Solve $\cos3x=\cos4x$

Question

I want to solve the equation $\cos3x=\cos4x$. The given solutions are $x= 0$, $2\pi/7$, $4\pi/7$ and $6\pi/7$.

My first approach was to write the whole thing in terms of $\cos x$ this gave,

$0=(\cos x - 1)(8\cos^3x + 4\cos^2x - 4\cos x - 1)$.

This gave me the obvious solution of $\cos x = 1$ and therefore $x=0$, however I don't know how to tackle the second set of brackets. I've also thought about writing in terms of exponentials, but didn't get to anything simpler.

Another way I tried was to say that $4x = \cos^{-1}(\cos3x)$

$\therefore 4x = 3x + (2n\pi)$.

However this just gives that $x$ is $2n\pi$, which ignores the 7 given in the solution!

There is no mention of an interval in the question and yet still only the 4 solutions given?

I would really appreciate any help, I'm studying for an exam in a few weeks time, and would hate to have an unsolved problem!

Thank you in advance.

Hint: $\cos a-\cos b=2\sin\frac{a-b}2\sin\frac{a+b}2$. – Tom-Tom May 28 '15 at 12:23 — Tom-Tom, May 28 '15 at 12:23

score 6 · Answer 1 · edited Apr 13 '17 at 12:20

6

HINT:

Using Basic trigonometry intuition, we have $4x=2m\pi\pm3x$

Considering the '+' sign, $x=2m\pi$

Considering the '-' sign, $x=2m\pi/7$ where $0\le m\le6$

But $\cos(2\pi-y)=\cos y$

So, $\cos\dfrac{(7-r)2\pi}7=\cos\left[2\pi-\dfrac{2r\pi}7\right]=\cos\dfrac{2r\pi}7$

Set $r=1,2,3$

edited Apr 13 '17 at 12:20

Community

1

answered May 28 '15 at 12:16

lab bhattacharjee

274,582

Thankyou! How do you narrow it down to just the 4 solutions given? – Grace May 28 '15 at 12:21
1

@Grace: You can't, because there are an infinite number of solutions. I assume that the original question asked for solutions in $[0,\pi]$, and that you forgot to supply us with this information.. – TonyK May 28 '15 at 12:24

absolute friend · Answer 2 · 2015-05-28T12:30:41.030

6

$\cos 4x=\cos 3x$

$\implies 4x=2n\pi\pm 3x$ where $n\in \mathbb{Z}$

Taknig positive sign we have

$4x=2n\pi + 3x $

$\implies x=2n\pi ~ ;n\in \mathbb{Z}$

Taking negative sign we have

$4x=2n\pi - 3x$

$\implies 7x=2n\pi$

$\implies x= \frac{2n\pi}{7} ~; n\in \mathbb{Z}$

edited May 28 '15 at 12:30

answered May 28 '15 at 12:26

absolute friend

1,308

nice answer (+1). – abel May 28 '15 at 12:29
That is completely true. you check it again – absolute friend May 28 '15 at 12:35
@5xum, $x=2n\pi\pm y\implies $ either $x=2n\pi+y$ or $x=2n\pi-y$ right? – lab bhattacharjee May 28 '15 at 12:36
Yeah ! what bhattacharjee is saying is true – absolute friend May 28 '15 at 12:38
1

@5xum If $x=\frac{\pi}2$ and $y=-\frac{\pi}2$ then $x=2n\pi\pm y$ for $n=0$ if you choose for minus. I don't see your point. – drhab May 28 '15 at 13:24
@drhab Yup. Sorry. By bad. Sorry for spamming. – 5xum May 28 '15 at 13:26
very nice answer........ – Jul 05 '15 at 15:57

score 2 · Answer 3 · answered May 28 '15 at 12:31

Looking at the solutions you gave, I assume that you are only interested in the interval $[0, \pi].$

Let $x \in [0, \pi].$ Then $\cos 3x - \cos 4x = 0$ iff $2\sin (-x/2)\sin(7x/2) = 0$ iff $\sin (-x/2) \sin(7x/2) = 0$ iff $x = 0, 2\pi/7, 4\pi/7, 6\pi/7$ (by inspection).

Solve $\cos3x=\cos4x$

3 Answers3