📚 Exercise 20A.2: Solving Trigonometric Equations | 练习 20A.2：三角方程求解
Trigonometric equations form a core part of IB Mathematics, requiring students to find all angles that satisfy a given equation within a specified domain. Exercise 20A.2 provides structured practice in solving such equations using algebraic manipulation, trigonometric identities, and a solid understanding of the unit circle. Mastering these skills is essential not only for examinations but also for advanced topics in calculus and physics.
三角方程是 IB 数学的核心内容，要求学生找出在指定范围内满足给定方程的所有角度。练习 20A.2 旨在通过系统的训练，帮助学生运用代数变形、三角恒等式和对单位圆的深入理解来求解这类方程。掌握这些技能不仅对考试至关重要，也为后续的微积分和物理学习打下坚实基础。
1. Understanding the Basics | 基础知识回顾
e must recall the definitions of the sine, cosine, and tangent functions. For any angle θ measured from the positive x‑axis, sin θ is the y‑coordinate of a point on the unit circle, cos θ is the x‑ e functions are periodic: sine and cosine repeat every 2π radians, while tangent repeats every π radians.
在着手练习 20A.2 之前，我们必须回顾正弦、余弦和正切函数的定义。对于从 x 轴正方向度量的任意角 θ，sin θ 是单位圆上点的 y 坐标，cos θ 是 x 坐标，tan θ = sin θ / cos θ。这些函数具有周期性：正弦和余弦每 2π 弧度重复一次，而正切每 π 弧度重复一次。
The exact values of trigonometric functions for key angles (0, π/6, π/4, π/3, π/2, etc.) must be memorised. These values can be derived from the special right‑angled triangles: the 45°–45°–90° triangle with sides 1 : 1 : √2, and the 30°–60°–90° triangle with sides 1 : √3 : 2. Recall that sin 30° = 1/2, cos 45° = 1/√2, tan 60° = √3, and so on.
必须熟记关键角度（0, π/6, π/4, π/3, π/2 等）的三角函数精确值。这些值可以从特殊直角三角形推导：45°–45°–90° 三角形的边长比为 1 : 1 : √2，30°–60°–90° 三角形的边长比为 1 : √3 : 2。请记住 sin 30° = 1/2，cos 45° = 1/√2，tan 60° = √3 等。
in the exercise (e.g., 0 ≤ x ≤ 2π or 0° ≤ x ≤ 360°) restricts the number of valid answers.
当求解形如 sin x = k 的方程时，我们要找出所有正弦值为 k 的角 x。由于单位圆的对称性，在一个完整的旋转圈内通常有两个解；而考虑到周期性，则有无穷多个解。练习中指定的区间（如 0 ≤ x ≤ 2π 或 0° ≤ x ≤ 360°）限制了有效答案的个数。
2. Using the Unit Circle | 利用单位圆
os θ = k corresponds to a vertical line x = k. The angles are measured from the positive x‑axis in anti‑clockwise direction.
单位圆是求解三角方程最强大的可视化工具。给定的 sin θ 值对应一条水平直线 y = k；该直线与单位圆的交点给出所求的角度。类似地，cos θ = k 对应一条垂直直线 x = k。角度均从 x 轴正方向按逆时针方向度量。
and one in the second quadrant (θ = π − π/6 = 5π/6). No other points on the circle have y‑ate 1/2. To generate all solutions, add integer multiples of 2π to these principal angles.
例如，若 sin θ = 1/2，画出直线 y = 1/2。它与单位圆相交于两点：一个在第一象限（θ = π/6），另一个在第二象限（θ = π − π/6 = 5π/6）。圆上没有其他点的 y 坐标等于 1/2。要得到所有解，只需在这些主值角上加上 2π 的整数倍即可。
In Exercise 20A.2, many problems will require you to find solutions within a given interval. Always sketch the unit circle quickly, mark the relevant horizontal or vertical line, and identify the quadrants where the solutions lie. This approach minimises algebraic errors and helps you keep track of the domain.
