Solving Simple Trigonometric Equations | 简单三角方程的求解

📚 Solving Simple Trigonometric Equations | 简单三角方程的求解

Trigonometric equations appear frequently in A-Level Mathematics, requiring students to find angles that satisfy a given equation involving sine, cosine, or tangent. Mastering these equations is essential for success in both pure mathematics and applications such as calculus and mechanics.

三角方程是 A-Level 数学中常见的一类问题，要求学生找出满足包含正弦、余弦或正切方程的角。熟练掌握这类方程，对纯数学以及微积分、力学等应用领域都至关重要。

1. Fundamental Concepts: Periodicity and General Solutions | 基本概念：周期性与通解

Trigonometric functions are periodic. For example, sin x and cos x repeat every 360° (or 2π radians), while tan x repeats every 180° (or π radians). This means an equation such as sin x = 0.5 has infinitely many solutions unless the domain is restricted.

三角函数具有周期性。例如，sin x 和 cos x 每 360°（或 2π 弧度）重复一次，而 tan x 每 180°（或 π 弧度）重复一次。这意味着，除非给定区间限制，否则像 sin x = 0.5 这样的方程会有无穷多个解。

When solving a trigonometric equation, we first find the principal value, which is the solution lying in a standard interval. Then we use the symmetry and periodicity of the function to generate all other solutions within the required range.

求解三角方程时，我们先找到主值，即落在标准区间内的解。然后利用函数的对称性和周期性，得出给定范围内的所有其他解。

sin(x + 360°n) = sin x, cos(x + 360°n) = cos x, tan(x + 180°n) = tan x

The general solution formulas are based on these periods, with n representing any integer.

通解公式正是基于这些周期，其中 n 表示任意整数。

2. Solving sin x = k | 求解 sin x = k

For an equation of the form sin x = k, where -1 ≤ k ≤ 1, the principal value is x = arcsin(k). Since sin x is positive in the first and second quadrants, a second solution in the range 0° to 360° is x = 180° – arcsin(k).

对于形如 sin x = k（其中 -1 ≤ k ≤ 1）的方程，主值为 x = arcsin(k)。由于 sin x 在第一、第二象限为正，在 0° 到 360° 范围内的第二个解是 x = 180° − arcsin(k)。

For example, solve sin x = 0.5 for 0° ≤ x < 360°.

例如，求解 sin x = 0.5，其中 0° ≤ x < 360°。

x = arcsin(0.5) = 30°, and x = 180° − 30° = 150°

Therefore the solutions are 30° and 150°. The general solution can be written as:

因此解为 30° 和 150°。通解可写成：

x = 30° + 360°n 或 x = 150° + 360°n

3. Solving cos x = k | 求解 cos x = k

For cos x = k, the principal value is x = arccos(k). Cosine is positive in the first and fourth quadrants, so in the range 0° to 360° the two solutions are x = arccos(k) and x = 360° – arccos(k).

对于 cos x = k，主值为 x = arccos(k)。余弦在第一、第四象限为正，因此在 0° 到 360° 范围内的两个解为 x = arccos(k) 和 x = 360° − arccos(k)。

Solve cos x = -0.5 for 0° ≤ x < 360°.

求解 cos x = −0.5，其中 0° ≤ x < 360°。

arccos(-0.5) = 120°, so x = 120° and x = 360° − 120° = 240°

The general solution is often compactly written as x = ±120° + 360°n.

通解通常可简洁地写为 x = ±120° + 360°n。

4. Solving tan x = k | 求解 tan x = k

Tangent is positive in the first and third quadrants, but because its period is 180°, the two solutions in 0° to 360° differ by exactly 180°.

正切在第一、第三象限为正，但由于其周期为 180°，在 0° 到 360° 内的两个解恰好相差 180°。

For tan x = k, the principal value is x = arctan(k). Then the general solution is simply x = arctan(k) + 180°n.

对于 tan x = k，主值为 x = arctan(k)。通解为 x = arctan(k) + 180°n。

Solve tan x = √3 for 0° ≤ x < 360°.

求解 tan x = √3，其中 0° ≤ x < 360°。

arctan(√3) = 60°, so x = 60° and x = 60° + 180° = 240°

Notice that once we have one solution, adding the period 180° immediately gives the next solution.

