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November 10th, 2018, 09:00 AM   #1
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Red face 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?
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November 10th, 2018, 12:04 PM   #2
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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
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November 10th, 2018, 08:43 PM   #3
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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.
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November 11th, 2018, 09:34 AM   #4
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reallllly thanks
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