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"