Sin(5x)=Sin(3x)

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

sin(5x) = sin(3x)

general solution in radians:

x = 0 + πn

x=π/8 + πn/4

Good work LambLamb,

I got exactly the same answer.

But

are you going to show how you did it? 

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

sin(5x) = sin(3x)

Thanks LambLamb CPhill and guest #4    

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

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