Solving Trigonometric Equations | 解三角方程

📚 Solving Trigonometric Equations | 解三角方程

Trigonometric equations appear throughout the Edexcel A Level Mathematics specification, especially in Pure Mathematics. They require you to combine algebraic manipulation with the periodic properties of sine, cosine and tangent. This article covers the key methods, common pitfalls and exam techniques for solving trigonometric equations confidently.

三角方程贯穿 Edexcel A Level 数学大纲，尤其在纯数学部分。解这类题需要把代数变形与正弦、余弦和正切的周期性质结合起来。本文介绍核心方法、常见易错点和考试技巧，帮助你自信地求解三角方程。

1. The unit circle and CAST diagram | 单位圆与 CAST 图

All trigonometric equations can be interpreted on the unit circle. The x-coordinate of a point on the circle is cos θ, the y-coordinate is sin θ, and the gradient from the origin is tan θ.

所有三角方程都可以在单位圆上解释。圆上点的 x 坐标是 cos θ，y 坐标是 sin θ，从原点出发的斜率是 tan θ。

In the Edexcel exam, a CAST diagram helps you remember which functions are positive in each quadrant: Cosine in the fourth, All in the first, Sine in the second, Tangent in the third.

在 Edexcel 考试中，CAST 图帮助你记住各象限中哪些函数为正：第四象限余弦为正，第一象限全为正，第二象限正弦为正，第三象限正切为正。

Once you find a principal value, the CAST diagram tells you which other quadrants produce the same sign for that ratio.

一旦求出主值，CAST 图会告诉你还有哪些象限具有相同的正负号。

2. Standard exact values and radian measure | 标准精确值与弧度制

Exact trigonometric values for 0, π/6, π/4, π/3, π/2 and their degree equivalents must be memorised. They allow calculator-free solutions and exact answers.

必须记住 0、π/6、π/4、π/3、π/2 及其角度制对应值的精确三角函数值。这些值帮助你脱离计算器求解并写出精确答案。

Since Edexcel questions often specify a range in radians, practise converting between degrees and radians using π = 180°.

由于 Edexcel 题目常以弧度给出区间，要熟练使用 π = 180° 进行角度与弧度的换算。

sin(π/6) = 1/2, cos(π/4) = √2/2, tan(π/3) = √3

These standard values are the building blocks for most exact trigonometric solutions.

这些标准值是大多数精确三角解的基础。

3. Solving basic sine equations: sin θ = k | 解基本正弦方程：sin θ = k

For sin θ = k with |k| ≤ 1, first find the principal value α = sin⁻¹ k, usually given in the range -π/2 ≤ α ≤ π/2.

对于 sin θ = k（|k| ≤ 1），先求主值 α = sin⁻¹ k，通常取值在 -π/2 ≤ α ≤ π/2。

The two solutions in one full cycle are α and π – α if α is positive, or more generally α and π – α adjusted by period. Then add multiples of 2π to cover the required interval.

在一个完整周期内，两个解通常是 α 和 π – α（当 α 为正时），更一般地可先找对应参考角再加 2π 的整数倍覆盖给定区间。

Example: Solve sin θ = 1/2 for 0 ≤ θ < 2π.

例：解 sin θ = 1/2，0 ≤ θ < 2π。

α = sin⁻¹(1/2) = π/6. The second solution is π – π/6 = 5π/6, so θ = π/6, 5π/6.

α = sin⁻¹(1/2) = π/6。第二个解是 π – π/6 = 5π/6，因此 θ = π/6, 5π/6。

4. Solving basic cosine and tangent equations | 解基本余弦与正切方程

For cos θ = k, the principal value α = cos⁻¹ k lies in [0, π]. The second solution in one full period is -α, or equivalently 2π – α in the interval [0, 2π].

对于 cos θ = k，主值 α = cos⁻¹ k 位于 [0, π]。一个完整周期内的第二个解是 -α，在 [0, 2π] 中等价于 2π – α。

Because tangent has period π, once you find α = tan⁻¹ k, all other solutions are α + nπ. There is only one solution per period.

因为正切函数的周期是 π，一旦求出 α = tan⁻¹ k，所有其他解就是 α + nπ。每个周期内只有一个解。

Example: Solve cos θ = 1/2 for 0 ≤ θ < 2π.

例：解 cos θ = 1/2，0 ≤ θ < 2π。

α = cos⁻¹(1/2) = π/3. Since cosine is positive in the first and fourth quadrants, θ = π/3 or θ = 2π – π/3 =

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