Exercise 12B.2: Solving Trigonometric Equations | 解三角方程

📚 Exercise 12B.2: Solving Trigonometric Equations | 解三角方程

Exercise 12B.2 in the IB Mathematics: Analysis and Approaches course, drawn from widely used textbooks such as Haese Mathematics, brings together the core skills needed to solve trigonometric equations. These problems require you to move beyond simple angle evaluation and to apply general solution formulae, unit circle reasoning, algebraic manipulation, and knowledge of trigonometric identities. Mastering this exercise builds a solid foundation for trig modelling, calculus with trig functions, and the IB exam itself.

IB数学《分析与方法》课程中广泛使用的教材（如 Haese Mathematics）中的练习12B.2，集中练习了解三角方程所需的核心技能。这些题目要求你不仅仅会求简单角的值，还要能够运用通解公式、单位圆推理、代数变形以及三角恒等式知识。掌握这一练习将为三角建模、三角函数的微积分以及IB考试本身打下坚实基础。

1. Understanding the Exercise Context | 理解练习背景

= k, cos 2θ = k, or tan(3x) = k, often within a specified domain such as 0° ≤ x ≤ 360° or – θ ≤ π. Some problems require the use of identities to reduce the equation to a basic form.

练习12B.2通常位于三角函数章节内，安排在你学习了正弦、余弦和正切图像之后。其目标是求解形如 sin x = k、cos 2θ = k 或 tan(3x) = k 的方程，通常限定在特定定义域内，例如 0° ≤ x ≤ 360° 或 –π ≤ θ ≤ π。有些题目需要利用恒等式将方程化简为基本形式。

equire a calculator when k is not a special ratio. The ability to generate all possible solutions using the periodic properties of trig functions is essential.

你会遇到精确值问题（如 sin x = ½）和当 k 不是特殊比值时需要计算器求解的混合题型。利用三角函数的周期性质生成所有可能解的能力至关重要。

2. Basic Trigonometric Equations | 基本三角方程

lution. Because sine is also positive in the second quadrant, the supplementary angle 180° – 30° = 150° is the second solution. Thus the solution set is {30°, 150°}.

考虑最简单的情形：在 0° ≤ θ ≤ 360° 内求解 sin θ = 0.5。你知道 sin 30° = 0.5，因此 θ = 30° 是一个解。由于正弦在第二象限也为正，其补角 180° – 30° = 150° 就是第二个解。因此解集为 {30°, 150°}。

quadrant solution is 60°, and cosine is also positive in the fourth quadrant, giving 360° – 60° = 300°. Hence θ = 60° or 300°.

对于同一区间内的 cos θ = 0.5，第一象限解为 60°，而余弦在第四象限也为正，得到 360° – 60° = 300°。因此 θ = 60° 或 300°。

cipal value is 45°, and since the tangent period is 180°, adding 180° gives 225° within 0°–360°. Always verify by substitution.

解 tan θ = 1 时，主值为 45°，正切周期为 180°，加 180° 得到 225° 落在 0°–360° 内。始终代入验证。

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3. Using the Unit Circle | 使用单位圆

line y = k; the intersection points with the circle give the angles in standard position.

单位圆是求解三角方程的强有力可视化工具。单位圆上点的 x 坐标代表 cos θ，y 坐标代表 sin θ。对于 sin θ = k，画出水平线 y = k；该线与圆的交点即给出标准位置角。

k. The symmetry of the circle immediately reveals all solutions within one revolution. This helps avoid sign errors and clarifies why some equations have 0, 1 or 2 solutions in a given interval.

对于 cos θ = k，画出垂直线 x = k。圆的对称性直接揭示一周内的所有解。这有助于避免符号错误，并说明为什么某些方程在给定区间内有0、1或2个解。

When using radians, the unit circle reinforces the key angles: π/6, π/4, π/3 and their multiples. The periodic nature of sine and cosine is visually represented by multiple revolutions.

