Simple Trigonometric Equations | 简单三角方程

📚 Simple Trigonometric Equations | 简单三角方程

osine, and tangent functions. You will learn to find all solutions within a given interval — often 0 ≤ x ≤ 2π or 0° ≤ x ≤ 360° — n²x - 1 = 0. Mastering these techniques now will prepare you for more advanced topics in calculus, vectors, and further pure mathematics.

在A-Level数学中，解简单三角方程是一项核心技能，它直接建立在你对正弦、余弦和正切函数的理解之上。你将学会如何在给定区间内（通常是 0 ≤ x ≤ 2π 或 0° ≤ x ≤ 360°）通过代数操作、单位圆和 CAST 图找出所有解。典型方程包括 sin x = k、cos (2x) = 0.5，或二次型如 3tan²x - 1 = 0。现在掌握这些技巧将为你后续学习微积分、向量和进阶纯数学做好准备。

1. Understanding Trigonometric Equations | 理解三角方程

sually have infinitely many solutions because the sine, cosine, and tangent functions are periodic. In the context of Edexcel A-Level, you are almost always asked to find all solutions within a specified interval, such as 0 ≤ x ≤ 360° or 0 ≤ x ≤ 2π. The simplest forms are sin x = k, cos x = k, and tan x = k, where k is a known constant within the function's range.

三角方程是任何包含未知角三角函数的方程。与标准代数方程不同，由于正弦、余弦和正切函数具有周期性，它们通常有无穷多个解。在Edexcel A-Level的考试中，你几乎总是被要求在指定区间内找出所有解，例如 0 ≤ x ≤ 360° 或 0 ≤ x ≤ 2π。最简单的形式是 sin x = k、cos x = k 和 tan x = k，其中 k 是函数值域内的已知常数。

Before attempting to solve, always check the domain: is it given in degrees or radians? ecause the sine function only takes values between -1 and 1. Identifying the range of possible k values is a vital first step.

在求解之前，务必检查定义域：题目是以角度制还是弧度制给出的？求解技巧完全相同，但最终答案必须以正确的单位表示。此外，要记住 sin x = 2 这样的方程无解，因为正弦函数的值域仅在 -1 到 1 之间。识别 k 的可能取值范围是关键的第一步。

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2. The Unit Circle and Exact Values | 单位圆与特殊值

The unit circle is a circle of radius 1 centred at the origin. It gives a geometric interpretation of the sine and cosine of any angle measured anticlockwise from the positive x-axis. For a point on the circle at angle θ, the coordinates are (cos θ, sin θ). This means the x-coordinate gives cos θ, and the y-coordinate gives sin θ. Tangent is then the ratio sin θ / cos θ. Understanding this helps you visualise why multiple angles can produce the same sine or cosine value.

单位圆是一个以原点为中心、半径为 1 的圆。它从几何上解释了从正 x 轴逆时针测量的任意角度 θ 的正弦和余弦。对于圆上角度 θ 处的点，其坐标为 (cos θ, sin θ)。这意味着 x 坐标给出 cos θ，y 坐标给出 sin θ。而正切则是比值 sin θ / cos θ。理解这一点有助于你直观地看到为什么多个角度可以产生相同的正弦或余弦值。

You must know the exact trigonometric values for key angles. These are summarised below and are essential when writing solutions without a calculator.

你必须熟记以下关键角度的精确三角函数值。在不用计算器写出解答时，这些值是必不可少的。

Angle (degrees)	Angle (radians)	sin	cos	tan
0°	0	0	1	0
30°	π/6	1/2	√3/2	1/√3
45°	π/4	1/√2	1/√2	1
60°	π/3	√3/2	1/2	√3
90°	π/2	1	0	undefined

other solutions follow from symmetry. Always actively recall these values rather than relying purely on a calculator, as exam questions often test exact form answers.

这些精确值让你能够立即写出主解。例如，如果 sin x = 1/2，那么一个显然的解是 x = 30° (或 π/6)。其他解则通过对称性得到。要主动回忆这些值，而不仅仅依赖计算器，因为考试题目常常要求以精确形式给出答案。

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3. Using the CAST Diagram | 使用 CAST 图

The CAST diagram is a memory aid that tells you which trigonometric functions are positive in each of the four quadrants. Starting from the top-right quadrant (0° to 90°): A for All (all functions positive); top-left (90° to 180°): S for Sine (only sine positive); bottom-left (180° to 270°): T for Tangent; bottom-right (270° to 360°): C for Cosine. In radians, the quadrants are (0, π/2), (π/2, π), (π, 3π/2), (3π/2, 2π).

