Solving Simple Trigonometric Equations | 简单三角方程

📚 Solving Simple Trigonometric Equations | 简单三角方程

Trigonometric equations are a core topic in A-Level mathematics. They require not only algebraic manipulation but also a deep understanding of the periodic nature of sine, cosine, and tangent functions. This article guides you step by step through solving simple trigonometric equations, from general solutions to solutions within a specified interval.

三角方程是 A-Level 数学的核心内容之一。它不仅考查代数运算能力，更要求我们深入理解正弦、余弦和正切函数的周期性。本文将从一般解到限定区间内的解，一步步带你掌握简单三角方程的解法。

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1. What Is a Trigonometric Equation? | 什么是三角方程？

A trigonometric equation is an equation that involves one or more trigonometric functions, such as sin x, cos x, or tan x. The unknown variable typically appears as the argument of the function, for example sin x = 0.5 or 2cos x + 1 = 0.

三角方程是含有一个或多个三角函数（如 sin x、cos x、tan x）的方程。未知数通常出现在函数的自变量位置，例如 sin x = 0.5 或 2cos x + 1 = 0。

Because trigonometric functions are periodic, a simple equation like sin x = 0.5 has infinitely many solutions unless a restricted interval is given. In A-Level exams, you are usually asked to find solutions in a specific range, such as 0° ≤ x ≤ 360° or 0 ≤ x ≤ 2π.

由于三角函数具有周期性，像 sin x = 0.5 这样的简单方程在无限制条件下有无数个解。在 A-Level 考试中，通常要求我们在特定范围内求解，例如 0° ≤ x ≤ 360° 或 0 ≤ x ≤ 2π。

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2. The Principal Value and Reference Angle | 主值与参考角

The principal value is the solution of a trigonometric equation that lies within the principal range of the inverse function. For example, the principal value of sin⁻¹(0.5) is 30°, because the range of sin⁻¹ is −90° to 90°.

主值（principal value）是三角方程在反函数主值范围内的解。例如，sin⁻¹(0.5) 的主值是 30°，因为 sin⁻¹ 的值域是 −90° 到 90°。

The reference angle is the acute angle between the terminal side of the angle and the x-axis. It is always positive and less than 90°. Once you find the reference angle, you can use the CAST diagram or the unit circle to find all solutions in the required interval.

参考角（reference angle）是角的终边与 x 轴之间所夹的锐角，它始终为正且小于 90°。求出参考角后，利用 CAST 图或单位圆即可在指定区间内找到所有解。

sin θ = k → reference angle α = sin⁻¹(k)

对于 cos θ = k 与 tan θ = k 同理

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3. Solving sin x = k | 解 sin x = k

For the equation sin x = k, where −1 ≤ k ≤ 1, the reference angle α is given by α = sin⁻¹(k). Since sine is positive in the first and second quadrants, the solutions within 0° to 360° are:

对于方程 sin x = k（其中 −1 ≤ k ≤ 1），参考角 α = sin⁻¹(k)。因为正弦函数在第一、第二象限为正，在 0° 到 360° 范围内的解为：

x = α and x = 180° − α

Example: Solve sin x = 0.5 for 0° ≤ x ≤ 360°.

例：解 sin x = 0.5，其中 0° ≤ x ≤ 360°。

α = sin⁻¹(0.5) = 30°. The two solutions are x = 30° and x = 180° − 30° = 150°.

α = sin⁻¹(0.5) = 30°。两个解为 x = 30° 和 x = 180° − 30° = 150°。

If the equation is sin x = −k, then the reference angle is still α, but since sine is negative in the third and fourth quadrants, the solutions are x = 180° + α and x = 360° − α.

若方程为 sin x = −k，参考角仍为 α，但由于正弦在第三、第四象限为负，解为 x = 180° + α 和 x = 360° − α。

sin x = 0.5	x = 30°, 150°
sin x = −0.5	x = 210°, 330°

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4. Solving cos x = k | 解 cos x = k

For the equation cos x = k, the reference angle α = cos⁻¹(k). Cosine is positive in the first and fourth quadrants, so the solutions in 0° to 360° are:

对于方程 cos x = k，参考角 α = cos⁻¹(k)。余弦函数在第一、第四象限为正，在 0° 到 360° 范围内的解为：

x = α and x = 360° − α

Example: Solve cos x = 0.5 for 0° ≤ x ≤ 360°.

例：解 cos x = 0.5，其中 0° ≤ x ≤ 360°。

α = cos⁻¹(0.5) = 60°. The solutions are x = 60° and x = 360° − 60° = 300°.

α = cos⁻¹(0.5) = 60°。解为 x = 60° 和 x = 360° − 60° = 300°。

For cos x = −k, since cosine is negative in the second and third quadrants, the solutions become:

对于 cos x = −k，由于余弦在第二、第三象限为负，解变为：

x = 180° − α and x = 180° + α

Example: cos x = −0.5 → α = 60°, solutions at x = 180° − 60° = 120° and x = 180° + 60° = 240°.

