SOLUTION: If sinx-3sin2x+sin3x=cosx-3cos2x+cos3x,then x=

sin(x)-3sin(2x)+sin(3x) = cos(x)-3cos(2x)+cos3x

Write x as 2x-x and 3x as 2x+x

sin(2x-x)-3sin(2x)+sin(2x+x) = cos(2x-x)-3cos(2x)+cos(2x+x)

Left hand side's 1st and 3rd terms:

sin(2x-x) = sin(2x)cos(x)-cos(2x)sin(x)
sin(2x+x) = sin(2x)cos(x)+cos(2x)sin(x)

So left hand side becomes

2sin(2x)cos(x) - 3sin(2x)

Right hand side's 1st and 3rd terms:

cos(2x-x) = cos(2x)cos(x)+sin(2x)sin(x)
cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x)

So right hand side becomes

2cos(2x)cos(x) - 3cos(2x)

So the equation is now:

2sin(2x)cos(x) - 3sin(2x) = 2cos(2x)cos(x) - 3cos(2x)

Factor out common factor:

sin(2x)[2cos(x) - 3] = cos(2x)[2cos(x) - 3]

sin(2x)[2cos(x) - 3] - cos(2x)[2cos(x) - 3] = 0

[2cos(x)-3][sin(2x)-cos(2x)] = 0

Use zero-factor property:

2cos(x)-3 = 0;    sin(2x)-cos(2x) = 0
  2cos(x) = 3                  sin(2x) = cos(2x)
   cos(x) = 3/2                 = 1
  (no solution to              tan(2x) = 1      
     this part)                     2x = 45° + 180°n
                                     x = 22.5° + 90°n

                              or in radians:
                                    2x = 
                                     x = 
                                     x = 
                                     x = 
                                     x = 

Edwin