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Sin(5x)=Sin(3x)

 Dec 4, 2015

Best Answer 

 #3
avatar+111321 
+15

sin(3x)  = sin(4x -x )  = sin4xcosx - sinxcos4x

 

sin(5x)  = sin(4x + x)  = sin4xcosx + sinxcos4x     .......so....

 

sin4xcosx + sinxcos4x  =  sin4xcosx - sinxcos4x

 

2sinxcos4x  = 0      divide both sides by 2

 

sinxcos4x = 0     

 

So either  

 

sinx  = 0     which happens at 0 + nPi    where n is an integer

 

or

 

cos(4x)  = 0

 

cos(x)  = 0   at   pi/2 and  3pi/2

 

So.....dividing each angle by 4 we have that

 

cos(4x)  =  0   at    pi/8 + n(pi/4)   where n is an integer

 

And   at    3pi/8 + n(pi/4).....however  the previous solution covers this one as well....so we have.......pi/8 + n(pi/4)

 

Here's the graph  of the intersection points [ in degrees].......https://www.desmos.com/calculator/jv5zqyexum

 

 

 

cool cool cool

 Dec 4, 2015
edited by CPhill  Dec 5, 2015
edited by CPhill  Dec 5, 2015
 #1
avatar+495 
+15

sin(5x) = sin(3x)

 

general solution in radians:

 

x = 0 + πn

 

x=π/8 + πn/4

 Dec 4, 2015
edited by LambLamb  Dec 4, 2015
 #2
avatar+109518 
+10

Good work LambLamb,

 

I got exactly the same answer.

But

are you going to show how you did it? 

 Dec 4, 2015
 #3
avatar+111321 
+15
Best Answer

sin(3x)  = sin(4x -x )  = sin4xcosx - sinxcos4x

 

sin(5x)  = sin(4x + x)  = sin4xcosx + sinxcos4x     .......so....

 

sin4xcosx + sinxcos4x  =  sin4xcosx - sinxcos4x

 

2sinxcos4x  = 0      divide both sides by 2

 

sinxcos4x = 0     

 

So either  

 

sinx  = 0     which happens at 0 + nPi    where n is an integer

 

or

 

cos(4x)  = 0

 

cos(x)  = 0   at   pi/2 and  3pi/2

 

So.....dividing each angle by 4 we have that

 

cos(4x)  =  0   at    pi/8 + n(pi/4)   where n is an integer

 

And   at    3pi/8 + n(pi/4).....however  the previous solution covers this one as well....so we have.......pi/8 + n(pi/4)

 

Here's the graph  of the intersection points [ in degrees].......https://www.desmos.com/calculator/jv5zqyexum

 

 

 

cool cool cool

CPhill Dec 4, 2015
edited by CPhill  Dec 5, 2015
edited by CPhill  Dec 5, 2015
 #4
avatar
+10

Solve for x:
sin(5 x) = sin(3 x)

Take the inverse sine of both sides:
5 x = pi-3 x+2 pi n_1  for  n_1  element Z
   or  5 x = 3 x+2 pi n_2  for  n_2  element Z

Add 3 x to both sides:
8 x = pi+2 pi n_1  for  n_1  element Z
   or  5 x = 3 x+2 pi n_2  for  n_2  element Z

Divide both sides by 8:
x = pi/8+(pi n_1)/4  for  n_1  element Z
   or  5 x = 3 x+2 pi n_2  for  n_2  element Z

Subtract 3 x from both sides:
x = pi/8+(pi n_1)/4  for  n_1  element Z
   or  2 x = 2 pi n_2  for  n_2  element Z

Divide both sides by 2:
Answer: | 
| x = pi/8+(pi n_1)/4  for  n_1  element Z
   or  x = pi n_2  for  n_2  element Z

 Dec 5, 2015
 #5
avatar+109518 
+10

sin(5x) = sin(3x)

Thanks LambLamb CPhill and guest #4    laugh

I think this answer is probably very similar to  guest     #4      

 

\(5x=3x+2\pi n\qquad or \qquad 5x=\pi-3x+2\pi n\\ 2x=2\pi n\qquad \qquad or \qquad 8x=\pi+2\pi n\\ x=\pi n\qquad \qquad \quad or \qquad x=\frac{\pi+2\pi n}{8}\qquad where\;\;n\in Z\\\)

.
 Dec 5, 2015

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