题目

Problem

(a) Use the substitution $t=\tan x$ to show that the equation

12\tan 2x+5\cot x\sec^2x=0

can be written in the form

5t^4-24t^2-5=0

(4)

(b) Hence solve, for $0\le x<360^\circ$ , the equation

12\tan 2x+5\cot x\sec^2x=0

Show each stage of your working and give your answers to one decimal place.

(4)

题目中文翻译

(a) 使用代换 $t=\tan x$ ，证明方程

12\tan 2x+5\cot x\sec^2x=0

可写成

5t^4-24t^2-5=0

的形式。

(b) 由此在 $0\le x<360^\circ$ 内解方程

12\tan 2x+5\cot x\sec^2x=0

写出每一步解题过程，并将答案精确到小数点后 1 位。

解答

Let

t=\tan x

We use

\tan2x=\frac{2\tan x}{1-\tan^2x}

\tan2x=\frac{2t}{1-t^2}

Also,

\cot x=\frac1t

and

\sec^2x=1+\tan^2x=1+t^2

Substitute these into the equation

12\tan2x+5\cot x\sec^2x=0

This gives

12\left(\frac{2t}{1-t^2}\right)+5\left(\frac1t\right)(1+t^2)=0

\frac{24t}{1-t^2}+\frac{5(1+t^2)}{t}=0

For the original equation to be defined, $t$ must be finite and $t\ne0,\pm1$ . Therefore we can multiply by $t(1-t^2)$ :

24t^2+5(1+t^2)(1-t^2)=0

Now

(1+t^2)(1-t^2)=1-t^4

24t^2+5(1-t^4)=0

Hence

24t^2+5-5t^4=0

Therefore

\boxed{5t^4-24t^2-5=0}

From part (a),

5t^4-24t^2-5=0

Let

u=t^2

Then

5u^2-24u-5=0

Factorise:

5u^2-24u-5=(5u+1)(u-5)

(5u+1)(u-5)=0

Therefore

u=-\frac15\quad \text{or}\quad u=5

But

u=t^2\geq 0

t^2=5

Hence

\tan^2x=5

Therefore

\tan x=\pm \sqrt5

Both values of $\tan x$ are finite and avoid $0,\pm1$ , so neither introduces an excluded angle. The acute angle is

\arctan\sqrt5=65.905\ldots^\circ

For

0\leq x<360^\circ

the solutions are

x=65.905\ldots^\circ,\quad 114.094\ldots^\circ,\quad 245.905\ldots^\circ,\quad 294.094\ldots^\circ

So, to one decimal place,

\boxed{x=65.9^\circ,\ 114.1^\circ,\ 245.9^\circ,\ 294.1^\circ}