使用弧度制时，单位圆强化了关键角：π/6、π/4、π/3 及其倍数。正弦和余弦的周期性通过多圈旋转得到可视化体现。

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4. General Solutions for sin θ = k | sin θ = k 的通解

For an equation sin θ = k where |k| α be the principal value (the acute reference angle). In degrees, the general solution is:

对于方程 sin θ = k，其中 |k| ≤ 1，设 α 为主值（锐参考角）。其度数制通解为：

θ = 180°n + (–1)ⁿ·α, n ∈ ℤ

In radians this becomes θ = nπ + (– solutions because the sine function repeats every 360° (2π) and has the symmetry sin(180° – θ) = sin θ.

在弧度制下，通解变为 θ = nπ + (–1)ⁿ·α，其中 α 以弧度表示。这一紧凑形式涵盖了所有解，因为正弦函数每360°(2π)重复一次，且具有对称性 sin(180° – θ) = sin θ。

If k = ½, α = 30°. Then θ = 180°n + (–1)ⁿ·30°. For n = 0, θ = 30°; n = 1 gives 180° – 30° = 150°; n = 2 gives 360° + 30° = 390°, and so on.

若 k = ½，α = 30°。则 θ = 180°n + (–1)ⁿ·30°。当 n = 0，θ = 30°；n = 1 得 180° – 30° = 150°；n = 2 得 360° + 30° = 390°，以此类推。

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5. General Solutions for cos θ = k | cos θ = k 的通解

For cos θ = k, let α be the reference angle such that cos α = |k| and α > 0. The general solution in degrees is:

对于 cos θ = k，令 α 为参考角，满足 cos α = |k| 且 α > 0。度数制通解为：

θ = 360°n ± α, n ∈ ℤ

In radians: θ = 2πn ± α. Because cosine is an even function, cos(–θ) = cos θ, the ± sign generates the symmetric solutions around 0° (or 0).

弧度制下：θ = 2πn ± α。由于余弦是偶函数，cos(–θ) = cos θ，± 号生成了关于 0°（或 0）对称的解。

If cos θ = –½, the reference angle α = 60°. Then solutions are θ = 360°n ± 120° because cos 120° = –½ and cos 240° = –½. The formula works elegantly.

若 cos θ = –½，参考角 α = 60°。则解为 θ = 360°n ± 120°，因为 cos 120° = –½ 且 cos 240° = –½。公式简洁有效。

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6. General Solutions for tan θ = k | tan θ = k 的通解

d), so the general solution for tan θ = k is much simpler. Let α be the acute angle tan⁻¹(|k|) adjusted for the sign of k to find a principal solution β. Then:

正切函数的周期为 180° (π rad)，因此 tan θ = k 的通解要简单得多。令 α 为 tan⁻¹(|k|) 的锐角，并根据 k 的符号调整得到一个主解 β。则：

θ = 180°n + β, n ∈ ℤ

tan repeats every 180°, adding integer multiples of 180° gives all solutions without the (–1)ⁿ or ± that appear for sine and cosine.

或弧度制：θ = nπ + β。由于 tan 每180°重复一次，加上 180° 的整数倍即可得到所有解，无需正弦和余弦所用的 (–1)ⁿ 或 ± 符号。

For tan θ = 1, β = 45° (or π/4). Solutions: 45°, 225°, 405°, etc. For tan θ = –√3, β = –60° (or –π/3). Then θ = 180°n – 60° covers values like 120°, 300°, etc.

对于 tan θ = 1，β = 45°（或 π/4）。解为：45°、225°、405° 等。对于 tan θ = –√3，β = –60°（或 –π/3）。则 θ = 180°n – 60° 涵盖了 120°、300° 等值。

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7. Equations Involving Multiple Angles | 涉及倍角的方程

Exercise 12B.2 regularly includes equations like sin 2x = 0.5 or cos 3θ = –1/√2. The strategy is to first solve for the multiple angle as if it were a single variable, then divide by the coefficient to recover x or θ.

练习12B.2经常包含像 sin 2x = 0.5 或 cos 3θ = –1/√2 这样的方程。解题策略是先将倍角当作单个变量求解，再除以系数还原出 x 或 θ。

sin 2x = 0.5, let 2x = 30°, 150°, 390°, 510°, ... Then x = 15°, 75°, 195°, 255° within 0°–360°. If the domain is wider, continue generating solutions.