在练习 20A.2 中，许多题目会要求你在给定区间内求解。务必快速画出单位圆草图，标出相关的水平线或垂直线，并确定解所在的象限。这种方法能将代数错误降至最低，并帮助你准确掌握定义域。
3. Solving sin x = k | 解 sin x = k 型方程
90°, 90°]). Given that sin(π − α) = sin α, the second solution in one full cycle is π − α.
方程 sin x = k 只有当 −1 ≤ k ≤ 1 时才有解。根据定义，主值解为 α = arcsin(k)，它位于区间 [−π/2, π/2]（或 [−90°, 90°]）内。由于 sin(π − α) = sin α，在一个完整周期内的第二个解为 π − α。
Thus, the general solutions are x = α + 2πn and x = (π − α) + 2πn, where n ∈ ℤ dents often forget to consider negative values of n when the domain includes negative angles.
因此，通解为 x = α + 2πn 和 x = (π − α) + 2πn，其中 n ∈ ℤ。当区间受限时，例如 0 ≤ x ≤ 2π，我们代入整数值 n，并只保留落在该范围内的解。当定义域包含负角度时，学生常常忘记考虑 n 取负值的情况。
Worked example: Solve sin x = √2/2 for 0 ≤ x ≤ 2π. Here α = π/4 (since sin π/4 = √2 3π/4. Adding 2π gives angles beyond the domain. Thus the answers are x = π/4 and x = 3π/4.
具体示例：在 0 ≤ x ≤ 2π 内求解 sin x = √2/2。此处 α = π/4（因为 sin π/4 = √2/2）。第二个解为 π − π/4 = 3π/4。加上 2π 后角度超出定义域。因此答案为 x = π/4 和 x = 3π/4。
4. Solving cos x = k | 解 cos x = k 型方程
− α) = cos α). However, many textbooks prefer to express the second solution as −α + 2πn.
对于 cos x = k，主值解 α = arccos(k) 位于区间 [0, π]（或 [0°, 180°]）内。由于余弦是偶函数，cos(−α) = cos α，但在一个 2π 周期内的第二个解实际上是 2π − α（因为 cos(2π − α) = cos α）。不过许多教材倾向于用 −α + 2πn 来表示第二个解。
The general solutions can be written compactly as x = ±α + 2πn, where n ∈ ℤ all solutions generated by this formula, not missing those that arise from negative signs.
通解可以简洁地写为 x = ±α + 2πn，其中 n ∈ ℤ。这一优雅的形式同时包含了 α 和 −α，然后加入了周期性。当在特定区间内求解时，你必须仔细列出由此公式产生的所有解，不能遗漏由负号产生的解。
The valid ones are −5π/3, −π/3, π/3, 5π/3. Always check your answers with a quick sketch.
例如，在 −2π ≤ x ≤ 2π 内求解 cos x = 1/2。此处 α = π/3。解为 x = π/3, −π/3, π/3 + 2π = 7π/3（超出区间），π/3 − 2π = −5π/3, −π/3 + 2π = 5π/3, −π/3 − 2π = −7π/3。有效解为 −5π/3, −π/3, π/3, 5π/3。务必用草图验证你的答案。
5. Solving tan x = k | 解 tan x = k 型方程
s in (−π/2, π/2). Because tan(x + π) = tan x, the general solution is simply x = α + πn, n ∈ ℤ.
正切函数有所不同：它的周期是 π 而非 2π，且其值 k 可为任意实数（不像正弦和余弦那样受限制）。主值解 α = arctan(k) 位于 (−π/2, π/2) 内。由于 tan(x + π) = tan x，通解就是 x = α + πn，n ∈ ℤ。
When the domain is given, finding all solutions involves adding or subtracting multiples of π until the values leave the interval. Students sometimes mistakenly add 2π instead of π; remember that tangent repeats every half‑revolution, not full revolution.
当给定了定义域时，要求出所有解，只需加减 π 的整数倍，直到值超出区间。学生有时会错误地加上 2π 而非 π；请记住正切每半个圆周重复一次，而非整个圆周。
Example: Solve tan x = √3 x = π/3 and 4π/3. Notice that these two angles differ by exactly π, confirming the period.