注意，只要找到一个解，加上周期 180° 就能立即得到下一个解。

5. Using the CAST Diagram | 使用 CAST 图

The CAST diagram is a quick way to determine the signs of trigonometric functions in each quadrant. It also helps generate all solutions systematically.

CAST 图是判断各象限三角函数正负号的快捷方法，也有助于系统地生成所有解。

C (Cosine positive): fourth quadrant, 270° to 360°
A (All positive): first quadrant, 0° to 90°
S (Sine positive): second quadrant, 90° to 180°
T (Tangent positive): third quadrant, 180° to 270°

C（余弦为正）：第四象限，270° 到 360°
A（全部为正）：第一象限，0° 到 90°
S（正弦为正）：第二象限，90° 到 180°
T（正切为正）：第三象限，180° 到 270°

Using the reference angle α, the solutions in each quadrant can be written as α, 180° − α, 180° + α, and 360° − α, depending on which quadrants satisfy the equation.

借助参考角 α，各象限解可写为 α、180° − α、180° + α 和 360° − α，具体取哪个由方程在哪个象限成立决定。

6. Equations with Multiple Angles | 复合角方程

Equations such as sin(2x) = 0.5 or cos(3x + 10°) = -0.8 require an extra initial step: let the compound angle be a new variable, solve for that variable, then divide by the coefficient of x.

对于像 sin(2x) = 0.5 或 cos(3x + 10°) = −0.8 的方程，需要先做一步换元：令复合角为新变量，先解出新变量，再除以 x 的系数。

Solve sin(2x) = 0.5 for 0° ≤ x < 180°.

求解 sin(2x) = 0.5，其中 0° ≤ x < 180°。

Let θ = 2x. Then 0° ≤ θ < 360°. The solutions for θ are 30° and 150°.

令 θ = 2x，则 0° ≤ θ < 360°。θ 的解为 30° 和 150°。

x = 30° ÷ 2 = 15° 或 x = 150° ÷ 2 = 75°

Because the domain for θ is doubled, there may be additional solutions obtained by adding the period before dividing. In this case, adding 360° to θ gives 390° and 510°, leading to x = 195° and x = 255°, both outside the required domain. Hence only 15° and 75° are valid.

因为 θ 的区间是 x 的两倍，所以有时需要先加上周期再除以系数，以得到额外解。本例中，θ 加 360° 得 390° 和 510°，对应 x = 195° 和 x = 255°，均超出所需范围。因此有效解只有 15° 和 75°。

7. Quadratic Trigonometric Equations | 二次型三角方程

Equations involving squared trigonometric functions are common. For example, 2 sin²x − sin x − 1 = 0 can be treated as a quadratic in sin x.

含三角函数平方的方程很常见。例如，2 sin²x − sin x − 1 = 0 可看作关于 sin x 的二次方程。

Factorise to obtain (2 sin x + 1)(sin x − 1) = 0. Therefore sin x = -1/2 or sin x = 1.

因式分解得 (2 sin x + 1)(sin x − 1) = 0。因此 sin x = −½ 或 sin x = 1。

For 0° ≤ x < 360°, sin x = -1/2 gives x = 210° and 330°. sin x = 1 gives x = 90°.

在 0° ≤ x < 360° 内，sin x = −½ 得 x = 210° 和 330°；sin x = 1 得 x = 90°。

Solutions: x = 90°, 210°, 330°

Sometimes the equation contains both sin²x and cos²x. Use the identity sin²x + cos²x = 1 to express everything in terms of one trigonometric function.

有时方程同时包含 sin²x 和 cos²x，可利用恒等式 sin²x + cos²x = 1 将各项化为同一个三角函数。

8. Equations Involving tan and sin/cos | 含 tan 与 sin/cos 的方程

An equation such as 2 sin x = cos x can be solved by dividing both sides by cos x, provided cos x ≠ 0. This gives tan x = 1/2.

对于 2 sin x = cos x 这类方程，可在 cos x ≠ 0 的前提下两边同除以 cos x，得到 tan x = ½。

Solve tan x = 1/2 for 0° ≤ x < 360°.

求解 tan x = ½，其中 0° ≤ x < 360°。

x = arctan(0.5) ≈ 26.57°, 以及 x ≈ 26.57° + 180° = 206.57°

Dividing by cos x is safe here because if cos x = 0, then sin x = ±1, which does not satisfy the original equation. However, dividing by a factor that could be zero may cause loss of solutions, so always check.

