My Math Forum help plz "equation"

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 November 10th, 2018, 09:00 AM #1 Newbie   Joined: Nov 2018 From: Iran Posts: 12 Thanks: 0 Math Focus: calculus help plz "equation" the problem is : cos(3x)sin(x)+sin3xcosx=0 I move one part to the other side cos(3x)sin(x)=-sin(3x)cos(x) and : cot3x=-cot(x) which is : cot3x=cot(π-x) 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?
 November 10th, 2018, 12:04 PM #2 Senior Member     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
 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
 November 11th, 2018, 09:34 AM #4 Newbie   Joined: Nov 2018 From: Iran Posts: 12 Thanks: 0 Math Focus: calculus reallllly thanks

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