📚 Exercise 8G: Solving Trigonometric Equations (IB Math SL/HL) | 练习8G：解三角方程（IB数学SL/HL）
Welcome to Exercise 8G. In this set of problems, you will develop the skills needed to solve trigonometric equations confidently — a core requirement in the IB Mathematics Analysis and Approaches syllabus. We will work exclusively in radians and cover fundamental identities, multiple angle equations, quadratic forms, and graphical interpretation. Mastery of these techniques will prepare you for Paper 1 and Paper 2 questions that demand precise algebraic manipulation and a deep understanding of periodicity.
欢迎来到练习8G。在这组题目中，你将掌握解三角方程的必要技能——这是IB数学分析与方法大纲的核心要求。我们将全程使用弧度制，并涵盖基本恒等式、多倍角方程、二次型方程以及图像解读。精通这些技巧将帮助你应对试卷一和试卷二中需要精准代数操作和深刻理解周期性的问题。
1. Radian Mode and the Unit Circle | 弧度模式与单位圆
he symmetry of the unit circle allows you to find all solutions within a given interval. Remember that sine is positive in the first and second quadrants, while cosine is positive in the first and fourth quadrants.
在IB课程中解三角方程时，务必确保计算器处于弧度模式。单位圆将正弦和余弦值表示为半径为1的圆上点的坐标。例如，sin(π/6) = 1/2，cos(π/3) = 1/2。理解单位圆的对称性能帮助你在给定区间内找出所有解。记住，正弦在第一和第二象限为正，余弦在第一和第四象限为正。
2. Solving Basic sin x = k and cos x = k | 解基本方程 sin x = k 与 cos x = k
xist within one full revolution. Always check the given domain, which might be in radians or degrees — but in IB, radian answers are expected unless stated otherwise.
对于像 sin x = 0.5 在区间 0 ≤ x ≤ 2π 内的方程，首先求出主值：x = arcsin(0.5) = π/6。然后利用对称性质 sin(π − x) = sin x 得出第二个解 x = π − π/6 = 5π/6。在一个完整旋转周期内不存在其他解。务必核对所给定义域，可能是弧度或角度——但在IB考试中，除非另有说明，答案通常要求用弧度表示。
3. General Solutions Using the Period of 2π | 利用周期 2π 写出通解
general solution for sin x = k is x = arcsin(k) + 2πn or x = π − arcsin(k) + 2πn, where n ∈ ℤ otation allows you to express all solutions elegantly and is often required in Paper 1 non‑calculator questions.
三角函数是周期性的，因此方程通常有无穷多个解。sin x = k 的通解为 x = arcsin(k) + 2πn 或 x = π − arcsin(k) + 2πn，其中 n ∈ ℤ。对于 cos x = k，通式是 x = ± arccos(k) + 2πn。这种简洁的记法能让你优雅地表达所有解，且常在试卷一非计算器题目中要求给出。
4. Applying the Pythagorean Identity | 应用毕达哥拉斯恒等式
− 1 = 0. This is a quadratic in cos x that you can solve by substitution or factorisation.
当方程中同时出现正弦和余弦时，利用恒等式 sin²θ + cos²θ = 1 将所有项转化为单一三角函数。例如，2sin²x + 3cos x − 3 = 0 可化为 2(1 − cos²x) + 3cos x − 3 = 0，简化后得 −2cos²x + 3cos x − 1 = 0。这是一个关于 cos x 的二次方程，可通过换元或因式分解来求解。
5. Solving Quadratic Trigonometric Equations | 解二次三角方程
t² + t − 1 = 0, which factors as (2t − 1)(t + 1) = 0. Thus cos x = 1/2 or cos x = −1. Back‑ d x = π for cos x = −1. Always check that each solution lies inside the required interval.
令 t = sin x、cos x 或 tan x，将方程转化为标准二次方程。在 0 ≤ x ≤ 2π 内解 2cos²x + cos x − 1 = 0。代入 t = cos x 得 2t² + t − 1 = 0，因式分解为 (2t − 1)(t + 1) = 0。因此 cos x = 1/2 或 cos x = −1。回代后，由 cos x = 1/2 得 x = π/3、5π/3，由 cos x = −1 得 x = π。务必检查每个解是否在指定区间内。
6. Equations Involving tan x and the Identity 1 + tan²θ = sec²θ | 含 tan x 的方程与恒等式 1 + tan²θ = sec²θ
iar solutions. Always be aware that tan x has period π, which affects the general solution pattern.
当出现 sec 或 cosec 时，利用倒数恒等式进行转化。方程 tan²x + sec x = 1 可利用 sec²x = 1 + tan²x 改写为 (sec²x − 1) + sec x = 1，简化得 sec²x + sec x − 2 = 0。令 u = sec x，则 u² + u − 2 = 0，解得 u = 1 或 u = −2。于是 cos x = 1 或 cos x = −1/2，从而导出熟悉的解。注意 tan x 的周期为 π，这将影响通解的形式。
7. Double Angle Equations (sin 2x, cos 2x) | 倍角方程（sin 2x、cos 2x）
2. Solve each factor separately within the given domain, remembering to cover all solutions for the original variable x — be careful not to double or halve the interval incorrectly.