CAST 图是一种记忆辅助工具，告诉你四个象限中哪些三角函数为正。从右上象限 (0° 至 90°) 开始：A 表示 All（全部为正）；左上象限 (90° 至 180°)：S 表示 Sine（仅正弦为正）；左下象限 (180° 至 270°)：T 表示 Tangent；右下象限 (270° 至 360°)：C 表示 Cosine。在弧度制下，象限依次为 (0, π/2)、(π/2, π)、(π, 3π/2)、(3π/2, 2π)。

th quadrants. The CAST diagram makes it easy to find all solutions from a related acute angle, often called the reference angle. You first calculate the acute angle α such that sin α = |k|, then place it in the appropriate quadrants based on sign.

要解 sin x = 正数，你立即知道该角可能位于第1或第2象限。若 a 为负，解则位于第3和第4象限。CAST 图能让你通过一个相关锐角（常称为参考角）轻松找到所有解。你先计算满足 sin α = |k| 的锐角 α，然后根据符号将其置于合适的象限中。

ns are x = 180° - 60° = 120° and x = 180° + 60° = 240°. Using CAST is faster and less error-prone than guessing with the calculator.

例如，解 cos x = -0.5，0° ≤ x ≤ 360°。余弦在第二和第三象限为负。余弦为 0.5 的锐角是 60°。因此解为 x = 180° - 60° = 120° 和 x = 180° + 60° = 240°。使用 CAST 比用计算器猜解更快且不易出错。

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4. Solving sin x = k | 解 sin x = k

lue using your calculator or exact values. In degree mode, if k is positive, the calculator gives an acute angle in the 1st quadrant. Let this be α. Since sine is also positive in the 2nd quadrant, the second solution in the 0° to 360° interval is 180° - α. If k is negative, the calculator returns a negative angle (in the 4th quadrant). You can add 360° to find the first positive solution, then use symmetry to find the second solution in the 3rd quadrant: 180° + α, where α is the absolute acute angle.

解 sin x = k 的一般方法是先通过计算器或精确值求出主值。在角度制下，若 k 为正，计算器给出第一象限的一个锐角，记作 α。由于正弦在第二象限也为正，在 0° 到 360° 区间内的第二个解就是 180° - α。如果 k 为负，计算器会返回一个负角（第四象限）。你可以加上 360° 得到第一个正解，然后利用对称性找到第三象限的第二个解：180° + α，其中 α 是绝对值对应的锐角。

are x = α and x = π - α for k > 0, and x = π + α and x = 2π - α for k < 0, where α = sin⁻¹|k| ate the domain and give all answers exactly unless the question specifies a decimal place.

在弧度制下，同样的逻辑适用：在 [0, 2π] 内 sin x = k 的解，当 k > 0 时为 x = α 和 x = π - α；当 k < 0 时为 x = π + α 和 x = 2π - α，其中 α = sin⁻¹|k|。符号 sin⁻¹ 表示反正弦函数，给出主值。始终说明定义域，并给出所有精确答案，除非题目明确要求保留小数。

Example: Solve sin x = √3/2 for 0 ≤ x ≤ 2π. α = sin⁻¹(√3/2) = π/3. Solutions: x = π/3, π - π/3 = 2π/3.

示例：在 0 ≤ x ≤ 2π 内解 sin x = √3/2。α = sin⁻¹(√3/2) = π/3。解：x = π/3，π - π/3 = 2π/3。

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5. Solving cos x = k | 解 cos x = k

n acute α = cos⁻¹(k) in the 1st quadrant. The second solution in [0, 2π] is 2π - α (or 360° - α). If k < 0, the calculator principal value is an obtuse angle in the 2nd quadrant (in radians, between π/2 and π). In that case, you can either use the calculator' the second solution is in the 3rd quadrant: 2π - α (which yields an angle larger than π).

对于 cos x = k，对称性有所不同：余弦在第一和第四象限为正，在第二和第三象限为负。若 k > 0，计算器会给出第一象限的锐角 α = cos⁻¹(k)。在 [0, 2π] 内的第二个解是 2π - α（或 360° - α）。若 k < 0，计算器主值是一个第二象限的钝角（弧度制下在 π/2 到 π 之间）。这时，你可以直接使用计算器输出作为一个解，然后利用对称性找到另一个解：两个解是 α 和 2π - α，但此时 α 已在第二象限，因此第二个解处于第三象限：2π - α（得到一个大于 π 的角）。

A simpler unified method is to set α = cos⁻¹|k| + α. This mirrors the CAST approach and avoids confusion with the obtuse principal value.

一种更简单的统一方法是设 α = cos⁻¹|k|，即锐角。那么对于 cos x = k（正数），解为 x = α, 360° - α。对于 cos x = -k（负数），解为 x = 180° - α, 180° + α。这反映了 CAST 方法，并避免了对钝角主值的混淆。

-1/2, 0 ≤ x ≤ 360°. α = cos⁻¹(1/2) = 60°. Solutions: 180° - 60° = 120°, 180° + 60° = 240°.