对于 sin 2x = 0.5，令 2x = 30°, 150°, 390°, 510°, … 然后得 x = 15°, 75°, 195°, 255° 落在 0°–360° 内。如果定义域更广，继续生成解。

Be careful: after dividing, check that all found values still satisfy the original equation and domain. Sometimes dividing restricts the solution set, so it's safer to list 2x solutions first and then map to x.

注意：相除之后，要检查所有求得的值是否仍满足原方程和定义域。有时除法会限制解集，因此更保险的做法是先列出 2x 的解，再映射到 x。

In radian mode, solve for 2x using general solutions, then x = (solution)/2. For cos 3θ = –1/√2, 3θ = 3π/4 + 2πn or 3θ = 5π/4 + 2πn, then divide by 3.

在弧度模式下，先用通解求出 2x，然后 x = 解/2。对于 cos 3θ = –1/√2，3θ = 3π/4 + 2πn 或 3θ = 5π/4 + 2πn，再除以3。

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8. Equations Requiring Factoring | 需要因式分解的方程

Many trigonometric equations appear in quadratic form, such as 2 sin² x – sin x – 1 = 0. Treat sin x as a variable y and factor: 2y² – y – 1 = (2y + 1)(y – 1) = 0. Then solve sin x = –½ or sin x = 1.

许多三角方程以二次形式出现，如 2 sin² x – sin x – 1 = 0。将 sin x 当作变量 y 进行因式分解：2y² – y – 1 = (2y + 1)(y – 1) = 0。然后解 sin x = –½ 或 sin x = 1。

Similarly, 3 cos² θ – 5 cos θ + 2 = 0 factors as (3 cos θ – 2)(cos θ – 1) = 0, giving cos θ = 2/3 or cos θ = 1. Make sure to check if the values are within [–1, 1] before proceeding.

类似地，3 cos² θ – 5 cos θ + 2 = 0 可分解为 (3 cos θ – 2)(cos θ – 1) = 0，得到 cos θ = 2/3 或 cos θ = 1。继续求解前务必检查数值是否在 [–1, 1] 内。

to rewrite in terms of a single function. For instance, 2 sin² x + 3 cos x = 0 becomes 2(1 – cos² x) + 3 cos x = 0, then factor.

如果方程包含不同三角函数，利用恒等式将其转化为同名函数。例如，2 sin² x + 3 cos x = 0 化为 2(1 – cos² x) + 3 cos x = 0，然后因式分解。

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9. Using Pythagorean Identities | 使用勾股恒等式

he identity sin² θ + cos² θ = 1 is indispensable. In Exercise 12B.2 you might see 2 sin² θ – cos θ = 1. Replace sin² θ with 1 – cos² θ to obtain –2 cos² θ – cos θ + 1 = 0, then multiply by –1: 2 cos² θ + cos θ – 1 = 0. Factor and solve.

恒等式 sin² θ + cos² θ = 1 不可或缺。在练习12B.2中，你可能会遇到 2 sin² θ – cos θ = 1。将 sin² θ 替换为 1 – cos² θ，得到 –2 cos² θ – cos θ + 1 = 0，乘以 –1 得 2 cos² θ + cos θ – 1 = 0。因式分解并求解。

gh less common in basic equation solving, they appear in simplification steps. Always check if an identity can reduce the complexity.

其他有用恒等式包括 1 + tan² θ = sec² θ 和 1 + cot² θ = csc² θ。虽然在基础方程求解中较少出现，但它们会出现在化简步骤中。务必检查能否通过恒等式降低复杂度。

If a problem contains both sin θ and cos θ, substituting an identity avoids solving simultaneous equations and ensures algebraic consistency.

若一道题同时包含 sin θ 和 cos θ，代入恒等式可避免解联立方程，并确保代数的一致性。

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10. Checking Domain Restrictions | 检查定义域限制

IB questions specify a domain such as 0 ≤ x ≤ 2π or – ll within the given interval. Report solutions in the requested unit (degrees or radians).

IB 题目会明确定义域，如 0 ≤ x ≤ 2π 或 –180° ≤ x ≤ 180°。得到通解后，必须仅选取落在给定区间内的值。按要求的单位（度或弧度）给出解。

s non-strict (≤, ≥). A solution exactly at the boundary is valid but may be easily missed.

