📚 Exercise 15B.1: Solving Trigonometric Equations | 练习15B.1：解三角方程
In IB Mathematics, Exercise 15B.1 typically appears in the chapter on trigonometric functions and equations. It challenges students to solve trigonometric equations within a specified interval or over the entire set of real numbers. A solid grasp of this topic is essential not only for Paper 1 and Paper 2 but also as a foundation for calculus and vector applications. This article provides a step-by-step walkthrough of the techniques, general solution formulas, common pitfalls, and practical tips you need to handle any Exercise 15B.1 question with confidence.
在IB数学中，练习15B.1通常出现在三角函数与方程章节。它要求学生求解指定区间内或整个实数范围内的三角方程。扎实掌握这一主题不仅对试卷一和试卷二的考试至关重要，也是微积分和向量应用的基础。本文将逐步讲解所需的方法、通解公式、常见错误和实用技巧，帮助你自信地应对任何练习15B.1中的问题。
1. Understanding the Problem | 理解问题
Before you begin any algebraic manipulation, read the question carefully. Identify the type of trigonometric function involved — sine, cosine, or tangent — and note whether the equation is in a standard form such as sin x = k or a more complex form like 2cos(3x) + 1 = 0. Pay close attention to the domain: sometimes it is given as an inequality, for example 0 ≤ x ≤ 2π or -π ≤ x ≤ π, while other problems ask for the general solution over all real numbers. A quick sketch of the function's graph can help you anticipate the number of solutions in a restricted interval.
在开始任何代数运算之前，请仔细阅读题目。确认涉及的三角函数类型——正弦、余弦或正切——并注意方程是标准形式如 sin x = k，还是更复杂的形式如 2cos(3x) + 1 = 0。密切关注定义域：有时以不等式给出，例如 0 ≤ x ≤ 2π 或 -π ≤ x ≤ π，而另一些问题则要求所有实数的通解。快速画出函数图像草图有助于预判在受限区间内解的个数。
2. Isolating the Trigonometric Function | 分离三角函数
The first algebraic step is to isolate the trigonometric term. Treat the trig function just like any variable and perform inverse operations. For 3sin x - 1 = 2, add 1 to both sides to get 3sin x = 3, then divide by 3 to obtain sin x = 1. If the equation has a leading coefficient or a constant term inside the argument, such as 2cos(2x) + √3 = 0, rearrange it to cos(2x) = -√3/2. Always reduce the equation to one of the three basic forms: sin(θ) = k, cos(θ) = k, or tan(θ) = k, where k is a real number within the function's range. If multiple trig ratios appear, you may need to apply identities later, but isolation is always the starting point.
代数运算的第一步是分离三角函数项。将三角函数视作普通变量，进行逆运算。对于 3sin x - 1 = 2，两边加 1 得到 3sin x = 3，然后除以 3 得到 sin x = 1。如果方程在自变量中有系数或常数项，例如 2cos(2x) + √3 = 0，则把它整理为 cos(2x) = -√3/2。务必将方程化简为三种基本形式之一：sin(θ) = k、cos(θ) = k 或 tan(θ) = k，其中 k 是一个在函数值域内的实数。如果出现了多个三角比，后续可能需要使用恒等式，但分离永远是第一步。
3. Finding the Principal Solution | 求主解
Once you have the basic form, use your calculator (in radian or degree mode as specified by the question) to find the principal value α. For sin x = k (with |k| ≤ 1), α = arcsin(k) and lies in the interval [-π/2, π/2]. For cos x = k, α = arccos(k) lies in [0, π]. For tan x = k, α = arctan(k) and lies in (-π/2, π/2). This principal value acts as the reference angle from which all other solutions are generated. Always write it down as an exact value when the answer is a familiar angle like π/6, π/4, or π/3; otherwise, round to the required decimal places.
一旦得到基本形式，使用计算器（根据题目要求设置为弧度或角度模式）求出主值 α。对于 sin x = k（满足 |k| ≤ 1），α = arcsin(k) 并且落在区间 [-π/2, π/2] 内。对于 cos x = k，α = arccos(k) 落在 [0, π] 内。对于 tan x = k，α = arctan(k) 并且落在 (-π/2, π/2) 内。这个主值是生成所有其他解的参考角。若结果是像 π/6、π/4 或 π/3 这样的特殊角，务必以精确值形式写出主值；否则根据要求保留相应的小数位数。
4. General Solution for Sine Equations | 正弦方程的通解
The sine function has a period of 2π and satisfies sin(π - α) = sin α. Therefore, the general solution of sin x = k can be compactly written as:
x = nπ + (-1)ⁿα, n ∈ ℤ
Alternatively, you can express it as two families: x = α + 2nπ and x = π - α + 2nπ. For example, if sin x = 1/2 and α = π/6, the solutions are x = π/6 + 2nπ and x = 5π/6 + 2nπ. The compact form is efficient for mathematical writing, but the two-family form often makes it easier to list solutions in a given interval.
正弦函数的周期为 2π，并且满足 sin(π - α) = sin α。因此，sin x = k 的通解可简洁地写成：
x = nπ + (-1)ⁿα， n ∈ ℤ
也可以写成两个解族：x = α + 2nπ 和 x = π - α + 2nπ。例如，若 sin x = 1/2 且 α = π/6，解为 x = π/6 + 2nπ 和 x = 5π/6 + 2nπ。紧凑形式在数学书写上较为高效，而两个解族的形式通常更便于在给定区间内列出全部解。
5. General Solution for Cosine Equations | 余弦方程的通解
Cosine has a period of 2π and is symmetric about 0: cos(-α) = cos α. Thus the general solution of cos x = k is:
x = 2nπ ± α, n ∈ ℤ
This single expression captures both x = α + 2nπ and x = -α + 2nπ (which is the same as 2π - α + 2nπ). For instance, solving cos x = √3/2 gives α = π/6, so the solutions are x = π/6 + 2nπ and x = -π/6 + 2nπ. When the domain is [0, 2π], the negative-angle version is often rewritten as 11π/6 + 2nπ for clarity.
余弦的周期为 2π，且关于 0 对称：cos(-α) = cos α。因此 cos x = k 的通解为：
x = 2nπ ± α， n ∈ ℤ
这个单一表达式涵盖了 x = α + 2nπ 和 x = -α + 2nπ（等价于 2π - α + 2nπ）。例如，求解 cos x = √3/2 得到 α = π/6，因此解为 x = π/6 + 2nπ 和 x = -π/6 + 2nπ。当定义域是 [0, 2π] 时，通常将负角度形式改写为 11&pi
Published by TutorHao | IB Mathematics Revision Series | aleveler.com 更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图