Solving Trigonometric Equations | 三角方程的求解

📚 Solving Trigonometric Equations | 三角方程的求解

Trigonometric equations are a core topic in A-Level Mathematics. They require a combination of algebraic manipulation, trigonometric identities, and a clear understanding of the unit circle and periodicity. This article provides a structured approach to solving them.

三角方程是 A-Level 数学的核心内容。求解三角方程需要综合运用代数变形、三角恒等式，以及对单位圆和周期性的清晰理解。本文将提供一种系统化的求解方法。

1. Basic Trigonometric Equations | 基本三角方程

The simplest equations involve a single trigonometric function, such as sin x = a, cos x = a, or tan x = a. For these, we first find the principal value using the inverse function, then use symmetry and periodicity to find all solutions.

最简单的方程只含一个三角函数，例如 sin x = a、cos x = a 或 tan x = a。对于这类方程，我们先用反三角函数求出主值，再利用对称性和周期性找出所有解。

For sin x = a, where |a| ≤ 1, the principal value is sin⁻¹ a (usually taken between −90° and 90° or −π/2 and π/2). A second solution in the range 0° to 360° is given by 180° − sin⁻¹ a.

对于 sin x = a（|a| ≤ 1），主值为 sin⁻¹ a（通常取在 −90° 到 90°，或 −π/2 到 π/2）。在 0° 到 360° 范围内的第二个解为 180° − sin⁻¹ a。

For cos x = a, the principal value is cos⁻¹ a (between 0° and 180°). The second solution in 0° to 360° is 360° − cos⁻¹ a.

对于 cos x = a，主值为 cos⁻¹ a（在 0° 到 180° 之间）。在 0° 到 360° 范围内的第二个解为 360° − cos⁻¹ a。

For tan x = a, the principal value is tan⁻¹ a (between −90° and 90°). Because tan has period 180°, all solutions differ by multiples of 180°.

对于 tan x = a，主值为 tan⁻¹ a（在 −90° 到 90° 之间）。由于 tan 的周期为 180°，所有解相差 180° 的整数倍。

sin x = a ⇒ x = nπ + (−1)ⁿ sin⁻¹ a, n ∈ ℤ

cos x = a ⇒ x = 2nπ ± cos⁻¹ a, n ∈ ℤ

tan x = a ⇒ x = nπ + tan⁻¹ a, n ∈ ℤ

2. General Solutions | 一般解

In A-Level, you are often asked to give the general solution, which expresses all possible values of x. The above formulas are the standard general solutions when angles are measured in radians.

在 A-Level 考试中，常要求写出一般解，即表示 x 的所有可能取值。当角以弧度为单位时，上述公式就是标准的一般解。

It is essential to state the domain of n explicitly, usually n ∈ ℤ (n is any integer). Some exam boards prefer the form that avoids the (−1)ⁿ sign, so be aware of both conventions.

必须明确说明 n 的取值范围，通常为 n ∈ ℤ（n 为任意整数）。有些考试局更倾向于不用 (−1)ⁿ 的写法，因此要注意两种约定。

For example, solve sin x = ½. The principal value is π/6, so the general solution is x = nπ + (−1)ⁿ (π/6). This gives x = π/6, 5π/6, 13π/6, etc. when n is chosen appropriately.

例如，解 sin x = ½。主值为 π/6，因此一般解为 x = nπ + (−1)ⁿ (π/6)。当 n 取适当整数时，可得到 x = π/6、5π/6、13π/6 等。

3. Solving on a Given Interval | 在给定区间内求解

Most exam questions specify a finite interval, such as 0° ≤ x ≤ 360° or 0 ≤ x ≤ 2π. In such cases, you must generate all solutions within that interval by adding or subtracting full periods from the general solution.

多数考题会给定有限区间，例如 0° ≤ x ≤ 360° 或 0 ≤ x ≤ 2π。此时你需要通过从一般解中加减整个周期，找出该区间内的所有解。

Step-by-step approach:

分步方法如下：

Find the principal value using a calculator or known exact values.

使用计算器或已知特殊值求出主值。
Determine all angles in the interval that have the same sine, cosine, or tangent by using symmetry in the appropriate quadrant.

利用相应象限中的对称性，确定区间内具有相同正弦、余弦或正切的所有角。
List the solutions in increasing order and check that each lies in the given interval.

按升序列出解，并检查每个解是否在给定区间内。

For example, solve cos x = −√2/2 for 0° ≤ x ≤ 360°. The principal value is 135°. Since cos is negative in the second and third quadrants, the solutions are x = 135° and x = 225°.