例：cos x = −0.5 → α = 60°，解为 x = 180° − 60° = 120° 和 x = 180° + 60° = 240°。

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5. Solving tan x = k | 解 tan x = k

For the equation tan x = k, the reference angle α = tan⁻¹(k). Tangent is positive in the first and third quadrants, and its period is 180°. Therefore, in the interval 0° to 360°, the solutions are:

对于方程 tan x = k，参考角 α = tan⁻¹(k)。正切函数在第一、第三象限为正，且其周期为 180°。因此在 0° 到 360° 区间内，解为：

x = α and x = α + 180°

Example: Solve tan x = 1 for 0° ≤ x ≤ 360°.

例：解 tan x = 1，其中 0° ≤ x ≤ 360°。

α = tan⁻¹(1) = 45°. The solutions are x = 45° and x = 45° + 180° = 225°.

α = tan⁻¹(1) = 45°。解为 x = 45° 和 x = 45° + 180° = 225°。

For tan x = −k, the reference angle is still α. Since tangent is negative in the second and fourth quadrants, the solutions are x = 180° − α and x = 360° − α.

对于 tan x = −k，参考角仍为 α。正切在第二、第四象限为负，解为 x = 180° − α 和 x = 360° − α。

Example: tan x = −1 → α = 45°, solutions at x = 180° − 45° = 135° and x = 360° − 45° = 315°.

例：tan x = −1 → α = 45°，解为 x = 180° − 45° = 135° 和 x = 360° − 45° = 315°。

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6. The General Solution | 一般解

A general solution expresses all possible solutions of a trigonometric equation using the period of the function. It is often required in higher-level questions.

一般解（general solution）利用函数的周期来表示三角方程的所有解，在进阶题目中经常见到。

For sin x = k, where α = sin⁻¹(k), the general solution is:

对于 sin x = k，其中 α = sin⁻¹(k)，一般解为：

x = α + 360°n or x = 180° − α + 360°n, n ∈ ℤ

For cos x = k, the general solution is:

对于 cos x = k，一般解为：

x = α + 360°n or x = −α + 360°n, n ∈ ℤ

For tan x = k, the general solution is:

对于 tan x = k，一般解为：

x = α + 180°n, n ∈ ℤ

In radians, replace 360°n with 2πn and 180°n with πn. The general solution is very useful when you need to check the number of solutions in a long interval or when deriving identities.

当使用弧度制时，将 360°n 替换为 2πn，将 180°n 替换为 πn。一般解在检验长区间内解的个数以及推导恒等式时非常有用。

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7. Solving Equations in a Given Interval | 在给定区间内求解

Most A-Level exam questions specify an interval, such as 0° ≤ x ≤ 360° or −180° ≤ x ≤ 180°. In such cases, first find the reference angle, then use the CAST diagram to determine which quadrants contain solutions, and finally list all solutions lying within the interval.

大多数 A-Level 考试题目会指定区间，如 0° ≤ x ≤ 360° 或 −180° ≤ x ≤ 180°。此时先求出参考角，再利用 CAST 图判断解所在的象限，最后列出区间内的所有解。

Example: Solve 2cos x + √3 = 0 for 0° ≤ x ≤ 360°.

例：解 2cos x + √3 = 0，其中 0° ≤ x ≤ 360°。

Rearrange: cos x = −√3/2. The reference angle is α = cos⁻¹(√3/2) = 30°. Since cosine is negative in quadrants II and III:

移项得：cos x = −√3/2。参考角 α = cos⁻¹(√3/2) = 30°。因余弦在第二、第三象限为负：

x = 180° − 30° = 150° and x = 180° + 30° = 210°

Both values lie in the given interval, so the final answer is x = 150°, 210°.

两个值均在给定区间内，因此最终答案为 x = 150°, 210°。

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8. Quadratic Trigonometric Equations | 二次三角方程

Quadratic trigonometric equations, such as 2sin²x − sin x − 1 = 0, are solved by treating the trigonometric function as a single variable. Let y = sin x, then solve the quadratic in y, and finally solve the resulting simple trigonometric equations.

二次三角方程，例如 2sin²x − sin x − 1 = 0，通过将三角函数视为单一变量来求解。令 y = sin x，先解关于 y 的二次方程，再解所得的简单三角方程。

Example: Solve 2sin²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°.

例：解 2sin²x − sin x − 1 = 0，其中 0° ≤ x ≤ 360°。

Let y = sin x. Then 2y² − y − 1 = 0. Factorise: (2y + 1)(y − 1) = 0, so y = −1/2 or y = 1.

令 y = sin x，则 2y² − y − 1 = 0。因式分解：(2y + 1)(y − 1) = 0，所以 y = −1/2 或 y = 1。

For sin x = 1: x = 90°. For sin x = −1/2: reference angle α = 30°, solutions in quadrants III and IV are x = 180° + 30° = 210° and x = 360° − 30° = 330°.

