📚 Worked Example 5.5.4: Solving Trigonometric Equations in Radians | 例题 5.5.4：弧度制下解三角方程
In this worked example we solve the trigonometric equation sin 2x = cos x for 0 ≤ x < 2π. The problem is typical of AQA A-Level Pure Mathematics questions on trigonometric identities and equations, where a double-angle identity must be used before factorising.
在本例题中，我们求解三角方程 sin 2x = cos x，其中 0 ≤ x < 2π。这是 AQA A-Level 纯数学中关于三角恒等式与三角方程的典型题目，需要先使用倍角公式，再进行因式分解。
We will develop the solution step by step, paying close attention to the use of radians, the correct application of the identity sin 2x ≡ 2 sin x cos x, and the selection of all solutions within the required interval.
我们将逐步推导解答，重点关注弧度的使用、恒等式 sin 2x ≡ 2 sin x cos x 的正确应用，以及在给定区间内选择所有解。
1. Problem Statement | 题目陈述
The equation we are asked to solve is shown below. The interval is given in radians, which is the default measure in AQA A-Level Mathematics unless the question states otherwise.
我们要求解的方程如下所示。区间以弧度给出，除非题目另有说明，弧度制是 AQA A-Level 数学中的默认度量方式。
Solve sin 2x = cos x for 0 ≤ x < 2π.
The notation 0 ≤ x < 2π means that x can take any value from 0 inclusive to 2π exclusive. In degree terms this is equivalent to 0° ≤ x < 360°.
记号 0 ≤ x < 2π 表示 x 可以取从 0 到 2π 的任意值，其中包含 0 但不包含 2π。用角度制表示，这等价于 0° ≤ x < 360°。
Because the equation contains both sin 2x and cos x, it cannot be solved immediately by reading values from a graph. We need to rewrite the left-hand side so that both sides are expressed in terms of the same angle and the same trigonometric functions.
由于方程同时含有 sin 2x 和 cos x，因此不能直接通过图像读数求解。我们需要重写左边，使两边用相同的角和相同的三角函数表示。
2. First Observations | 初步观察
There is no simple value of x for which sin 2x and cos x are obviously equal, so a substitution method or an identity is required. The presence of sin 2x strongly suggests using the double-angle identity for sine.
没有简单的 x 值能使 sin 2x 和 cos x 显然相等，因此需要使用代换法或恒等式。sin 2x 的出现强烈提示使用正弦的倍角公式。
It is important not to divide both sides by cos x at this stage. Division by a trigonometric expression can remove valid solutions because that expression might equal zero for some values of x in the interval.
现阶段不要将方程两边同时除以 cos x。除以一个三角表达式可能会丢失有效解，因为该表达式在区间内的某些 x 值处可能为零。
Instead, we should rearrange the equation so that one side equals zero, then factorise. This approach preserves all solutions and is the method expected in AQA mark schemes.
相反，我们应重排方程使一边等于零，然后进行因式分解。这种方法能保留所有解，也是 AQA 评分标准中期望使用的方法。
3. Applying the Double-Angle Identity | 应用倍角公式
The double-angle identity for sine states that sin 2x ≡ 2 sin x cos x. This identity is valid for all real values of x, so we can use it to rewrite the equation without changing its solution set.
正弦的倍角公式为 sin 2x ≡ 2 sin x cos x。该恒等式对所有实数 x 都成立，因此我们可以用它来重写方程而不改变其解集。
sin 2x ≡ 2 sin x cos x
Substituting this expression into the left-hand side of the original equation gives the following equivalent equation.
将该表达式代入原方程的左边，得到如下等价方程。
2 sin x cos x = cos x
Now both terms involve cos x, so we can bring all terms to one side and look for a common factor. This is a standard step in solving trigonometric equations at A-Level.
现在两项都含有 cos x，因此我们可以将所有项移到一边并寻找公因式。这是 A-Level 解三角方程的标准步骤。
4. Rearranging and Factorising | 重排与因式分解
Subtract cos x from both sides to set the equation equal to zero. This is essential before factorising, because a product can be zero only when one of its factors is zero.
