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Solve Sin(3x)+sin(4x)=0 

 Apr 15, 2019
edited by GermanManny  Apr 15, 2019
edited by GermanManny  Apr 15, 2019
edited by GermanManny  Apr 15, 2019
edited by GermanManny  Apr 15, 2019
edited by GermanManny  Apr 15, 2019
 #2
avatar+101746 
+2

Not so easy.

 

Here is the graph of y=sin(3x)+sin(4x)

The solutions are the intersections of this graph with y=0    (the x axis)

 

You can see that there are 9 solutions.    (including x=0 and x=2pi)

 

 Apr 15, 2019
 #3
avatar
+3

Trig identity

 \(\displaystyle \sin(A)+\sin(B)=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})\).

So,

\(\displaystyle \sin(3x)+\sin(4x)=2\sin(\frac{7x}{2})\cos(\frac{x}{2})\),

so

\(\displaystyle \sin(\frac{7x}{2})=0,\text{ or }\cos(\frac{x}{2})=0\)

etc..

 Apr 15, 2019
 #5
avatar+101746 
0

Thanks guest,

I should have remembered to use that identity. 

Tiggsy used it recnently and I spent a lot of time making full sense of his answer. 

Thanks for reminding me of it again!

Melody  Apr 15, 2019
 #4
avatar+22290 
+4

Solve for

\( 0 \text{ to } 2\pi\)

\(\sin(3x)+\sin(4x)=0 \)

 

\(\begin{array}{|lrcll|} \hline & \sin(3x)+\sin(4x) &=& 0 \quad | \quad -\sin(3x) \\ & \sin(4x) &=& -\sin(3x) \quad | \quad -\sin(3x) = \sin(-3x) \\ & \sin(4x) &=& \sin(-3x) \quad | \quad \arcsin() \\\\ (1) & 4x + 2\pi n &=& -3x + 2\pi m \quad | \quad +3x \\ & 7x &=& 2\pi z \\ & \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{2\pi}{7} z } \qquad z \in \mathbb{Z} \\\\ (2) & \sin(4x) &=& \sin(-3x) \quad | \quad \sin(-3x) = \sin(\pi -(-3x) ) \\ & \sin(4x ) &=& \sin(\pi + 3x ) \quad | \quad \arcsin() \\ & 4x +2\pi n &=& \pi+3x + 2\pi m \quad | \quad -3x \\ & x &=& \pi + 2\pi z \\ & \mathbf{x} &\mathbf{=}& \mathbf{ \pi + 2\pi z } \qquad z \in \mathbb{Z} \\ \hline \end{array}\)

 

 

\(\begin{array}{|r|r|r|} \hline & 0\le x\le2\pi & 0\le x\le2\pi \\ z & x=\dfrac{2\pi}{7} z & x = \pi + 2\pi z \\ \hline 0 & 0 & \pi \\ 1 & \dfrac{2\pi}{7} & \\ 2 & \dfrac{4\pi}{7} & \\ 3 & \dfrac{6\pi}{7} & \\ 4 & \dfrac{8\pi}{7} & \\ 5 & \dfrac{10\pi}{7} & \\ 6 & \dfrac{12\pi}{7} & \\ 7 & \dfrac{14\pi}{7}=2\pi & \\ \hline \end{array}\)

 

\(x = \left\{ 0,\ \dfrac{2\pi}{7},\ \dfrac{4\pi}{7},\ \dfrac{6\pi}{7},\ \pi ,\ \dfrac{8\pi}{7},\ \dfrac{10\pi}{7},\ \dfrac{12\pi}{7},\ 2\pi \right\}\)

 

 

laugh

 Apr 15, 2019
edited by heureka  Apr 15, 2019
edited by heureka  Apr 15, 2019
 #6
avatar+101746 
+3

Thanks Heureka,

Another great answer from you.  laugh cool laugh

Melody  Apr 15, 2019
 #7
avatar+22290 
+3

Thank you, Melody !

 

laugh

heureka  Apr 15, 2019

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