例如，在 0° ≤ x ≤ 360° 内解 cos x = −√2/2。主值为 135°。由于余弦在第二、三象限为负，所以解为 x = 135° 和 x = 225°。

4. Quadratic Trigonometric Equations | 二次三角方程

Equations of the form a sin² x + b sin x + c = 0 (or with cos or tan) are quadratic in the trigonometric function. Substitute y = sin x (or the relevant function) to reduce it to an ordinary quadratic equation.

形如 a sin² x + b sin x + c = 0（或含 cos、tan）的方程是关于三角函数的二次方程。令 y = sin x（或相应函数），将其化为普通二次方程。

Solve the quadratic first, then solve the resulting simple trigonometric equations for each possible value of the function.

先解二次方程，然后针对函数的每个可能值，求解相应的简单三角方程。

2 sin² x − sin x − 1 = 0 ⇒ (2 sin x + 1)(sin x − 1) = 0

Thus sin x = 1 or sin x = −½. These give the same set of solutions as the basic equations.

于是 sin x = 1 或 sin x = −½。这些与基本方程给出相同的解集。

Be careful not to lose solutions: if the quadratic has no real roots, the equation has no real solutions. Also, remember that the sine and cosine functions are bounded between −1 and 1, so values outside this range must be rejected.

注意不要丢解：如果二次方程没有实根，则原方程无实数解。另外，正弦和余弦函数的值域在 −1 到 1 之间，因此超出此范围的取值必须舍弃。

5. Using Identities to Simplify | 用恒等式化简

Many equations contain both sin x and cos x, or powers higher than 2. The key is to use trigonometric identities to reduce everything to a single function or a single angle.

许多方程同时含有 sin x 和 cos x，或者幂次高于 2。关键是利用三角恒等式将所有项化为单一函数或单一角。

Common identities include:

常用恒等式包括：

sin² x + cos² x = 1

sin² x + cos² x = 1
tan x = sin x / cos x

tan x = sin x / cos x
sin 2x = 2 sin x cos x

sin 2x = 2 sin x cos x
cos 2x = cos² x − sin² x = 2 cos² x − 1 = 1 − 2 sin² x

cos 2x = cos² x − sin² x = 2 cos² x − 1 = 1 − 2 sin² x

Example: Solve 2 cos² x + 3 sin x = 0 for 0° ≤ x ≤ 360°. Replace cos² x with 1 − sin² x:

例：在 0° ≤ x ≤ 360° 内解 2 cos² x + 3 sin x = 0。用 1 − sin² x 替换 cos² x：

2(1 − sin² x) + 3 sin x = 0 ⇒ 2 sin² x − 3 sin x − 2 = 0

This becomes a quadratic in sin x. Factor to get (2 sin x + 1)(sin x − 2) = 0, so sin x = −½ (since sin x = 2 is impossible). The solutions are x = 210° and x = 330°.

这就变成了关于 sin x 的二次方程。因式分解得 (2 sin x + 1)(sin x − 2) = 0，所以 sin x = −½（因为 sin x = 2 不可能）。解为 x = 210° 和 x = 330°。

6. The Auxiliary Angle Method | 辅助角法

Expressions of the form a sin x + b cos x can be rewritten as R sin(x + α) or R cos(x + α). This is extremely useful when solving equations that mix sine and cosine with equal arguments.

形如 a sin x + b cos x 的表达式可以改写为 R sin(x + α) 或 R cos(x + α)。这在求解同时包含相同角的正弦和余弦的方程时非常有用。

Given a sin x + b cos x = R sin(x + α), we have:

设 a sin x + b cos x = R sin(x + α)，则有：

R = √(a² + b²), tan α = b/a

The angle α is chosen in the correct quadrant based on the signs of a and b. Alternatively, use R cos(x − α) with tan α = a/b.

根据 a 和 b 的符号，选择 α 所在的合适象限。或者使用 R cos(x − α)，此时 tan α = a/b。

Example: Solve 3 sin x + 4 cos x = 5 for 0° ≤ x ≤ 360°. Here R = √(3² + 4²) = 5, and tan α = 4/3, so α ≈ 53.13°. The equation becomes 5 sin(x + 53.13°) = 5, i.e. sin(x + 53.13°) = 1. Thus x + 53.13° = 90° + 360°n, giving x ≈ 36.87° within the interval.

例：在 0° ≤ x ≤ 360° 内解 3 sin x + 4 cos x = 5。这里 R = √(3² + 4²) = 5，且 tan α = 4/3，所以 α ≈ 53.13°。方程变为 5 sin(x + 53.13°) = 5，即 sin(x + 53.13°) = 1。因此 x + 53.13° = 90° + 360°n，在区间内得 x ≈ 36.87°。

7. Equations with Multiple Angles | 倍角方程

When the equation contains expressions like sin 2x or cos(3x + 1), the interval must be adjusted. For example, if x lies in [0, 2π], then 2x lies in [0, 4π], so we must list all solutions for 2x over two full periods.

