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

\cos\theta\left(3\tan\theta+\frac{2}{\tan\theta}\right) \equiv \sin\theta+\frac{2}{\sin\theta}, \qquad \theta\ne\frac{n\pi}{2}

(4)

(b) Hence solve, for $0<x<2\pi$ , the equation

\cos x\left(3\tan x+\frac{2}{\tan x}\right)=4\sin x-5

giving your answers to 3 significant figures.

(4)

解答

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把 $\tan\theta$ 写成 $\frac{\sin\theta}{\cos\theta}$ 。目标式只含 $\sin\theta$ ，所以还要用 $\cos^2\theta=1-\sin^2\theta$ 。

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Starting from the left-hand side,

\begin{align*} &\cos\theta\left(3\tan\theta+\frac{2}{\tan\theta}\right)\\ =&\,\cos\theta\left( 3\frac{\sin\theta}{\cos\theta} +2\frac{\cos\theta}{\sin\theta} \right)\\ =&\,3\sin\theta+\frac{2\cos^2\theta}{\sin\theta}. \end{align*}

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

\begin{align*} 3\sin\theta+\frac{2\cos^2\theta}{\sin\theta} =&\,3\sin\theta+\frac{2(1-\sin^2\theta)}{\sin\theta}\\ =&\,3\sin\theta+\frac{2}{\sin\theta} -\frac{2\sin^2\theta}{\sin\theta}\\ =&\,3\sin\theta+\frac{2}{\sin\theta}-2\sin\theta\\ =&\,\sin\theta+\frac{2}{\sin\theta}. \end{align*}

Therefore,

\cos\theta\left(3\tan\theta+\frac{2}{\tan\theta}\right) \equiv \sin\theta+\frac{2}{\sin\theta}.

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用第 (a) 题把左边换成 $\sin x+\frac{2}{\sin x}$ ，然后令 $s=\sin x$ 解二次方程。最后要回到 $0<x<2\pi$ 找所有角。

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Using part (a), the equation becomes

\begin{align*} \sin x+\frac{2}{\sin x}=&\,4\sin x-5. \end{align*}

Multiply by $\sin x$ :

\begin{align*} \sin^2x+2=&\,4\sin^2x-5\sin x\\ 3\sin^2x-5\sin x-2=&\,0. \end{align*}

Let $s=\sin x$ . Then

\begin{align*} 3s^2-5s-2=&\,0\\ (3s+1)(s-2)=&\,0. \end{align*}

\begin{align*} s=2\quad\text{or}\quad s=-\frac13. \end{align*}

Since $\sin x=2$ is impossible,

\begin{align*} \sin x=-\frac13. \end{align*}

For $0<x<2\pi$ , this gives

\begin{align*} x=&\,\pi+\sin^{-1}\left(\frac13\right),\\ x=&\,2\pi-\sin^{-1}\left(\frac13\right). \end{align*}

Therefore,

\begin{align*} x=3.48,\quad 5.94 \end{align*}

to 3 significant figures.