示例：在 0 ≤ x ≤ 2π 内求解 tan x = √3。α = π/3。加上 π 的倍数得到：π/3，4π/3。再加 π 得到 7π/3 > 2π。因此解为 x = π/3 和 4π/3。注意这两个角恰好相差 π，印证了周期。
6. Equations with Multiple Angles | 含倍角的方程
ikely includes equations such as sin 2x = 1/2 or cos 3θ = 0. The key is to first solve the “inner” n on x, determine the corresponding range for u. Step 3: Solve u = ... and then x = u / 2.
练习 20A.2 很可能包含诸如 sin 2x = 1/2 或 cos 3θ = 0 这样的方程。关键在于首先对倍角求解“内部”方程，然后再除以系数。步骤一：令 u = 2x，求解 sin u = 1/2 得到所有可能的 u。步骤二：根据 x 的限制，确定 u 的对应范围。步骤三：解出 u = ...，然后 x = u / 2。
Be extremely careful to multiply the domain by the coefficient when converting x‑restrictions to u‑ This doubling can introduce extra solutions that some students overlook. Always list all u‑values first, then divide by the coefficient to obtain x‑values and filter those that naturally fall into the original domain.
在将 x 的限制转换为 u 的限制时，要极其小心地将定义域乘以系数。若原定义域为 0 ≤ x ≤ 2π，则对于 u = 2x，定义域变为 0 ≤ u ≤ 4π。这种倍增会引入一些额外的解，有些学生会忽略。务必先列出所有 u 值，然后除以系数得到 x 值，并筛选出自然落在原定义域内的解。
Illustrative problem: 330°, 690°, 1050°. Dividing by 3: θ = 70°, 190°, 310°, 110°, 230°, 350°. All are within 0°–360°.
示例问题：在 0° ≤ θ ≤ 360° 内求解 sin 3θ = −½。令 u = 3θ。sin u = −½ 的主值为 u = 210°（或 330°）。通解：u = 210° + 360°k 或 u = 330° + 360°k。u 的定义域：0 ≤ u ≤ 1080°。列出 u 值：210°, 570°, 930°, 330°, 690°, 1050°。除以 3 得：θ = 70°, 190°, 310°, 110°, 230°, 350°。所有值均在 0°–360° 内。
7. Using Pythagorean Identities | 利用毕达哥拉斯恒等式
r. For the given equation, sin² x = 1 − cos² x, transforming it into a quadratic in cos x.
许多三角方程并非直接是 sin x = k 的形式。它们可能同时包含正弦和余弦，如 sin² x + cos x = 1。这时我们利用毕达哥拉斯恒等式 sin² x + cos² x = 1，将一种函数替换为另一种。对于上述方程，sin² x = 1 − cos² x，从而将其转化为关于 cos x 的二次方程。
sec² x, especially when the equation involves tangent and secant. In IB Mathematics, both identities are standard tools. After substitution, you typically obtain a quadratic (or sometimes cubic) that must be solved like an algebraic equation, keeping in mind the allowed range for sin and cos values.
另一个有用的恒等式是 tan² x + 1 = sec² x，尤其是当方程包含正切和正割时。在 IB 数学中，这两个恒等式都是标准工具。代换后，通常会得到一个二次（有时是三次）方程，你必须像解代数方程一样解它，并始终牢记正弦和余弦值的允许范围。
x + 3 cos x = 0 for 0 ≤ x ≤ 2π. Replace sin² x by 1 − cos² x: 2(1 − cos² x) + 3 cos x = 0 ⇒ 2 − 2 cos² x + 3 cos x = 0 ⇒ + 1)(cos x − 2) = 0. cos x = 2 is impossible. So cos x = −1/2. Solutions: x = 2π/3, 4π/3.
