Example 6.12.2: Solving Trigonometric Equations | 例题6.12.2：解三角方程

📚 Example 6.12.2: Solving Trigonometric Equations | 例题6.12.2：解三角方程

This worked example, labelled Example 6.12.2, is typical of an AQA A-Level Mathematics question that asks you to solve a trigonometric equation using a Pythagorean identity. It tests your ability to transform a quadratic expression, factorise, and then find all solutions in a specified interval.

本例题（例题6.12.2）是AQA A-Level数学中的典型题目，要求你运用毕达哥拉斯恒等式来解三角方程。它考察你将二次表达式变形、因式分解，并在给定区间内求出所有解的能力。

1. Problem Statement | 问题陈述

Solve the equation 2 cos² x − 3 sin x = 3 for 0 ≤ x < 2π. Give your answers in terms of π.

解方程 2 cos² x − 3 sin x = 3，其中 0 ≤ x < 2π，答案用含 π 的形式表示。

You may assume that x is measured in radians. This interval is a full revolution, which is standard for AQA questions unless otherwise stated.

你可以假设 x 以弧度为单位。这个区间是一个完整周期，除非另有说明，这是AQA题目的标准设定。

2. Key Identity | 关键恒等式

The most useful identity for this problem is the Pythagorean identity:

解决本题最常用的恒等式是毕达哥拉斯恒等式：

sin² x + cos² x = 1

Rearranging this gives cos² x = 1 − sin² x. This allows us to rewrite the equation entirely in terms of sin x.

将上式变形可得 cos² x = 1 − sin² x。这样我们就能把原方程完全改写为关于 sin x 的形式。

3. Step 1: Rewrite the Equation | 第一步：改写方程

Start with the original equation:

从原方程出发：

2 cos² x − 3 sin x = 3

Replace cos² x with 1 − sin² x:

将 cos² x 替换为 1 − sin² x：

2(1 − sin² x) − 3 sin x = 3

Expand the bracket and simplify:

展开括号并化简：

2 − 2 sin² x − 3 sin x = 3

Move all terms to the left-hand side:

将所有项移到等号左边：

−2 sin² x − 3 sin x − 1 = 0

Multiply through by −1 to obtain a cleaner quadratic:

等式两边同乘 −1，得到一个更简洁的二次方程：

2 sin² x + 3 sin x + 1 = 0

4. Step 2: Factor the Quadratic | 第二步：因式分解二次式

Let u = sin x. Then the quadratic becomes:

令 u = sin x，则这个二次式变为：

2u² + 3u + 1 = 0

Look for two numbers that multiply to 2 × 1 = 2 and add to 3. The numbers 2 and 1 work. Therefore:

寻找两个数，它们的乘积等于 2 × 1 = 2，且和为 3。数字 2 和 1 满足条件。因此：

(2u + 1)(u + 1) = 0

Substitute u = sin x back into the factorised form:

将 u = sin x 代回因式分解后的形式：

(2 sin x + 1)(sin x + 1) = 0

5. Step 3: Solve Each Factor | 第三步：分别解每个因子

For the product to be zero, at least one factor must be zero. This gives two cases:

要使乘积为零，至少有一个因子为零。这给出两种情况：

Case 1: 2 sin x + 1 = 0 ⇒ sin x = −1/2
情况一：2 sin x + 1 = 0 ⇒ sin x = −1/2
Case 2: sin x + 1 = 0 ⇒ sin x = −1
情况二：sin x + 1 = 0 ⇒ sin x = −1

Now solve each equation separately within the interval 0 ≤ x < 2π.

现在分别在区间 0 ≤ x < 2π 内解每个方程。

6. Step 4: Find All Solutions | 第四步：求出所有解

For sin x = −1/2, recall that the reference angle is π/6 because sin(π/6) = 1/2. Since sine is negative in the third and fourth quadrants, the solutions are:

对于 sin x = −1/2，因为 sin(π/6) = 1/2，所以参考角为 π/6。由于正弦在第三和第四象限为负，因此解为：

x = π + π/6 = 7π/6

x = 2π − π/6 = 11π/6

For sin x = −1, the sine function equals −1 at exactly one point in the interval 0 ≤ x < 2π:

对于 sin x = −1，在区间 0 ≤ x < 2π 内，正弦函数取 −1 的点只有一个：

x = 3π/2

Therefore the complete solution set is:

因此完整的解集为：

x = 7π/6, 3π/2, 11π/6

7. Step 5: Verify the Solutions | 第五步：检验解

It is good practice to substitute the solutions back into the original equation to check for errors. For x = 7π/6:

将解代回原方程进行检验是一个好习惯，这样可以发现计算错误。对于 x = 7π/6：

sin(7π/6) = −1/2, cos(7π/6) = −√3/2

Then 2 cos² x − 3 sin x = 2 × (3/4) − 3 × (−1/2) = 3/2 + 3/2 = 3. Correct.

