📚 Solving Trigonometric Equations | 求解三角方程
Trigonometric equations lie at the heart of Edexcel A-Level Mathematics, bridging the gap between pure algebraic manipulation and the periodic behaviour of sine, cosine and tangent. A confident approach to solving these equations requires an understanding of the unit circle, CAST diagram, fundamental identities, and the ability to recognise hidden quadratics or factorisations. This revision guide walks you through a systematic method, from the very first step of isolating a trigonometric function to writing general solutions in radians or degrees.
三角方程是Edexcel A-Level数学的核心内容，它将纯代数运算与正弦、余弦和正切的周期性行为连接起来。要自信地解决这类方程，需要理解单位圆、CAST图、基本恒等式，并能识别隐藏的二次结构或因式分解。本复习指南将带你系统地走过每一个步骤：从分离三角函数的第一步，一直到用弧度或角度写出通解。
1. What Are Trigonometric Equations? | 什么是三角方程
own is an angle (usually denoted θ or x). Simple examples include sin θ = 0.5, cos 2θ = 1/√2 π, and sometimes to write the general solution that describes all possible angles satisfying the equation.
三角方程是任何包含三角函数 sin θ、cos θ、tan θ 或其倒数，且未知数为角度（通常记为 θ 或 x）的方程。简单的例子有 sin θ = 0.5、cos 2θ = 1/√2 或 tan (θ−30°) = −1。在Edexcel考试中，通常会要求找出在指定区间内的所有解，例如 0° ≤ θ ≤ 360° 或 0 ≤ θ < 2π，有时还要求写出描述所有可能角度的通解。
eriodicity of the graphs of y = sin x, y = cos x and y = tan x to generate all solutions from a principal value. Understanding the CAST diagram is therefore essential.
由于三角函数具有周期性，一个方程往往有无穷多个解。关键技能是利用 y = sin x、y = cos x 和 y = tan x 图像的对称性和周期性，从主值出发生成所有解。因此，理解CAST图至关重要。
2. The CAST Diagram and Quadrant Angles | CAST图与象限角
The CAST diagram (or ASTC rule) helps you remember which trigonometric ratios are positive in each quadrant of the coordinate plane. Starting from the first quadrant (top right) and moving anticlockwise: A (All positive), S (Sin positive only), T (Tan positive only), C (Cos positive only). When a calculator gives you a principal value, you must use the CAST diagram to find the other angles in the given interval that have the same sine, cosine or tangent value.
CAST图（或ASTC规则）帮助你记住在坐标平面的每个象限中哪些三角比是正的。从第一象限（右上角）开始逆时针移动：A（全正）、S（仅正弦为正）、T（仅正切为正）、C（仅余弦为正）。当计算器给出一个主值时，你必须使用CAST图在给定区间内找出具有相同正弦、余弦或正切值的其他角度。
d quadrant (S), the other angle in the range 0° to 360° is 180° − 30° = 150°. If cos θ = −√3 is 360° − 150° = 210°. Always sketch the relevant quadrant positions to avoid sign errors.
例如，若 sin θ = 0.5，计算器给出 θ = 30°（主值）。因为正弦在第二象限（S）也是正值，所以在 0° 到 360° 范围内的另一个角是 180° − 30° = 150°。若 cos θ = −√3/2，主值是 150°（余弦为负）。余弦在第二和第三象限为负。利用CAST，在 0° 到 360° 中的另一个解是 360° − 150° = 210°。总是画出相关象限的位置，以避免符号错误。
3. Using Basic Trigonometric Identities | 使用基本三角恒等式
Many equations cannot be solved immediately unless they are simplified using identities. The two most fundamental identities are:
tan θ = sin θ / cos θ
sin²θ + cos²θ = 1
许多方程不能立即求解，除非先用恒等式进行化简。两个最基本的恒等式是 tan θ = sin θ / cos θ 和 sin²θ + cos²θ = 1。
when we want to express everything in terms of one function to create a quadratic. For instance, the equation 3 sin²θ + 2 cos θ − 2 = 0 can be rewritten using sin²θ = 1 − cos²θ to obtain a quadratic in cos θ.
由 sin²θ + cos²θ = 1 还可导出变形：sin²θ = 1 − cos²θ 以及 cos²θ = 1 − sin²θ。当方程混合了 sin 和 cos，或想将所有项表示为同一种函数以构成二次型时，这些变形非常有用。例如，方程 3 sin²θ + 2 cos θ − 2 = 0 可以利用 sin²θ = 1 − cos²θ 改写为关于 cos θ 的二次方程。
Always check if dividing by a trigonometric function is legitimate. Dividing both sides by cos θ or sin θ can cause loss of solutions if the function equals zero. Factorising is safer when possible.
