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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 Tags equation, plz Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zzzhhh Abstract Algebra 4 July 14th, 2017 10:33 AM SedaKhold Calculus 0 February 13th, 2012 11:45 AM The Chaz Calculus 1 August 5th, 2011 09:03 PM katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM

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