📚 Example 4.3.2: Solving a Trigonometric Equation | 例题4.3.2：求解三角方程
In this worked example, we solve a typical AQA A-Level Mathematics problem that involves using a Pythagorean identity to reduce a trigonometric equation to a quadratic equation. Understanding this process is essential for Paper 1 or Paper 2 of the Pure Mathematics examination.
在本例题中，我们求解一道典型的AQA A-Level数学题目：通过毕达哥拉斯恒等式将三角方程化为二次方程。掌握这一过程对纯数学Paper 1或Paper 2至关重要。
1. The Problem | 题目
Solve the equation 2cos²x = 1 − sin x for 0° ≤ x < 360°.
求解方程 2cos²x = 1 − sin x，其中 0° ≤ x < 360°。
The equation mixes sine and cosine, so we first try to rewrite it in terms of only one trigonometric function.
该方程同时含正弦和余弦，因此我们首先尝试用单一的三角函数来重写它。
2. Why This Example Matters | 本例题的重要性
This type of question appears frequently in AQA exam papers because it tests several core skills: manipulation of trigonometric identities, solving quadratic equations, and finding all solutions in a given interval.
这类问题在AQA考试中频繁出现，因为它考查多项核心技能：三角恒等式的变形、二次方程的求解，以及在给定区间内求全部解。
It also reminds students to be careful with algebra signs and to use the symmetry of the sine graph to generate all valid angles.
它还提醒学生注意代数符号，并利用正弦图像的对称性产生所有有效角度。
3. Key Identity | 核心恒等式
Recall the Pythagorean identity: sin²x + cos²x = 1.
回顾毕达哥拉斯恒等式：sin²x + cos²x = 1。
Rearranging this identity gives cos²x = 1 − sin²x.
变形该恒等式可得 cos²x = 1 − sin²x。
This substitution is valid for every real value of x, because the identity holds for all angles.
这一代入对任意实数x都有效，因为该恒等式对所有角度都成立。
4. Transforming the Equation | 转化方程
Substitute cos²x = 1 − sin²x into the original equation.
将 cos²x = 1 − sin²x 代入原方程。
The equation becomes 2(1 − sin²x) = 1 − sin x.
方程变为 2(1 − sin²x) = 1 − sin x。
Expand the left-hand side: 2 − 2sin²x = 1 − sin x.
展开左边：2 − 2sin²x = 1 − sin x。
Bring all terms to one side: 2 − 2sin²x − 1 + sin x = 0.
移项到等号一边：2 − 2sin²x − 1 + sin x = 0。
Simplify the constants: −2sin²x + sin x + 1 = 0.
合并常数项：−2sin²x + sin x + 1 = 0。
Multiplying both sides by −1 gives the standard quadratic form:
两边同乘 −1 得到标准二次形式：
2sin²x − sin x − 1 = 0
5. Factorising the Quadratic | 因式分解二次式
We aim to factorise the quadratic expression 2sin²x − sin x − 1.
我们目标是因式分解二次表达式 2sin²x − sin x − 1。
Find two numbers that multiply to 2 × (−1) = −2 and add to −1 (the coefficient of sin x).
寻找两个数，使它们的乘积为 2 × (−1) = −2，并且和为 sin x 的系数 −1。
These numbers are −2 and 1, so we split the middle term:
这两个数是 −2 和 1，因此我们裂项：
2sin²x − 2sin x + sin x − 1 = 0.
2sin²x − 2sin x + sin x − 1 = 0。
Factorise in pairs: 2sin x(sin x − 1) + 1(sin x − 1) = 0.
分组因式分解：2sin x(sin x − 1) + 1(sin x − 1) = 0。
Hence we obtain:
因此我们得到：
(2sin x + 1)(sin x − 1) = 0
6. Solving the Linear Factors | 解线性因式
Using the zero product property, we set each factor equal to zero:
根据零积性质，令每个因式等于零：
2sin x + 1 = 0 or sin x − 1 = 0.
2sin x + 1 = 0 或 sin x − 1 = 0。
From the first equation, we get sin x = −½.
由第一个方程得 sin x = −½。
From the second equation, we get sin x = 1.
由第二个方程得 sin x = 1。
Both values are valid because −1 ≤ sin x ≤ 1.
两个值都有效，因为 −1 ≤ sin x ≤ 1。
7. Finding Angles from sin x = 1 | 从 sin x = 1 求角
We solve sin x = 1 for 0° ≤ x < 360°.
在 0° ≤ x < 360° 中求解 sin x = 1。
On the unit circle, sin x = 1 occurs at the top of the circle, which corresponds to x = 90°.
在单位圆上，sin x = 1 出现在圆的最顶部，对应 x = 90°。
Within one revolution, this is the only solution.
在旋转一周内，这是唯一解。
8. Finding Angles from sin x = −½ | 从 sin x = −½ 求角
We now solve sin x = −½ for 0° ≤ x < 360°.
现在在 0° ≤ x < 360° 中求解 sin x = −½。
Since sine is negative in the third and fourth quadrants, there will be two solutions.
由于正弦在第三、四象限为负，因此将有两个解。
The reference angle is the acute angle whose sine is ½: that is 30°.
参考角是正弦值为 ½ 的锐角，即 30°。
For the third quadrant, the angle is 180° + 30° = 210°.
对于第三象限，角度为 180° + 30° = 210°。
For the fourth quadrant, the angle is 360° − 30° = 330
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