例如，在 0 ≤ x ≤ 2π 内求解 2 sin² x + 3 cos x = 0。将 sin² x 替换为 1 − cos² x：2(1 − cos² x) + 3 cos x = 0 ⇒ 2 − 2 cos² x + 3 cos x = 0 ⇒ 2 cos² x − 3 cos x − 2 = 0。因式分解：(2 cos x + 1)(cos x − 2) = 0。cos x = 2 不可能成立。故 cos x = −1/2。解为 x = 2π/3, 4π/3。
8. Quadratic Trigonometric Equations | 二次三角方程
0. After finding t, solve the basic trigonometric equation for x. Always check that the t‑values lie in the interval [−1, 1].
当方程可写为 a sin² x + b sin x + c = 0（或类似地用 cos x）时，它就是二次三角方程。方法是令 t = sin x（或 cos x），解 a t² + b t + c = 0。求出 t 后，再解关于 x 的基本三角方程。务必检查 t 值是否在区间 [−1, 1] 内。
y look cubic but is actually quadratic in sin x if you use the identity. The discriminant of the quadratic can hint at the number of real solutions: negative discriminant means no real t, hence no solution.
有时，方程需要重新整理才能变成二次形式。例如，2 cos x = 3 sin² x 看起来像三次，但如果利用恒等式，它实际上是关于 sin x 的二次方程。二次方程的判别式可以提示实数解的个数：判别式为负意味着没有实数 t，因此无解。
eads to four solutions per 2π cycle instead of two. Exercise 20A.2 will test this nuance repeatedly.
一个常见的错误是在开平方时忘记考虑正负号。如果你得到 sin² x = 1/4，那么 sin x = ±1/2，而不仅仅是 +1/2。这会导致每 2π 周期内有四个解，而非两个。练习 20A.2 会反复测试这一细微之处。
9. Equations Requiring Factoring | 需要因式分解的方程
cos x − 1) = 0. Then set each factor equal to zero: sin x = 0 or cos x = 1. Solve each sub‑equation separately and combine all solutions within the domain.
并非所有三角方程都是二次的；有些需要提出公因式。例如，sin x cos x − sin x = 0 可因式分解为 sin x (cos x − 1) = 0。然后令每个因式等于零：sin x = 0 或 cos x = 1。分别求解每个子方程，再合并定义域内的所有解。
x = 0 or cos x = 1/2. This approach is much quicker than solving sin 2x = sin x directly.
另一类涉及倍角恒等式。例如，sin 2x − sin x = 0。利用恒等式 sin 2x = 2 sin x cos x，得到 2 sin x cos x − sin x = 0 → sin x (2 cos x − 1) = 0。解来自 sin x = 0 或 cos x = 1/2。这种方法比直接解 sin 2x = sin x 快得多。
bring all terms to one side and factor. Missing solutions is a common source of lost marks in IB exams.
在做因式分解时，切勿将两边同时除以可能为零的三角表达式，否则可能丢失解。例如，对于 2 sin x cos x = sin x，不要约去 sin x；而应将所有项移到一边然后因式分解。丢失解是 IB 考试中常见的失分原因。
10. General Solutions vs. Specific Domain | 通解与特定区间
In some IB problems, you are asked to give the general solution (all real numbers), while in others you must list solutions within a given interval. The general solution is expressed using an integer parameter n (or k), and it shows the infinite family of angles. For sin and cos, you use +2πn; for tan, you use +πn.
在一些 IB 题目中，你被要求给出通解（所有实数），而在另一些题目中则需列出给定区间内的解。通解用整数参数 n（或 k）来表达，它展示了一组无穷多个角度。对于 sin 和 cos，使用 +2πn；对于 tan，使用 +πn。
olution compactly is a skill. For example, the general solution for sin x = 0 is x = nπ, n ∈ ℤ ees if the domain is given in degrees. Usually general solutions are expressed in radians.