当方程含有 sin 2x 或 cos(3x + 1) 之类的表达式时，必须调整区间。例如，若 x 在 [0, 2π] 内，则 2x 在 [0, 4π] 内，因此需要列出 2x 在两个完整周期内的所有解。

For an equation like sin 2x = ½, with 0 ≤ x ≤ 360°, first solve θ = 2x in the range 0° ≤ θ ≤ 720°. The solutions in that range are θ = 30°, 150°, 390°, 510°, so x = 15°, 75°, 195°, 255°.

对于方程 sin 2x = ½，在 0° ≤ x ≤ 360° 内，先令 θ = 2x，在 0° ≤ θ ≤ 720° 内求解。该范围内的解为 θ = 30°、150°、390°、510°，因此 x = 15°、75°、195°、255°。

Do not forget to divide the period by the coefficient of x when plotting or listing solutions. The period of sin kx is 360°/k.

在作图或列解时，不要忘记将周期除以 x 的系数。sin kx 的周期为 360°/k。

8. Using Graphs to Find Solutions | 用图像找解

Graphical methods are often used to approximate solutions or to verify the number of solutions. Sketching y = f(x) and y = g(x) shows the intersection points, which correspond to the solutions of f(x) = g(x).

图像法常用于近似求解或验证解的个数。画出 y = f(x) 和 y = g(x) 的草图，交点即为 f(x) = g(x) 的解。

For example, the equation cos x = x² has no simple algebraic solution, but a graph shows one intersection in the range 0 to 1, so there is one real solution.

例如，方程 cos x = x² 没有简单的代数解法，但图像显示在 0 到 1 之间有一个交点，因此存在一个实数解。

You should be able to sketch the basic graphs of sin x, cos x, and tan x, including their key features: maximum, minimum, zeros, and asymptotes for tan x.

你应该能画出 sin x、cos x 和 tan x 的基本图像，包括关键特征：最大值、最小值、零点，以及 tan x 的渐近线。

When using a calculator, make sure it is in the correct mode (degrees or radians) and use the inverse functions wisely, as they only give principal values.

使用计算器时，确保处于正确的模式（角度或弧度），并合理使用反函数，因为反函数只给出主值。

9. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧

The following mistakes are frequently seen in exams:

以下错误在考试中经常出现：

Forgetting to add the period when the interval is wider than 360° or 2π.

当区间超过 360° 或 2π 时，忘记加上周期。
Losing solutions when dividing by a trigonometric function instead of factoring.

除以某个三角函数而非因式分解，导致丢解。
Using the calculator in the wrong angle mode.

计算器使用了错误的角度模式。
Ignoring that sin x and cos x lie in [−1, 1], leading to invalid roots.

忽略 sin x 和 cos x 的值域为 [−1, 1]，导致出现无效根。

To avoid these, always factor rather than divide, check your solutions in the original equation, and write down the general solution before narrowing to the interval.

为避免这些问题，应优先因式分解而非直接相除，在原始方程中检验解，并在区间缩限前先写下一般解。

Remember that the inverse trigonometric functions output an angle in a restricted range. For tan⁻¹, the range is (−90°, 90°), for sin⁻¹ it is [−90°, 90°], and for cos⁻¹ it is [0°, 180°].

记住，反三角函数的输出范围受限：tan⁻¹ 的范围是 (−90°, 90°)，sin⁻¹ 是 [−90°, 90°]，cos⁻¹ 是 [0°, 180°]。

10. Summary | 总结

Solving trigonometric equations involves a clear sequence: simplify using identities, rewrite as a basic equation, find the general solution, then restrict to the given interval. Mastery of the unit circle and special angles is essential.

求解三角方程有一清晰流程：用恒等式化简，改写为基本方程，求一般解，再限制到给定区间。熟练掌握单位圆和特殊角至关重要。

Always check for extraneous solutions and be careful with domain restrictions. Practice with a variety of equations, including those requiring the auxiliary angle method, to build confidence.

务必检查增根，并注意定义域的限制。通过练习多种类型的方程（包括需要辅助角法的题目）来增强信心。

With a systematic approach, trigonometric equations become a reliable source of marks in the A-Level examination.

采用系统化的方法，三角方程就会成为 A-Level 考试中稳定的得分点。

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Solving Trigonometric Equations | 三角方程的求解