Solve sin2x+sin4x+sin6x=0

 Hello

Here is the solution to your answer:

Given: 

sin2x + sin 4x + sin6x=0

(sin 2x + sin 6x) + sin 4x = 0

2.sin[ (2x+6x)/2 ].cos[ (2x-6x)/2 ] + sin 4x = 0

2sin4x.cos2x + sin 4x = 0

sin 4x (2cos 2x +1) = 0

sin 4x = 0 ............... or..............  2cos2x + 1 = 0

4x = n(pi) ................or ..............  cos2x = -1/2

x = [ n(pi)/4 ] .............or ............. cos 2x = cos [ 2(pi)/3 ] 

.................  2x = 2n(pi) +/- [ 2(pi)/3 ]

................. x = n(pi) +/- [ (pi)/3 ]  >>>>>>>>>>>  " +/- " means "plus or minus"