常见错误：忘记限定解集会导致多余答案。对于非严格不等式（≤, ≥），务必测试边界值。恰好位于边界的解是有效的，但容易被忽略。

multiple-angle equations like sin 2x = 0, the number of solutions doubles. A domain of 0°–360° for x yields solutions for 2x in 0°–720°, effectively producing four x-values. Systematically list and filter.

当处理倍角方程如 sin 2x = 0 时，解的数量会翻倍。x 的定义域 0°–360° 使得 2x 在 0°–720° 区间内，实际上会产生四个 x 值。要系统地列出并筛选。

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11. Common Mistakes & Tips | 常见错误与提示

Mistake 1: Forgetting the period of tangent is 180°, not 360°. Many students incorrectly use 360°n for tan, resulting in missing solutions. Use tan's specific general solution formula.

错误一：忘记正切函数的周期是 180° 而非 360°。许多学生错误地对 tan 使用 360°n，导致遗漏解。要使用正切专用的通解公式。

= sin x, do not cancel sin x; instead bring all terms to one side and factor: sin x (cos x – 1) = 0.

错误二：方程两边同除以可能为零的三角函数。例如，由 sin x cos x = sin x，不要约去 sin x；而应将所有项移到一边因式分解：sin x (cos x – 1) = 0。

Mistake 3: Misidentifying the reference angle for negative ratios. Always use the absolute value of k to find the acute reference angle α, then place it in the correct quadrants according to the sign.

错误三：对负比值错误识别参考角。始终用 |k| 求锐参考角 α，然后根据符号将其放置在正确的象限内。

Tip: Always draw a quick sketch of the unit circle or the function graph to confirm the number and positions of solutions. Use the CAST diagram to remember which trig functions are positive in each quadrant.

提示：始终快速画一个单位圆或函数图像草图来确认解的个数和位置。使用 CAST 图记忆各象限内三角函数的正负。

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12. Practice Problems from Exercise 12B.2 | 练习12B.2典型题目

Here are representative problems to illustrate the skills needed. Work through them step by step.

以下为代表性题目，展示所需技能。请逐步练习。

Problem 1: Solve for 0° ≤ x ≤ 360°: 2 sin x – 1 = 0.

问题1： 在 0° ≤ x ≤ 360° 内求解：2 sin x – 1 = 0。

Solution: sin x = ½ → x = 30°, 150°.

解答：sin x = ½ → x = 30°, 150°。

Problem 2: Find all solutions for –π ≤ θ ≤ π: cos 2θ = –½.

问题2： 在 –π ≤ θ ≤ π 内求所有解：cos 2θ = –½。

olution: 2θ = 2π/3 + 2πn or 2θ = 4π/3 + 2πn → θ = π/3 + πn or θ = 2π/3 + πn. Filtering for –π ≤ θ ≤ π gives θ = –2π/3, –π/3, π/3, 2π/3.

解答：2θ = 2π/3 + 2πn 或 2θ = 4π/3 + 2πn → θ = π/3 + πn 或 θ = 2π/3 + πn。筛选 –π ≤ θ ≤ π 得 θ = –2π/3, –π/3, π/3, 2π/3。

Problem 3: Solve tan² x – 3 = 0 for 0 ≤ x ≤ 2π.

问题3： 在 0 ≤ x ≤ 2π 内解 tan² x – 3 = 0。

Solution: tan x = ±√3 → x = π/3, 2π/3, 4π/3, 5π/3.

解答：tan x = ±√3 → x = π/3, 2π/3, 4π/3, 5π/3。

Problem 4: Solve 2 cos² x – 3 sin x = 0 for 0° ≤ x ≤ 360°.

问题4： 在 0° ≤ x ≤ 360° 内解 2 cos² x – 3 sin x = 0。

Solution: Replace cos² x = 1 – sin² x → 2 – 2 sin² x – 3 sin x = 0 → 2 sin² x + 3 sin x – 2 = 0 → (2 sin x – 1)(sin x + 2) = 0 → sin x = ½ (since sin x = –2 impossible) → x = 30°, 150°.

解答：代入 cos² x = 1 – sin² x → 2 – 2 sin² x – 3 sin x = 0 → 2 sin² x + 3 sin x – 2 = 0 → (2 sin x – 1)(sin x + 2) = 0 → sin x = ½（sin x = –2 不可能）→ x = 30°, 150°。

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