin x for clarity. Then the quadratic becomes 2u² − 3u − 2 = 0.
为清晰起见，令 u = sin x。于是二次方程变为 2u² − 3u − 2 = 0。
) = −4 and add to −3. These numbers are −4 and 1, because (−4) × 1 = −4 and (−4) + 1 = −3.
我们寻找两个数字，它们相乘等于 2 × (−2) = −4，相加等于 −3。这两个数字是 −4 和 1，因为 (−4) × 1 = −4，且 (−4) + 1 = −3。
Rewrite the middle term −3u as −4u + 1u:
将中间项 −3u 改写为 −4u + 1u：
2u² − 4u + u − 2 = 0
Factor by grouping. Group the first two terms and the last two terms:
分组因式分解。将前两项和后两项分别分组：
2u(u − 2) + 1(u − 2) = 0
Notice that (u − 2) appears in both groups. Factor it out:
注意 (u − 2) 出现在两组中。将它提出：
(2u + 1)(u − 2) = 0
ing the zero-product property, either 2u + 1 = 0 or u − 2 = 0. Therefore, u = −½ or u = 2.
根据零积性质，要么 2u + 1 = 0，要么 u − 2 = 0。因此，u = −½ 或 u = 2。
5. Solving sin x = −½ | 求解 sin x = −½
Returning to x, the first equation is sin x = −½.
回到 x，第一个方程是 sin x = −½。
nd the related acute angle, take the inverse sine of the positive value ½: sin⁻¹(½) = 30°.
在区间 0° ≤ x ≤ 360° 内，正弦函数在第三和第四象限为负。要找到对应的锐角，先对正值 ½ 取反正弦：sin⁻¹(½) = 30°。
10°. In the fourth quadrant, the angle is 360° − 30° = 330°. Therefore, the solutions are:
使用 CAST 图或正弦图像的对称性，我们知道第三象限的角为 180° + 30° = 210°，第四象限的角为 360° − 30° = 330°。因此，解为：
x = 210° 或 x = 330°
Both angles lie within the required interval. Notice that these two solutions are symmetric about the x-axis in the sine graph, which provides a good visual check.
这两个角都在所需区间内。注意在正弦图像中，这两个解关于 x 轴有对称关系，这提供了一个很好的可视化检查。
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