两边同时减去 cos x，使方程右边为零。这在因式分解之前是必要的，因为只有当一个因式为零时，乘积才可能为零。
2 sin x cos x − cos x = 0
The common factor is cos x, so we can factorise the left-hand side as follows.
公因式为 cos x，因此我们可以将左边因式分解如下。
cos x (2 sin x − 1) = 0
This factorised form is much easier to solve because it splits the original equation into two simpler equations. We now use the zero product property: if the product equals zero, then at least one factor must equal zero.
这种因式分解后的形式更容易求解，因为它将原方程拆分为两个较简单的方程。现在我们使用零乘积性质：如果乘积为零，那么至少有一个因式必须为零。
5. Solving the First Factor cos x = 0 | 解第一个因式 cos x = 0
The first factor gives the equation cos x = 0. We need to find all angles x in the interval 0 ≤ x < 2π for which the cosine function is zero.
第一个因式给出方程 cos x = 0。我们需要找出在区间 0 ≤ x < 2π 内所有使余弦函数为零的角 x。
From the unit circle or the graph of y = cos x, the cosine function equals zero at the top and bottom of the unit circle, that is at x = π/2 and x = 3π/2.
根据单位圆或 y = cos x 的图像，余弦函数在单位圆的顶部和底部为零，即 x = π/2 和 x = 3π/2。
cos x = 0 → x = π/2, 3π/2
These two values are inside the required interval, so they are both valid solutions of the first factor. We do not reject either one, and we do not need to add 2π to them because 2π is not included.
这两个值都在给定区间内，因此它们都是第一个因式的有效解。我们不应舍弃其中任何一个，也不需要给它们加上 2π，因为 2π 不包含在区间内。
6. Solving the Second Factor 2 sin x − 1 = 0 | 解第二个因式 2 sin x − 1 = 0
The second factor gives the equation 2 sin x − 1 = 0. Rearranging this equation gives sin x = 1/2.
第二个因式给出方程 2 sin x − 1 = 0。重排该方程得到 sin x = 1/2。
2 sin x − 1 = 0 → sin x = 1/2
The sine function is positive in the first quadrant and the second quadrant. The first-quadrant reference angle for which sin x = 1/2 is x = π/6.
正弦函数在第一象限和第二象限为正。满足 sin x = 1/2 的第一象限参考角是 x = π/6。
In the second quadrant, the corresponding angle is π − π/6 = 5π/6. Both of these values lie inside the interval 0 ≤ x < 2π, so both are valid solutions of the second factor.
在第二象限，对应的角为 π − π/6 = 5π/6。这两个值都位于区间 0 ≤ x < 2π 内，因此都是第二个因式的有效解。
sin x = 1/2 → x = π/6, 5π/6
If we had worked in degrees, these answers would be 30° and 150°. The AQA specification expects fluency in converting between degrees and radians, but the final answer should match the measure used in the question.
如果我们使用角度制，这些答案应为 30° 和 150°。AQA 考试要求熟练地在角度制与弧度制之间转换，但最终答案应与题目所用的度量方式一致。
7. Collecting All Solutions | 汇总所有解
We now combine the solutions from both factors. The complete solution set consists of four values, all expressed in radians and lying in the required interval.
现在我们合并两个因式的解。完整的解集由四个值组成，全部用弧度表示，并且都在给定区间内。
x = π/6, π/2, 5π/6, 3π/2
It is helpful to arrange the solutions in increasing order. This makes it easier to check that none are outside the interval and that none have been duplicated.
将解按递增顺序排列会很有帮助。这样可以更容易检查是否有解超出区间，或者是否有重复。
| Factor | Equation | Solutions in 0 ≤ x < 2π |
|---|---|---|
| First factor | cos x = 0 | π/2, 3π/2 |
| Second factor | sin x = 1/2 | π/6, 5π/6 |
The table shows that the equation has exactly four solutions in the given interval. No additional solutions arise from periodicity because 2π is not included in the interval.
