IB Mathematics: Solving Quadratic Trigonometric Equations | IB数学：二次型三角方程的解法

📚 IB Mathematics: Solving Quadratic Trigonometric Equations | IB数学：二次型三角方程的解法

Quadratic trigonometric equations are equations in which a trigonometric function appears squared, such as a sin²x + b sin x + c = 0. They are a standard topic in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses, across both Standard Level and Higher Level papers.

二次型三角方程是指某个三角函数以平方形式出现的方程，例如 a sin²x + b sin x + c = 0。这是 IB 数学中的标准考点，出现在分析与方法（AA）和应用与解释（AI）两门课程中，也出现在标准级（SL）和高级级（HL）试卷里。

To succeed with these equations, you need a combination of algebraic skills and trigonometric knowledge. You must be able to factorise or apply the quadratic formula, use fundamental identities, respect the domain, and convert solutions back into angles using the unit circle or a calculator.

要顺利解决这类方程，你需要同时具备代数和三角知识。你必须能够因式分解或使用求根公式、运用基本恒等式、注意定义域，并借助单位圆或计算器将解转化为角度。

1. Recognizing the Quadratic Form | 识别二次形式

A quadratic trigonometric equation has the general structure a f(x)² + b f(x) + c = 0, where f(x) is one trigonometric function such as sin x, cos x, or tan x. The coefficient a is not zero.

二次型三角方程的一般结构为 a f(x)² + b f(x) + c = 0，其中 f(x) 是某一个三角函数，例如 sin x、cos x 或 tan x，且二次项系数 a 不为零。

Here are some clear examples:

以下是一些典型的例子：

2 sin²x – sin x – 1 = 0

cos²θ + cos θ = 0

tan²x – 3 tan x + 2 = 0

In each case, the variable inside the trigonometric function is the same expression, and the trigonometric function itself is squared. This is what makes the equation “quadratic in sin x” or “quadratic in cos x”.

在上述每种情况中，三角函数内部的变量都是同一个表达式，而三角函数本身出现了平方。这正是方程“关于 sin x 为二次”或“关于 cos x 为二次”的原因。

Sometimes the quadratic form is disguised. For example, an equation like 1 – cos²x = cos x can be rearranged into cos²x + cos x – 1 = 0. You should always look for opportunities to rearrange and reveal the quadratic structure.

有时二次形式并不明显。例如，方程 1 – cos²x = cos x 可以变形为 cos²x + cos x – 1 = 0。你应该始终寻找重新整理方程、从而暴露出二次结构的机会。

2. The Substitution Method | 换元法

The most reliable first step is to introduce a new variable. Let u = sin x, u = cos x, or u = tan x, depending on which function appears in the equation.

最可靠的第一步是引入新变量。根据方程中出现的函数类型，令 u = sin x、u = cos x 或 u = tan x。

For example, consider the equation:

例如，考虑方程：

2 sin²x – sin x – 1 = 0

Let u = sin x. Then sin²x becomes u², and the equation becomes:

令 u = sin x，则 sin²x 变成 u²，原方程变为：

2u² – u – 1 = 0

This is now a normal quadratic equation in u. Once u is found, replace u with sin x and solve the resulting trigonometric equation for x.

现在这是一个关于 u 的普通一元二次方程。求出 u 之后，将 u 替换为 sin x，并解对应的三角方程。

The substitution also reminds us that there is a hidden restriction: since sin x and cos x always lie between -1 and 1, any value of u outside this interval must be rejected immediately.

换元还提醒我们一个隐藏条件：由于 sin x 和 cos x 始终在 -1 到 1 之间，任何超出该区间的 u 值都必须立即舍去。

3. Factoring the Quadratic | 因式分解法

After substitution, many quadratic expressions can be factored. This is often the fastest and cleanest method, especially when the coefficients are small integers.

换元之后，许多二次式都可以因式分解。当系数是比较小的整数时，因式分解往往是最快、最干净的方法。

Using the example above:

继续使用上面的例子：

2u² – u – 1 = (2u + 1)(u – 1) = 0

To check this factorisation, expand (2u + 1)(u – 1):

验证这个因式分解：展开 (2u + 1)(u – 1)：

2u² – 2u + u – 1 = 2u² – u – 1

Hence the solutions in u are:

因此 u 的解为：

u = -1/2 或 u = 1

Returning to the trigonometric equation gives:

回到三角方程：

sin x = -1/2 或 sin x = 1

Next, solve each equation on the required domain. For example, on x ∈ [0, 2π), sin x = -1/2 gives x = 7π/6 and x = 11π/6, while sin x = 1 gives x = π/2.

接下来，在题目要求的定义域上分别解方程。例如，在 x ∈ [0, 2π) 上，sin x = -1/2 给出 x = 7π/6 和 x = 11π/6，而 sin x = 1 给出 x = π/2。

4. Using the Quadratic Formula | 使用求根公式

When the quadratic expression cannot be factored easily, use the quadratic formula:

当二次式不容易因式分解时，使用求根公式：

u = (-b ± √(b² – 4ac)) / (2a)

Here a, b, and c are the coefficients from a u² + b u + c = 0. The discriminant Δ = b² – 4ac tells us the number of real solutions:

其中 a、b、c 是方程 a u² + b u + c = 0 的系数。判别式 Δ = b² – 4ac 告诉我们实数解的个数：

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IB Mathematics: Solving Quadratic Trigonometric Equations | IB数学：二次型三角方程的解法