题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(a) Show that

3\cos\theta\,(\tan\theta\sin\theta + 3) = 11 - 5\cos\theta

may be written as

3\cos^2\theta - 14\cos\theta + 8 = 0

(3)

(b) Hence solve, for $0° < x < 360°$ ,

3\cos 2x\,(\tan 2x\sin 2x + 3) = 11 - 5\cos 2x

giving your answers to one decimal place.

(4)

题目中文翻译

本题必须展示全部计算步骤。完全依赖计算器技术的解答不予接受。

(a) 证明

3\cos\theta(\tan\theta\sin\theta+3)=11-5\cos\theta

可写成 $3\cos^2\theta-14\cos\theta+8=0$ 。（3 分）

(b) 由此在 $0<x<360^\circ$ 范围内解方程

3\cos2x(\tan2x\sin2x+3)=11-5\cos2x,

答案保留一位小数。（4 分）

解答

展开

将正切写为正弦除以余弦，展开后使用平方恒等式，统一成关于余弦的二次式。

展开

Expand the left side:

\begin{align*} &\,3\cos\theta(\tan\theta\sin\theta + 3)\\ =&\,3\cos\theta \cdot \frac{\sin\theta}{\cos\theta} \cdot \sin\theta\\ &\,\hspace{2pt}+9\cos\theta\\ =&\, 3\sin^2\theta + 9\cos\theta \end{align*}

Using $\sin^2\theta = 1 - \cos^2\theta$ :

\begin{align*} 3(1 - \cos^2\theta) + 9\cos\theta =&\, 11 - 5\cos\theta \\ 3 - 3\cos^2\theta + 9\cos\theta =&\, 11 - 5\cos\theta \\ -3\cos^2\theta + 14\cos\theta - 8 =&\, 0 \\ 3\cos^2\theta - 14\cos\theta + 8 =&\, 0 \end{align*}

\boxed{3\cos^2\theta - 14\cos\theta + 8 = 0}

展开

使用 (a) 的二次式，排除超出余弦取值范围的根。令角为 $2x$ ，在 $0<2x<720^\circ$ 内列出四个角，最后除以 2 并舍入。

展开

Let $\theta = 2x$ . From part (a):

(3\cos\theta - 2)(\cos\theta - 4) = 0

Since $|\cos\theta| \leqslant 1$ , $\cos\theta = 4$ is impossible.

\cos\theta = \frac{2}{3}

Let $\beta=\cos^{-1}(2/3)=48.189685104\ldots^\circ$ . Keep this value unrounded until the final answers.

For $0° < x < 360°$ , we have $0° < 2x < 720°$ , so:

\begin{align*} 2x=&\,\beta,\quad360^\circ-\beta,\\ &\,360^\circ+\beta,\quad720^\circ-\beta. \end{align*}

Dividing the unrounded angles by 2 gives

\begin{align*} x\approx&\,24.094842552^\circ,\quad155.905157448^\circ,\\ &\,204.094842552^\circ,\quad335.905157448^\circ. \end{align*}

\boxed{x = 24.1°, \quad 155.9°, \quad 204.1°, \quad 335.9°}