则 2 cos² x − 3 sin x = 2 × (3/4) − 3 × (−1/2) = 3/2 + 3/2 = 3。正确。

For x = 3π/2, sin x = −1 and cos x = 0, so 2(0) − 3(−1) = 3. Correct.

对于 x = 3π/2，sin x = −1，cos x = 0，所以 2(0) − 3(−1) = 3。正确。

For x = 11π/6, sin x = −1/2 and cos x = √3/2, giving 2 × (3/4) − 3 × (−1/2) = 3. Correct.

对于 x = 11π/6，sin x = −1/2，cos x = √3/2，得到 2 × (3/4) − 3 × (−1/2) = 3。正确。

8. Common Mistakes | 常见错误

A frequent mistake is forgetting to change the sign when moving the constant term. In this problem, moving “3” to the left and multiplying by −1 must be done carefully. Another common error is missing the solution 3π/2 because students forget that sin x can equal −1 as well as −1/2.

常见的错误是移动常数项时忘记变号。在本题中，将“3”移到左边并乘以 −1 时必须非常小心。另一个常见错误是漏解 3π/2，因为学生只注意到 sin x = −1/2，而忽略了 sin x 也可能等于 −1。

When using the interval 0 ≤ x < 2π, always sketch the sine graph or use the ASTC quadrant rule. This helps you avoid missing solutions in the third or fourth quadrants.

在使用区间 0 ≤ x < 2π 时，最好画出正弦函数草图或使用ASTC象限法则。这有助于避免漏掉第三或第四象限的解。

9. Exam Tip | 考试提示

In AQA A-Level Mathematics, when you see a trigonometric equation involving both sin x and cos² x, your first step should be to use sin² x + cos² x = 1 to express everything in terms of one trigonometric function. Then treat it as a quadratic equation.

在AQA A-Level数学考试中，当你看到同时含有 sin x 和 cos² x 的三角方程时，第一步应当使用 sin² x + cos² x = 1 将所有项化为同一个三角函数，然后将其当作二次方程来处理。

Always check that your final answers lie within the given interval. If the interval is 0 ≤ x < 2π, write your answers in radians with π. If the interval is in degrees, convert the reference angle accordingly.

始终检查最终答案是否位于给定区间内。如果区间是 0 ≤ x < 2π，答案要用含 π 的弧度制表示。如果区间使用角度制，则相应地转换参考角。

10. Practice Question | 练习题

Try this similar question on your own: Solve 2 cos² x + 5 sin x = 4 for 0 ≤ x < 2π. Give your answers in radians.

请自己尝试一个类似问题：解方程 2 cos² x + 5 sin x = 4，其中 0 ≤ x < 2π，答案用弧度制表示。

Solution guidance: rewrite cos² x as 1 − sin² x, simplify to 2 sin² x − 5 sin x + 2 = 0, factor to (2 sin x − 1)(sin x − 2) = 0. Since sin x cannot equal 2, only sin x = 1/2 is valid, giving x = π/6 and x = 5π/6.

解题提示：将 cos² x 改写为 1 − sin² x，化简得 2 sin² x − 5 sin x + 2 = 0，因式分解为 (2 sin x − 1)(sin x − 2) = 0。因为 sin x 不可能等于 2，所以只有 sin x = 1/2 可取，得到 x = π/6 和 x = 5π/6。

11. Summary | 总结

Example 6.12.2 demonstrates a complete strategy for solving a trigonometric quadratic equation. The essential steps are: apply the Pythagorean identity, rearrange into a quadratic, factorise, solve the two linear trigonometric equations, and select all solutions that fall inside the required interval.

例题6.12.2演示了求解三角二次方程的完整策略。关键步骤是：运用毕达哥拉斯恒等式、化成二次方程、因式分解、求解两个一次三角方程，并选出落在要求区间内的所有解。

Remember to work in radians, use the unit circle or ASTC rule, and verify your answers whenever time permits. Mastering this technique will help you gain secure marks in exam questions involving trigonometric equations.

记住使用弧度制，借助单位圆或ASTC法则，并在时间允许时检验答案。掌握这一技巧将帮助你在涉及三角方程的考试题目中稳妥拿分。

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Example 6.12.2: Solving Trigonometric Equations | 例题6.12.2：解三角方程