涉及 sin 2x 或 cos 2x 的方程常需使用倍角恒等式。对于 sin 2x = cos x，利用 sin 2x = 2sin x cos x 得到 2sin x cos x = cos x。将所有项移到一边：cos x (2sin x − 1) = 0。由此产生 cos x = 0 或 sin x = 1/2。在给定定义域内分别求解每个因式，牢记要覆盖原变量 x 的所有解——注意不要错误地加倍或减半区间。
8. Solving Equations with Phase Shifts and Transformations | 解含相位平移和变换的方程
An equation like sin(x + π/4) = √2 solved by first treating the argument as a single variable. Let u = x + π/4, then sin u = √2 n the domain is specified, list all values of x that lie in that interval. This method works equally well for cos(x − a) and tan(x + b).
像 sin(x + π/4) = √2/2 这样的方程可先将整个角度视为单一变量。令 u = x + π/4，则 sin u = √2/2 给出 u = π/4 + 2πn 或 u = 3π/4 + 2πn。代回 u 得 x = 2πn 或 x = π/2 + 2πn。若指定了定义域，列出落在该区间内的所有 x 值。此方法同样适用于 cos(x − a) 和 tan(x + b)。
9. Graphical Interpretation and the Number of Solutions | 图像解读与解的个数
sections because the frequency increases. Graphical thinking helps avoid missing solutions and is particularly useful in multiple‑choice sections where you might need to identify the correct set of solutions quickly.
在同一坐标系中绘制 y = sin x、y = cos x 和 y = k 的图像，可以直观检验解的个数。方程 sin x = 0.3 在 [0, 2π] 内有两个解，cos x = −0.4 同样有两个。更复杂的方程如 2sin 3x = 1 在 [0, 2π] 内会有更多交点，因为频率增大。图像思维有助于避免遗漏解，在选择题部分特别有用，因为你可能需要快速识别正确解集。
10. Common Pitfalls and Verification Strategies | 常见错误与验证策略
lways substitute your final answers back into the original equation to verify them. If you obtain an answer such as x = 7π/6, check that sin(7π/6) indeed matches the right‑hand side.
一个常见错误是在解 sin(kx) = a 时忘记对周期进行除法。正确的做法是先按常规方法求出 kx 的所有解，列出所有解之后再除以 k。另一个陷阱是两边同时平方，这可能引入增根。务必把最终答案代回原始方程进行验证。若你得到 x = 7π/6 这样的答案，请检查 sin(7π/6) 是否确实等于右边。
11. Exercise 8G Practice: Linking Techniques | 练习8G实战：综合技巧
x = π/4 + πn. In [0, π] this yields x = π/4, 5π/4 (but 5π/4 > π, so only π/4). For sin 2x = 1/2, 2x = π/6 + 2πn or 5π/6 + 2πn → x = π/12 + πn or 5π/12 + πn. Within [0, π]: x = π/12, 5π/12, 13π/12 (discard as > π), 17π/12 (discard). Final solutions: π/12, π/4, 5π/12. This systematic approach uses identities, substitution and careful domain handling.
现在综合运用所有技巧。在 0 ≤ x ≤ π 内解方程 2cos² 2x + 3sin 2x − 3 = 0。利用恒等式 cos² 2x = 1 − sin² 2x 将方程改写为 2(1 − sin² 2x) + 3sin 2x − 3 = 0 → −2sin² 2x + 3sin 2x − 1 = 0。乘以 −1 得 2sin² 2x − 3sin 2x + 1 = 0。令 u = sin 2x，得到 2u² − 3u + 1 = 0 → u = 1 或 u = 1/2。那么 sin 2x = 1 给出 2x = π/2 + 2πn → x = π/4 + πn。在 [0, π] 内得 x = π/4（5π/4 > π 舍去）。对于 sin 2x = 1/2，2x = π/6 + 2πn 或 5π/6 + 2πn → x = π/12 + πn 或 5π/12 + πn。在 [0, π] 内：x = π/12、5π/12（其余超出区间舍去）。最终解为 π/12、π/4、5π/12。这种系统方法运用了恒等式、换元以及严谨的定义域处理。
12. Looking Ahead to Exam-Style Questions | 展望考试风格问题
IB exam questions frequently ask you to find all solutions in a given interval and then write the general solution. They may also embed trigonometric equations within a modelling context, such as the height of a tide or the position of a Ferris wheel. In such problems, you will need to interpret the solutions in the context of the real‑world situation, discarding values that do not make physical sense. Always read the question carefully and show clear steps, including the substitution and the period handling, to achieve full marks.
IB考试题目经常要求先找出给定区间内的所有解，然后写出通解。它们也可能将三角方程嵌入建模情境中，例如潮汐高度或摩天轮的位置。在这类问题里，你需要结合实际意义来解读解，剔除物理上不合理的值。务必仔细审题，并展示清晰的步骤，包括换元和对周期的处理，以获得满分。
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