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"

  • 27
any reason behind why we use sin 2x + sin 6x and no other combination? if i used sin 2x + sin 4x , how could i go about this sum?
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thanks
  • 3
we used sin 2x and sin 6x to make the solution easy . there is no hard and fast rule that you have to use these two only . it totally depends upon you how you solve it but just to make the solution easy we try to choose and pair up the smallest and the largest angles in these type of questions.
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