简洁地写出通解是一项技能。例如，sin x = 0 的通解是 x = nπ, n ∈ ℤ。cos x = 1 的通解是 x = 2πn。tan x = 1 的通解是 x = π/4 + πn。如果定义域以度为单位给出，请确保你能在弧度和度之间灵活转换。通常通解用弧度表示。
When the domain is a specific interval such as [0, 2π), the number of solutions is finite. Always count how many solutions you expect based on the period and the number of base solutions. A quick verification with the unit circle can confirm you have not missed or added any.
当定义域为特定区间如 [0, 2π) 时，解的个数是有限的。务必根据周期和基本解的个数，估算出你预期的解的数量。利用单位圆快速验证，可以确认你没有遗漏或增加任何解。
11. Common Mistakes to Avoid | 常见错误避免
n for x by 2 to get the domain for (x/2). This often leads to missing solutions or including extraneous ones.
一个常见的错误是把 sin² x 当作 sin(x²)，或者忘记平方是作用于函数而非角度本身。另一个错误是在求解 sin(x/2) = k 时，忘记将 x 的定义域除以 2 来得到 (x/2) 的定义域。这常常导致遗漏解或包含无关解。
Mishandling the sign of a trigonometric value when determining quadrants is another pitfall. Always remember the CAST diagram or the mnemonic All Students Take Calculus: in Quadrant I, all are positive; II, only sine; III, only tangent; IV, only cosine. Use this to determine the correct angles for a given positive or negative value of sin, cos, or tan.
在确定象限时弄错三角函数值的正负号是另一个陷阱。务必牢记 CAST 图或记忆口诀“All Students Take Calculus”：第一象限全正；第二象限仅正弦为正；第三象限仅正切为正；第四象限仅余弦为正。利用它来确定给定正或负的 sin、cos、tan 值所对应的正确角度。
Finally, not simplifying expressions by rationalising denominators or reducing fractions can lead to incorrectly identifying the standard angles. For instance, 1/√2 is recognised as √2/2, which corresponds to π/4, but if left as 1/√2 some students might not see the connection immediately. Always simplify exact answers.
最后，不通过有理化分母或约分来简化表达式，可能导致无法正确识别标准角。例如，1/√2 应视为 √2/2，对应 π/4，但如果保留为 1/√2，有些学生可能无法立即看出关联。务必简化精确答案。
12. Practice Problems and Solutions | 练习题与解答
owing representative problems. Solve each for 0 ≤ x ≤ 2π, giving exact answers in radians.
为巩固练习 20A.2 所需的技能，请尝试以下代表性题目。每题均在 0 ≤ x ≤ 2π 内求解，用弧度给出精确答案。
Problem 1: 2 sin x + 1 = 0. Solution: n x = −1/2. Reference angle π/6. sin is negative in III and IV quadrants. x = 7π/6, 11π/6.
问题 1：2 sin x + 1 = 0。解答：sin x = −1/2。参考角 π/6。正弦在第三、四象限为负。x = 7π/6, 11π/6。
Problem 2: cos 2x = cos x. Hint: → x = 0, 2π (but 2π is endpoint, usually included). Combined solutions: 0, 2π/3, 4π/3, 2π.
问题 2：cos 2x = cos x。提示：利用恒等式 cos 2x = 2 cos² x − 1。然后 2 cos² x − cos x − 1 = 0 → (2 cos x + 1)(cos x − 1) = 0。cos x = −1/2 → x = 2π/3, 4π/3；cos x = 1 → x = 0, 2π（2π 是端点，通常包含在内）。合并解：0, 2π/3, 4π/3, 2π。
Problem 3: 0. sin x = 1/2 → x = π/6, 5π/6; sin x = −1 → x = 3π/2. Full solution set: π/6, 5π/6, 3π/2.
问题 3：2 sin² x + sin x − 1 = 0。因式分解：(2 sin x − 1)(sin x + 1) = 0。sin x = 1/2 → x = π/6, 5π/6；sin x = −1 → x = 3π/2。完整解集：π/6, 5π/6, 3π/2。
By practising these patterns, you will build the fluency needed to tackle the entire Exercise 20A.2 with confidence.
通过练习这些模式，你将建立起自信，从而能从容应对整个练习 20A.2。
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