表格显示该方程在给定区间内恰好有四个解。由于 2π 不包含在区间内，周期性的加入不会产生额外的解。
8. Checking Solutions in the Original Equation | 代回原方程检验
Substituting each solution into the original equation is an excellent way to verify the answers. We show two of the checks here, but all four can be verified in the same way.
将每个解代回原方程是验证答案的一种极好方法。我们在这里展示其中两个检验，其余四个都可以用相同方式验证。
For x = π/6, we find sin 2x = sin π/3 = √3/2 and cos x = cos π/6 = √3/2, so the two sides are equal.
当 x = π/6 时，得到 sin 2x = sin π/3 = √3/2，而 cos x = cos π/6 = √3/2，因此两边相等。
For x = π/2, we find sin 2x = sin π = 0 and cos x = cos π/2 = 0, so the equation is also satisfied.
当 x = π/2 时，得到 sin 2x = sin π = 0，而 cos x = cos π/2 = 0，因此方程也成立。
Similar substitutions for x = 5π/6 and x = 3π/2 confirm that all four values are correct. This checking step is not always required for full marks, but it is useful for catching sign errors or incorrect quadrant choices.
对 x = 5π/6 和 x = 3π/2 进行类似代换可以确认所有四个值都是正确的。这一检验步骤并不总是得到满分所必需，但对于发现符号错误或象限选择错误非常有用。
9. Common Mistakes and Misconceptions | 常见错误与误区
A very common mistake is to divide both sides of the equation by cos x. This gives 2 sin x = 1, which leads only to x = π/6 and x = 5π/6, and the solutions from cos x = 0 are lost.
一个很常见的错误是将方程两边同时除以 cos x。这样得到 2 sin x = 1，只能得出 x = π/6 和 x = 5π/6，而 cos x = 0 的解则被丢失。
Another common mistake is to solve sin x = 1/2 by writing only x = π/6. Since the sine function is also positive in the second quadrant, the solution x = 5π/6 is required as well.
另一个常见错误是在解 sin x = 1/2 时只写出 x = π/6。由于正弦函数在第二象限也为正，因此还需要 x = 5π/6 这个解。
Some students also confuse the solutions of cos x = 0 with those of sin x = 0. Remember that cos x = 0 at π/2 and 3π/2, whereas sin x = 0 at 0, π and 2π.
有些学生还会将 cos x = 0 的解与 sin x = 0 的解混淆。请记住 cos x = 0 出现在 π/2 和 3π/2，而 sin x = 0 出现在 0、π 和 2π。
Finally, mixing radians and degrees in the final answer is not acceptable. If the interval is given as 0 ≤ x < 2π, the solutions must be given in radians.
最后，在最终答案中混用弧度和角度是不可接受的。如果区间以 0 ≤ x < 2π 给出，解就必须以弧度表示。
10. AQA Exam Technique and Marking Points | AQA 考试技巧与得分点
In an AQA mark scheme, the first mark is usually awarded for applying the double-angle identity correctly. Write sin 2x = 2 sin x cos x explicitly before substituting.
在 AQA 评分标准中，第一个得分点通常是正确应用倍角公式。请先明确写出 sin 2x = 2 sin x cos x，然后再代入。
The next mark is for rearranging to zero and factorising. You should show the line cos x (2 sin x − 1) = 0 clearly, because this demonstrates the method.
下一个得分点是移项使右边为零并进行因式分解。你应清楚地写出 cos x (2 sin x − 1) = 0 这一行，因为这展示了方法。
Marks are then awarded for solving each factor and for giving all four solutions. Using a unit-circle sketch or graph can help you avoid missing the second-quadrant solution.
之后的得分点用于解每个因式并给出所有四个解。绘制单位圆草图或图像可以帮助你避免遗漏第二象限的解。
When writing the final answer, present it as a solution set or a list: x = π/6, π/2, 5π/6, 3π/2. Under AQA conventions, these exact values are preferred over decimal approximations.
书写最终答案时，请以解集或列表的形式呈现：x = π/6, π/2, 5π/6, 3π/2。按照 AQA 惯例，这些精确值优于小数近似值。
11. Extension: General Solution and Degrees | 拓展：通解与角度制
If the question asked for all real solutions rather than solutions
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