My Math Forum  

Go Back   My Math Forum > High School Math Forum > Trigonometry

Trigonometry Trigonometry Math Forum

Thanks Tree3Thanks
  • 2 Post By romsek
  • 1 Post By v8archie
LinkBack Thread Tools Display Modes
November 10th, 2018, 09:00 AM   #1
Joined: Nov 2018
From: Iran

Posts: 12
Thanks: 0

Math Focus: calculus
Red face help plz "equation"

the problem is :

I move one part to the other side

and :
which is :
but my book suggests a complete diffrent solution of
sin(3x+x)=0 ;sin(4x)=0
and it is solve with trigometric identities

now plz help is my solution wrong?
shadow dancer is offline  
November 10th, 2018, 12:04 PM   #2
Senior Member
romsek's Avatar
Joined: Sep 2015
From: USA

Posts: 2,529
Thanks: 1389

well let's see

I would solve this as

$\cos(3x)\sin(x) + \sin(3x)\cos(x) = 0$

$\sin(x+3x) = 0$

$\sin(4x) = 0$

$\sin(4x) = 0$

$4x = k \pi,~k \in \mathbb{Z}$

$x = \dfrac{k\pi}{4},~k \in \mathbb{Z}$

If you want to restrict the domain to say $[0,~2\pi)$ we have

$x = \dfrac{k \pi}{4},~k=0, 1, \dots, 7$

Now let's look at your solution

$\cos(3x)\sin(x) + \sin(3x)\cos(x) = 0$

$\cos(3x)\sin(x) = -\sin(3x)\cos(x)$

$\forall x \ni \sin(3x) \neq 0 \wedge \sin(x) \neq 0$
$\cot(3x) = -\cot(x) = \cot(-x) = \cot(-x + k \pi),~k\in \mathbb{Z}$

now we take inverse cotangents of both sides

$3x = -x + k\pi$

$4x = k\pi$

$x = \dfrac{k \pi}{4},~k \in \mathbb{Z}$

which is identical to the first solution.

There are still the corner cases of $\sin(x) = \sin(3x) = 0$ but you can show that those don't affect the solution.

So both methods are equivalent but the first way is a bit cleaner in my opinion.
Thanks from topsquark and shadow dancer
romsek is offline  
November 10th, 2018, 08:43 PM   #3
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,674
Thanks: 2654

Math Focus: Mainly analysis and algebra
The problem with the cotangent solution is that at $x=n\pi$, the contangent functions are undefined and cannot therefore be equal. In the sine solution, $x = n\pi$ is a valid solution.

If you are going to use the cotangent derivation, then you have to recognise that you have divided by both $\sin x$ and $\sin 3x$. This is not a valid operation when these expressions are equal to zero, so you have to validate those values of $x$ separately.
Thanks from shadow dancer
v8archie is offline  
November 11th, 2018, 09:34 AM   #4
Joined: Nov 2018
From: Iran

Posts: 12
Thanks: 0

Math Focus: calculus
reallllly thanks
shadow dancer is offline  

  My Math Forum > High School Math Forum > Trigonometry

equation, plz

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Does "free" mean "finitely generated"? zzzhhh Abstract Algebra 4 July 14th, 2017 10:33 AM
A "simple" application of dirac delta "shift theorem" SedaKhold Calculus 0 February 13th, 2012 11:45 AM
"separate and integrate" or "Orangutang method" The Chaz Calculus 1 August 5th, 2011 09:03 PM
sample exeriment-need help finding "statistic" and "result" katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM

Copyright © 2019 My Math Forum. All rights reserved.