题目

Problem

In this question you must show detailed reasoning. Solutions relying entirely on calculator technology are not acceptable.

(a) Show that the equation

2\tan\theta=3\cos\theta

can be written as

3\sin^2\theta+2\sin\theta-3=0

(3)

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

2\tan\left(2x+\frac{\pi}{3}\right) = 3\cos\left(2x+\frac{\pi}{3}\right)

giving your answers to 3 significant figures.

(4)

解答

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把 $\tan\theta$ 写成 $\frac{\sin\theta}{\cos\theta}$ ，然后用 $\cos^2\theta=1-\sin^2\theta$ 。

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Starting with

\begin{align*} 2\tan\theta=3\cos\theta, \end{align*}

use $\tan\theta=\frac{\sin\theta}{\cos\theta}$ :

\begin{align*} 2\cdot\frac{\sin\theta}{\cos\theta}=&\,3\cos\theta. \end{align*}

Multiply by $\cos\theta$ :

\begin{align*} 2\sin\theta=&\,3\cos^2\theta. \end{align*}

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

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

This is the required form.

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令 $\theta=2x+\frac{\pi}{3}$ ，直接用 (a) 得到关于 $\sin\theta$ 的二次方程。然后把所有 $\theta$ 的解转回 $x$ ，并筛选 $-\pi<x<\pi$ 。

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Let

\begin{align*} \theta=2x+\frac{\pi}{3}. \end{align*}

Using part (a),

\begin{align*} 3\sin^2\theta+2\sin\theta-3=0. \end{align*}

Solving this quadratic,

\begin{align*} \sin\theta =&\,\frac{-2\pm\sqrt{2^2-4(3)(-3)}}{2(3)}\\ =&\,\frac{-2\pm\sqrt{40}}{6}\\ =&\,\frac{-1\pm\sqrt{10}}{3}. \end{align*}

Only

\begin{align*} \sin\theta=\frac{-1+\sqrt{10}}{3} \end{align*}

is possible, since the other root is less than $-1$ .

Solving

\sin\left(2x+\frac{\pi}{3}\right) =\frac{-1+\sqrt{10}}{3}

for $-\pi<x<\pi$ gives

\begin{align*} x=-2.50,\quad -0.121,\quad 0.645,\quad 3.02 \end{align*}

to 3 significant figures.