Solving Quadratic-Type Trigonometric Equations | IB数学：二次型三角方程的求解方法

📚 Solving Quadratic-Type Trigonometric Equations | IB数学：二次型三角方程的求解方法

Quadratic-type trigonometric equations are a common topic in IB Mathematics, appearing in both Analysis and Approaches and Applications and Interpretation. These equations involve a trigonometric function raised to the second power, such as sin²θ, or a product of two different trig functions, and they often require substitution, factoring, or identities to solve. In this guide, we systematically break down the strategies.

二次型三角方程是IB数学中的常见考点，出现在分析与方法（AA）和应用与解释（AI）中。这类方程涉及三角函数的二次幂，如 sin²θ，或两个不同三角函数的乘积，通常需要通过换元、因式分解或恒等式来求解。在本指南中，我们将系统地梳理各类求解策略。

1. What Is a Quadratic-Type Trigonometric Equation? | 什么是二次型三角方程？

A quadratic-type trigonometric equation can be written in the form a·f(θ)² + b·f(θ) + c = 0, where f(θ) is a trigonometric function such as sinθ, cosθ, or tanθ. It may also involve products like sinθ·cosθ, which are quadratic in degree but not in a single function. For example: 2sin²θ – sinθ – 1 = 0, cos²θ – 3cosθ + 2 = 0, tan²θ + tanθ = 0.

二次型三角方程可以写成 a·f(θ)² + b·f(θ) + c = 0 的形式，其中 f(θ) 是 sinθ、cosθ 或 tanθ 等三角函数。也可能包含像 sinθ·cosθ 这样的乘积，虽然是二次的，但不是单一函数。例如：2sin²θ – sinθ – 1 = 0，cos²θ – 3cosθ + 2 = 0，tan²θ + tanθ = 0。

2. The Substitution Method | 换元法基础

The substitution method is the most direct approach. Let t = f(θ), solve the quadratic at² + bt + c = 0, then solve f(θ) = t for θ. For instance, in 2sin²θ – sinθ – 1 = 0, set t = sinθ to obtain:

换元法是最直接的方法。令 t = f(θ)，解二次方程 at² + bt + c = 0，然后解 f(θ) = t 得到 θ。例如，在 2sin²θ – sinθ – 1 = 0 中，设 t = sinθ，得到：

2t² – t – 1 = 0

Factor: (2t + 1)(t – 1) = 0, so t = -1/2 or t = 1. Then solve sinθ = -1/2 and sinθ = 1.

因式分解得 (2t + 1)(t – 1) = 0，因此 t = -1/2 或 t = 1。然后求解 sinθ = -1/2 和 sinθ = 1。

3. Checking the Range | 值域检查

After solving for t, you must check that each solution lies within the range of f(θ). For sinθ and cosθ, t ∈ [-1, 1]; for tanθ, t ∈ ℝ. If t is outside the range, discard it immediately. This step is crucial for avoiding impossible solutions.

解得 t 后，必须检查每个解是否落在 f(θ) 的值域内。对于 sinθ 和 cosθ，t ∈ [-1, 1]；对于 tanθ，t ∈ ℝ。如果 t 超出值域，应马上舍去。这一步对于排除无效解至关重要。

For example, if solving cos²θ + 2cosθ + 3 = 0 gives cosθ = -1 or cosθ = -3, the latter is rejected because -3 is outside [-1, 1].

例如，解 cos²θ + 2cosθ + 3 = 0 得到 cosθ = -1 或 cosθ = -3，后者因 -3 不在 [-1, 1] 内而被舍去。

4. Factoring the Quadratic | 因式分解法

When the quadratic in a single trig function factors nicely, factoring is faster than substitution. For example, cos²θ – 2cosθ + 1 = (cosθ – 1)² = 0, so cosθ = 1. Always look for common factors first, such as:

当单一三角函数的二次式容易因式分解时，因式分解比换元更快。例如，cos²θ – 2cosθ + 1 = (cosθ – 1)² = 0，所以 cosθ = 1。始终先寻找公因式，例如：

sinθ·tanθ – sinθ = sinθ(tanθ – 1) = 0

This gives sinθ = 0 or tanθ = 1, which is much simpler than expanding.

这样得到 sinθ = 0 或 tanθ = 1，远比展开更简单。

5. Using the Pythagorean Identity | 利用毕达哥拉斯恒等式

If the equation mixes sin²θ and cosθ (or cos²θ and sinθ), use sin²θ + cos²θ = 1 to express everything in terms of one function. For example:

如果方程同时含有 sin²θ 和 cosθ（或 cos²θ 和 sinθ），使用 sin²θ + cos²θ = 1 将所有项化为一个函数。例如：

sin²θ – cosθ + 1 = 0

Substitute sin²θ = 1 – cos²θ:

代入 sin²θ = 1 – cos²θ：

(1 – cos²θ) – cosθ + 1 = 0 → -cos²θ – cosθ + 2 = 0 → cos²θ + cosθ – 2 = 0

Now factor: (cosθ + 2)(cosθ – 1) = 0, hence cosθ = 1 (the solution cosθ = -2 is rejected).