始终要检查除以一个三角函数是否合理。如果该函数可以等于零，两边同除以 cos θ 或 sin θ 会导致遗漏解。在可能的情况下，因式分解更为稳妥。
4. Factorising Trigonometric Equations | 因式分解三角方程
= 0 should be approached by taking out the common factor sin θ: sin θ (2 cos θ − 1) = 0. Then apply the zero-product rule: either sin θ = 0 or 2 cos θ − 1 = 0, i.e. cos θ = 1/2.
因式分解是求解三角方程时最可靠的技术之一，且不会丢失答案。像 2 sin θ cos θ − sin θ = 0 这样的方程，应提取公因式 sin θ：sin θ (2 cos θ − 1) = 0。然后应用零乘积规则：要么 sin θ = 0，要么 2 cos θ − 1 = 0，即 cos θ = 1/2。
Each resulting simple equation can then be solved using the CAST diagram. This method naturally produces all solutions within the interval. Avoid the temptation to divide the original equation by sin θ, because sin θ could be zero, and you would miss those solutions entirely.
然后可使用CAST图分别求解每个简单方程。这种方法自然地得出区间内的所有解。要抵制将原方程两边除以 sin θ 的冲动，因为 sin θ 可能为零，那样就会完全丢失这部分解。
θ)(cos θ + sin θ) = 0, or equations where a double-angle identity reveals a common factor.
其他常见的因式分解包括平方差，例如 cos²θ − sin²θ = (cos θ − sin θ)(cos θ + sin θ) = 0，或者倍角恒等式能揭示公因式。
5. Quadratic Forms in Sine and Cosine | 正弦与余弦的二次型
rms it into 2u² − 3u + 1 = 0, which factorises to (2u − 1)(u − 1) = 0. Hence cos θ = 1/2 or cos θ = 1, and each case is solved separately.
三角方程经常以二次型出现。例如，2 cos²θ − 3 cos θ + 1 = 0 是关于 cos θ 的二次方程。令 u = cos θ，将其转化为 2u² − 3u + 1 = 0，因式分解得 (2u − 1)(u − 1) = 0。于是 cos θ = 1/2 或 cos θ = 1，再分别求解。
er that taking the square root introduces both positive and negative cases, doubling the number of solution families. Always apply CAST after determining the basic reference angle.
像 4 sin²θ − 1 = 0 这样的方程，可先分离出 sin²θ = 1/4，然后开平方根得到 sin θ = ±1/2。注意，开平方会引入正负两种情况，使解的族数加倍。在确定基本参考角后，务必应用CAST图。
−1. The principal values and the period of tangent (180° or π) are used to list all solutions.
如果出现关于 tan θ 的二次方程，如 tan²θ − tan θ − 2 = 0，同样处理：因式分解为 (tan θ − 2)(tan θ + 1) = 0，然后分别求解 tan θ = 2 和 tan θ = −1。利用主值和正切函数的周期（180° 或 π）列出所有解。
6. Equations Involving Double Angles | 倍角方程
Double-angle identities are frequently tested. The three forms for cos 2θ are especially important:
cos 2θ = cos²θ − sin²θ
cos 2θ = 2cos²θ − 1
cos 2θ = 1 − 2sin²θ
倍角恒等式经常出现在考题中。cos 2θ 的三种形式尤为重要：cos 2θ = cos²θ − sin²θ，cos 2θ = 2cos²θ − 1，cos 2θ = 1 − 2sin²θ。
urns the equation into 2cos²θ + cos θ − 1 = 0, a quadratic in cos θ. Similarly, sin 2θ = 2 sin θ cos θ allows factorisation when combined with other terms.
当方程混合了 θ 和 2θ，如 cos 2θ + cos θ = 0，将所有项用同一个角度表示会很有帮助。用 cos 2θ = 2cos²θ − 1 代入，方程变为 2cos²θ + cos θ − 1 = 0，这是关于 cos θ 的二次方程。类似地，sin 2θ = 2 sin θ cos θ 在与其他项结合时可以实现因式分解。
olutions from sin θ = 0 and cos θ = 1/2. Never cancel sin θ without considering the zero case.
对于 sin 2θ = sin θ 这样的方程，改写为 2 sin θ cos θ = sin θ，将所有项移到一边并因式分解：sin θ (2 cos θ − 1) = 0。这样能清晰地得出 sin θ = 0 和 cos θ = 1/2 的解。绝不要在不考虑零值的情况下约去 sin θ。
7. Reciprocal Trigonometric Functions | 倒数三角函数
Edexcel A-Level also tests sec θ, cosec θ and cot θ. These are defined as:
sec θ = 1 / cos θ, cosec θ = 1 / sin θ, cot θ = cos θ / sin θ
Edexcel A-Level 考试还会涉及 sec θ、cosec θ 和 cot θ。它们的定义是：sec θ = 1 / cos θ，cosec θ = 1 / sin θ，cot θ = cos θ / sin θ。
iting in terms of sin and cos or using the relevant identities. Since 1 + cot²θ = cosec²θ, the equation can be transformed into a quadratic in cosec θ: (cosec²θ − 1) + cosec θ = 1 ⇒ cosec²θ + cosec θ − 2 = 0.