此处除以 cos x 是安全的，因为若 cos x = 0，则 sin x = ±1，不满足原方程。但除以可能为零的因子可能导致丢解，因此必须检查。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Students often make predictable mistakes when solving trigonometric equations. Being aware of these can save many marks.

学生在解三角方程时常犯一些可预见的错误。了解这些错误可以避免失分。

Mistake \| 错误	Correction \| 正确做法
Forgetting the second solution in 0° to 360°	Always use CAST diagram or symmetry to find both solutions
Dividing by sin x or cos x without checking loss of solutions	Factorise instead of dividing, or verify the zero cases separately
Forgetting to adjust the domain when solving sin(2x) = k	First solve for the compound angle, then divide by the coefficient
Using the wrong period for tan	Remember tan has period 180° or π, not 360°
Confusing degrees and radians in general solutions	Use consistent units and include units in your answer
在 0° 到 360° 内漏掉第二个解	始终用 CAST 图或对称性找出两个解
除以 sin x 或 cos x 时不检查是否丢解	改用因式分解，或单独验证零解情况
解 sin(2x) = k 时忘记调整定义域	先解复合角，再除以系数
对 tan 使用错误的周期	记住 tan 的周期是 180° 或 π，而不是 360°
通解中角度与弧度混用	单位保持一致，并在答案中写明单位

10. Worked Examples | 综合例题

Example 1: Solve cos(2x) = 0.3 for 0° ≤ x ≤ 180°, giving answers to 1 decimal place.

例 1：求解 cos(2x) = 0.3，其中 0° ≤ x ≤ 180°，答案保留一位小数。

Let θ = 2x, so 0° ≤ θ ≤ 360°. The principal value is arccos(0.3) ≈ 72.5°. Hence θ = 72.5° and θ = 360° − 72.5° = 287.5°.

令 θ = 2x，则 0° ≤ θ ≤ 360°。主值 arccos(0.3) ≈ 72.5°。因此 θ = 72.5°，以及 θ = 360° − 72.5° = 287.5°。

x ≈ 72.5° ÷ 2 = 36.3° 或 x ≈ 287.5° ÷ 2 = 143.8°

Example 2: Solve 3 tan x + 1 = 0 for 0° ≤ x < 360°.

例 2：求解 3 tan x + 1 = 0，其中 0° ≤ x < 360°。

Rearrange to get tan x = -1/3. The principal value is arctan(-1/3) ≈ -18.4°, which lies outside the required range. Add 180° to obtain 161.6°, then add another 180° to obtain 341.6°.

移项得 tan x = −⅓。主值 arctan(−⅓) ≈ −18.4°，不在所需区间内。加上 180° 得 161.6°，再加 180° 得 341.6°。

x ≈ 161.6°, 341.6°

11. Summary and Examination Tips | 总结与考试建议

To solve simple trigonometric equations successfully, first identify the type of equation, find the principal value, use symmetry or the CAST diagram to locate all solutions in the specified domain, and always check whether solutions are within range.

成功求解简单三角方程的第一步是判断方程类型，然后找到主值，再利用对称性或 CAST 图确定指定区间内的所有解，并始终检查解是否在范围内。

Here are some key tips for exams:

以下是几条考试关键建议：

Set your calculator to the correct mode (degrees or radians) before starting.
Write down the general solution before substituting integer values for n.
When the domain is given, generate solutions systematically using the period.
Never divide by sin x or cos x unless you have verified that it cannot be zero.
Check answers by substituting back into the original equation.

开始前将计算器设为正确的模式（角度制或弧度制）。
先写出通解，再代入整数 n 求具体解。
题目给定区间时，利用周期系统地生成所有解。
除非确认 sin x 或 cos x 不可能为零，否则绝不能除以它们。
将解代回原方程进行检验。

With consistent practice, solving trigonometric equations becomes a reliable skill. Understanding the underlying periodic nature is far better than memorising formulas blindly.

通过持续练习，解三角方程会成为一项稳定可靠的技能。理解背后的周期性，远比盲目记忆公式更有价值。

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Solving Simple Trigonometric Equations | 简单三角方程的求解