然后因式分解：(cosθ + 2)(cosθ – 1) = 0，因此 cosθ = 1（解 cosθ = -2 被舍去）。

6. Solving Homogeneous Equations | 齐次方程的求解

A homogeneous trigonometric equation has the form a sin²θ + b sinθ cosθ + c cos²θ = 0. To solve, divide every term by cos²θ (provided cosθ ≠ 0) to obtain a tan²θ + b tanθ + c = 0:

齐次三角方程的形式为 a sin²θ + b sinθ cosθ + c cos²θ = 0。求解时，每一项除以 cos²θ（前提 cosθ ≠ 0），得到 a tan²θ + b tanθ + c = 0：

a tan²θ + b tanθ + c = 0

Then solve the quadratic in tanθ. Do not forget to check the case cosθ = 0 separately, because dividing by cos²θ may lose solutions. For example, if cosθ = 0, then sinθ = ±1, and you must verify whether these values satisfy the original equation.

然后解关于 tanθ 的二次方程。不要忘记单独检查 cosθ = 0 的情况，因为除以 cos²θ 可能会丢失解。例如，若 cosθ = 0，则 sinθ = ±1，必须验证这些值是否满足原方程。

7. Using Double-Angle Identities | 二倍角降次法

Some quadratic-type equations are better handled with double-angle identities. Recall:

有些二次型方程用二倍角公式处理更好。回顾：

sin²θ = (1 – cos2θ)/2, cos²θ = (1 + cos2θ)/2

This transforms a quadratic equation into a linear equation in cos2θ, which is often easier to solve. Example:

这样将二次方程转化为关于 cos2θ 的线性方程，通常更容易求解。示例：

sin²θ – cos²θ = 0 → -cos2θ = 0 → cos2θ = 0

Similarly, sinθ cosθ = (1/2)sin2θ can reduce products to a single trig function.

类似地，sinθ cosθ = (1/2)sin2θ 可将乘积转化为单一三角函数。

8. Writing General Solutions | 一般解的写法

When solving trig equations in IB, you must give general solutions unless a specific interval is stated. The general solution patterns are:

在IB中求解三角方程时，除非题目给出特定区间，否则必须给出一般解。一般解的格式为：

For sinθ = k: θ = arcsin(k) + 2πn or θ = π – arcsin(k) + 2πn, n ∈ ℤ.

对于 sinθ = k：θ = arcsin(k) + 2πn 或 θ = π – arcsin(k) + 2πn，其中 n ∈ ℤ。
For cosθ = k: θ = arccos(k) + 2πn or θ = -arccos(k) + 2πn, n ∈ ℤ.

对于 cosθ = k：θ = arccos(k) + 2πn 或 θ = -arccos(k) + 2πn，其中 n ∈ ℤ。
For tanθ = k: θ = arctan(k) + πn, n ∈ ℤ.

对于 tanθ = k：θ = arctan(k) + πn，其中 n ∈ ℤ。

Remember to state the domain if required, e.g., 0 ≤ θ < 2π.

记住如果题目要求，需指出定义域，例如 0 ≤ θ < 2π。

9. Common Mistakes and Verification | 常见错误与检查

Common mistakes include failing to check the range of the substituted variable, losing solutions when dividing by sinθ or cosθ, and ignoring the periodic nature of trig functions. Always verify your answers by substituting them back into the original equation. Also be careful when using inverse functions: calculators often give only the principal value, so you must manually find all solutions in the required interval.

常见错误包括忘记检查换元变量的值域，除以 sinθ 或 cosθ 时丢失解，以及忽略三角函数的周期性。始终将答案代回原方程验证。此外，使用反函数时要小心：计算器通常只给出主值，因此你必须手动找出所需区间内的所有解。

A good practice is to sketch the graph of the function to confirm the number of solutions in the given interval.

一个好的做法是画出函数图像，以确认给定区间内解的个数。

10. Worked Example (IB-Style) | 综合例题（IB风格）

Solve 2sin²θ + cosθ – 1 = 0 for 0 ≤ θ < 2π.

解方程 2sin²θ + cosθ – 1 = 0，其中 0 ≤ θ < 2π。

Using sin²θ = 1 – cos²θ:

使用 sin²θ = 1 – cos²θ：

2(1 – cos²θ) + cosθ – 1 = 0 → -2cos²θ + cosθ + 1 = 0 → 2cos²θ – cosθ – 1 = 0

Factor: (2cosθ + 1)(cosθ – 1) = 0, so cosθ = -1/2 or cosθ = 1.

因式分解得 (2cosθ + 1)(cosθ – 1) = 0，所以 cosθ = -1/2 或 cosθ = 1。

For 0 ≤ θ < 2π, cosθ = -1/2 gives θ = 2π/3 or 4π/3; cosθ = 1 gives θ = 0.

在 0 ≤ θ < 2π 内，cosθ = -1/2 给出 θ = 2π/3 或 4π/3；cosθ = 1 给出 θ = 0。

Thus the full solution set is θ = 0, 2π/3, 4π/3.

因此完整的解集为 θ = 0，2π/3，4π/3。

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Solving Quadratic-Type Trigonometric Equations | IB数学：二次型三角方程的求解方法