对于 sin x = 1：x = 90°。对于 sin x = −1/2：参考角 α = 30°，第三、第四象限的解为 x = 180° + 30° = 210° 和 x = 360° − 30° = 330°。

Final answer: x = 90°, 210°, 330°

最终答案：x = 90°, 210°, 330°

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9. Equations Involving Multiple Angles | 含倍角的方程

Equations such as cos 2x = 0.5 or sin(2x + 10°) = 0.3 require a change of variable before applying the standard method.

对于 cos 2x = 0.5 或 sin(2x + 10°) = 0.3 这类方程，需要先进行换元，再应用标准方法。

Example: Solve cos 2x = 0.5 for 0° ≤ x ≤ 360°.

例：解 cos 2x = 0.5，其中 0° ≤ x ≤ 360°。

Let θ = 2x. Since 0° ≤ x ≤ 360°, we have 0° ≤ θ ≤ 720°. Solve cos θ = 0.5:

令 θ = 2x。因为 0° ≤ x ≤ 360°，所以 0° ≤ θ ≤ 720°。解 cos θ = 0.5：

θ = 60°, 300°, 60° + 360° = 420°, 300° + 360° = 660°.

θ = 60°, 300°, 60° + 360° = 420°, 300° + 360° = 660°。

Now divide by 2: x = 30°, 150°, 210°, 330°.

再除以 2：x = 30°, 150°, 210°, 330°。

It is essential to extend the interval during the substitution step; otherwise, solutions will be missed.

换元时务必扩充区间，否则会遗漏解。

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10. Using Identities to Simplify Equations | 利用恒等式化简方程

Some equations appear to involve two different trigonometric functions, but can be simplified using basic identities. The most commonly used identity is sin²x + cos²x = 1.

有些方程看似含有两个不同的三角函数，但可以通过基本恒等式化简。最常用的是 sin²x + cos²x = 1。

Example: Solve 2cos²x + 3sin x − 3 = 0 for 0° ≤ x ≤ 360°.

例：解 2cos²x + 3sin x − 3 = 0，其中 0° ≤ x ≤ 360°。

Replace cos²x with 1 − sin²x:

将 cos²x 替换为 1 − sin²x：

2(1 − sin²x) + 3sin x − 3 = 0

= 2 − 2sin²x + 3sin x − 3 = −2sin²x + 3sin x − 1 = 0, which gives 2sin²x − 3sin x + 1 = 0.

= 2 − 2sin²x + 3sin x − 3 = −2sin²x + 3sin x − 1 = 0，即 2sin²x − 3sin x + 1 = 0。

Factorise: (2sin x − 1)(sin x − 1) = 0, so sin x = 1/2 or sin x = 1.

因式分解：(2sin x − 1)(sin x − 1) = 0，所以 sin x = 1/2 或 sin x = 1。

sin x = 1 → x = 90°. sin x = 1/2 → x = 30°, 150°. Final answer: x = 30°, 90°, 150°.

sin x = 1 → x = 90°。sin x = 1/2 → x = 30°, 150°。最终答案：x = 30°, 90°, 150°。

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

One of the most common mistakes is forgetting to extend the interval when solving equations with multiple angles. Always replace the original interval before applying the inverse trigonometric function.

最常见的错误之一是解倍角方程时忘记将区间同步扩大。在套用反三角函数之前，务必将原区间按系数放大。

Another frequent error is confusing the quadrants for negative values. Make sure you memorise the CAST diagram: C (cos positive) in quadrant IV, A (all positive) in quadrant I, S (sin positive) in quadrant II, and T (tan positive) in quadrant III.

另一个常见错误是混淆负值对应象限。务必熟记 CAST 图：第四象限 C（cos 为正）、第一象限 A（全为正）、第二象限 S（sin 为正）、第三象限 T（tan 为正）。

Finally, always check whether the solutions lie within the given interval. It is good practice to substitute each solution back into the original equation as a verification.

最后，始终检查解是否落在给定区间内。建议将每个解代回原方程进行验算。

错误	修正方法
忘记扩展区间	如果解 2x，则将区间乘以 2
负值象限判断错误	反复使用 CAST 图训练
丢失周期解	列出区间内所有符合条件的角度

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12. Summary and Exam Tips | 总结与备考建议

To solve simple trigonometric equations, follow these steps: (1) isolate the trigonometric function; (2) find the reference angle; (3) use CAST to determine the quadrants; (4) write all solutions in the given interval; and (5) check your answers.

解简单三角方程时，按以下步骤操作：(1) 将三角函数单独移出；(2) 求出参考角；(3) 利用 CAST 判断象限；(4) 写出给定区间内的全部解；(5) 检查答案。

In exams, pay special attention to the unit (degrees or radians), the required precision, and the interval boundaries. When in doubt, substitute your final answers back into the original equation.

考试中要特别注意单位（角度制或弧度制）、精度要求以及区间端点。若有疑问，将最终答案代回原方程验证即可。

Practice is the key to mastering trigonometric equations. Once you are comfortable with the CAST diagram and the standard solution patterns, even the most complex-looking equations become manageable.

熟能生巧是掌握三角方程的关键。一旦你熟练使用 CAST 图和标准解的规律，即使表面复杂的方程也能迎刃而解。

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