示例：在 0 ≤ x ≤ 360° 内解 cos x = -1/2。α = cos⁻¹(1/2) = 60°。解：180° - 60° = 120°，180° + 60° = 240°。

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6. Solving tan x = k | 解 tan x = k

¹(k) using your calculator. In degrees, for any k, one solution is α (or α + 180° if α is negative to bring it into 0°– given interval. Tangent is positive in the 1st and 3rd quadrants, negative in the 2nd and 4th.

正切函数的周期为 π (180°)，这意味着它的解每半个圆周重复一次。要解 tan x = k，先用计算器求出主值 α = tan⁻¹(k)。在角度制下，对于任意 k，其中一个解就是 α（若 α 为负，则可加 180° 使其进入 0°–180°）。区间内的下一个解就是 α + 180°，你可以根据需要在给定区间内继续加或减 180° 来找到所有解。正切在第一和第三象限为正，在第二和第四象限为负。

For example, tan x = 1 gives α = π/4, so solutions are π/4 and 5π/4. If k is negative, say tan x = -√3 d then α + 2π gives 5π/3 to cover 0 to 2π. Always ensure that all final answers lie strictly within the required interval.

在弧度制下，若需在 [0, 2π] 内求解，两个解为 α 和 α + π，其中 α 是调整后落在第一或第四象限的主值。例如，tan x = 1 得到 α = π/4，因此解为 π/4 和 5π/4。若 k 为负，例如 tan x = -√3，则 α = -π/3；加上 π 得到 2π/3，然后 α + 2π 得到 5π/3，以覆盖 0 到 2π。请务必确保所有最终答案严格位于要求的区间内。

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7. General Solutions for Sine and Cosine | 正弦与余弦的通解

e angles that satisfy the equation. For sin x = sin α, the general solution in radians is:

当未指定区间时，你需要给出通解，它包含了所有满足方程的角。对于 sin x = sin α，弧度制下的通解为：

x = α + 2πn or x = (π - α) + 2πn, n ∈ ℤ

x = α + 2πn 或 x = (π - α) + 2πn，n ∈ ℤ

general solution is simpler: cos x = cos α gives x = ±α + 2πn, or in degrees, x = ±α + 360°n. This compact form works because cosine is an even function and repeats every 2π.

该公式源于 2π 的周期性和正弦波的对称性。在角度制下，它变为 x = α + 360°n 或 x = (180° - α) + 360°n。余弦的通解更为简单：cos x = cos α 得到 x = ±α + 2πn，或在角度制下 x = ±α + 360°n。这个紧凑形式之所以有效，是因为余弦是偶函数且每 2π 重复一次。

ll solutions of cos x = 1/2 in [0, 4π], set α = π/3, then x = π/3 + 2πn or -π/3 + 2πn. Taking n = 0 gives π/3 and 5π/3; n = 1 gives 7π/3 and 11π/3; all lie within [0, 4π].

你可以利用这些通解，通过代入 n 的整数值来生成特定区间内的解。例如，要在 [0, 4π] 内找到 cos x = 1/2 的所有解，设 α = π/3，则 x = π/3 + 2πn 或 -π/3 + 2πn。取 n = 0 得到 π/3 和 5π/3；n = 1 得到 7π/3 和 11π/3；这些解全部落在 [0, 4π] 内。

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8. General Solution for Tangent | 正切的通解

general solution is elegantly simple. For tan x = tan α, the complete set of solutions is:

由于正切函数的周期是 π，其通解简洁优雅。对于 tan x = tan α，完整的解集为：

x = α + πn, n ∈ ℤ

x = α + πn，n ∈ ℤ

is no second branch because adding π already switches the quadrant and covers both positive and negative tangent values. When using this formula, remember that α is usually taken as the principal value from the calculator, which lies between -π/2 and π/2 (-90° and 90°).

在角度制下，这就是 x = α + 180°n。没有第二分支，因为加上 π 本身就切换了象限并覆盖了正切值的正负情况。使用该公式时，请记住 α 通常取计算器给出的主值，其范围在 -π/2 到 π/2 (即 -90° 到 90°) 之间。

xtremely useful when solving equations that lead to multiple cycles. You must simply list all n values that produce answers within the required bounds.

例如，tan x = 1 的通解是 x = π/4 + πn。在角度制下，x = 45° + 180°n。当求解涉及多个周期的方程时，这个简洁的格式极为有用。你只需列出所有能在指定范围内产生答案的 n 值即可。

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9. Solving Equations with Angle Transformations | 解角度变换的方程

mation inside the trigonometric function, such as sin (2x - 30°) = 1/2 or cos (3x + π/4) = √2/2. The strategy is to treat the entire argument as a single

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