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IAL 2020 Oct Q1

A Level / Edexcel / P3

IAL 2020 Oct Paper · Question 1

题目

Problem

Solve, for 0x<3600\le x<360^\circ, the equation

2cos2x=7cosx2\cos 2x=7\cos x

giving your solutions to one decimal place.

(Solutions based entirely on graphical or numerical methods are not acceptable.)

(5)
题目中文翻译

0x<3600\le x<360^\circ 内解方程

2cos2x=7cosx2\cos 2x=7\cos x

答案精确到小数点后 1 位。

(不接受完全基于图像法或数值法的解答。)

解答

We need to solve

2cos2x=7cosx2\cos2x=7\cos x

Use the identity

cos2x=2cos2x1\cos2x=2\cos^2x-1

Then

2(2cos2x1)=7cosx2(2\cos^2x-1)=7\cos x

Expand:

4cos2x2=7cosx4\cos^2x-2=7\cos x

Rearrange:

4cos2x7cosx2=04\cos^2x-7\cos x-2=0

Let

u=cosxu=\cos x

Then

4u27u2=04u^2-7u-2=0

Factorise:

(4u+1)(u2)=0(4u+1)(u-2)=0

So

u=14oru=2u=-\frac14\quad \text{or}\quad u=2

Since

1cosx1-1\leq \cos x\leq 1

we reject

u=2u=2

Thus

cosx=14\cos x=-\frac14

For

0x<3600\leq x<360^\circ

the solutions are in the second and third quadrants.

So

x=104.477x=104.477\ldots^\circ

or

x=255.522x=255.522\ldots^\circ

Therefore, to one decimal place,

x=104.5, 255.5\boxed{x=104.5^\circ,\ 255.5^\circ}