要求解像 sec θ = 2 这样的方程，只需将其改写为 cos θ = 1/2 然后按常规方法计算。更复杂的方程，如 cot²θ + cosec θ = 1，需要用 sin 和 cos 表示，或利用相关恒等式。由于 1 + cot²θ = cosec²θ，该方程可变为关于 cosec θ 的二次方程：(cosec²θ − 1) + cosec θ = 1 ⇒ cosec²θ + cosec θ − 2 = 0。
n cos θ = 0, so check that your solutions do not fall on the vertical asymptotes of sec θ.
永远记住倒数函数有定义域限制。例如，当 cos θ = 0 时 sec θ 无定义，所以要检查所得的解是否落在 sec θ 的垂直渐近线上。
8. General Solutions and Periodicity | 通解与周期性
When the question asks for “all possible solutions”, you must give the general solution using the periodic nature of trig functions. The standard forms are:
- If sin θ = sin α, then θ = 180°n + (−1)ⁿ α, or in radians: θ = πn + (−1)ⁿ α.
- If cos θ = cos α, then θ = 360°n ± α, or in radians: θ = 2πn ± α.
- If tan θ = tan α, then θ = 180°n + α, or in radians: θ = πn + α, where n ∈ ℤ.
当题目要求“所有可能的解”时，就必须利用三角函数的周期性给出通解。标准形式为：若 sin θ = sin α，则 θ = 180°n + (−1)ⁿ α（弧度制为 πn + (−1)ⁿ α）；若 cos θ = cos α，则 θ = 360°n ± α（2πn ± α）；若 tan θ = tan α，则 θ = 180°n + α（πn + α），其中 n ∈ ℤ。
Here α is a particular solution (usually the principal value or a reference angle). The integer n generates the infinite family. Exam questions often specify the set notation or ask for solutions in a given interval before moving to the general form.
这里 α 是一个特解（通常是主值或参考角）。整数 n 可生成无限的解族。考试题目通常会指定集合符号，或要求先给出特定区间内的解，再过渡到通解形式。
Be careful to express α correctly when the equation involves transformations like sin(θ+30°) or cos(2θ). Solve for the inner bracket first, then adjust for the transformation.
当方程含有 sin(θ+30°) 或 cos(2θ) 等变换时，要注意正确表示 α。先对内部括号求解，然后再针对变换进行调整。
9. Common Pitfalls and How to Avoid Them | 常见误区及避免方法
both the positive and negative square roots when solving sin²θ = k. Always write sin θ = ±√k and find the corresponding angles. Another common error is dividing by a trigonometric function instead of factorising, which silently discards valid solutions.
最常见的错误之一是在求解 sin²θ = k 时忘记考虑正负平方根。务必写出 sin θ = ±√k 并找出相应的角度。另一个常见误区是用三角函数去除方程而不是因式分解，这会无声地舍弃有效的解。
Misapplying the CAST diagram is also typical: students might place an angle in the wrong quadrant or forget the sign of the function. A rough sketch of the trigonometric graph or the unit circle can dramatically reduce such mistakes.
误用CAST图也很典型：学生可能将角度放错象限，或忘记函数的符号。快速画出三角函数图像或单位圆的草图，可以显著减少此类错误。
from 0° to 720°. Failing to expand the interval appropriately leads to missing solutions.
在处理倍角时，变量的区间会改变。例如，若 0° ≤ θ ≤ 360°，而你在解 sin 2θ = 0.5，则 2θ 的范围是 0° 到 720°。未能适当地扩展区间会导致漏解。
10. Exam-Style Practice Tips | 考试练习技巧
In an Edexcel exam, always show clear step-by-step reasoning: state the identity used, show the factorisation, reference the CAST diagram, and write solutions in the required range. When a question asks for exact values, give answers in fractions or surds, and use degrees or radians as specified.
在Edexcel考试中，始终要展示清晰的分步推理：写出所用的恒等式、展示因式分解、引用CAST图，并在要求的范围内写出解。当题目要求精确值时，答案应以分数或根式形式给出，并按说明使用角度或弧度。
Practice with a variety of equations: linear, quadratic, those requiring identity substitution, and reciprocal functions. Use past paper questions to build speed and accuracy. Time yourself while solving sets of trigonometric equations — the skill becomes automatic with regular drilling.
练习各类方程：线性的、二次的、需要恒等式代入的，以及倒数函数。利用历年真题提高速度和准确率。在求解多组三角方程时计时——通过定期训练，这一技能会变得自动化。
Finally, check your solutions by substituting a few back into the original equation, especially if you have spare time. This quick verification can catch sign errors or misinterpreted quadrants.
最后，如果有时间，将几个解代回原方程进行检验。这种快速验证能够发现符号错误或对象限的误解。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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