Question

Solve sin 2x +sin4x+sin6x=0

Answer 1

Sin2x-sin4x+sin6x=0
(sin6x+sn2x)-sin4x=0
We know that sinx +siny=2sin(x+Y/2)cos(x-y/2)
Thereofre ,
=2sin(6x+2x/2)cos(6x-2x/2)-sin4x=0
=2sin8x/2cos4x/2-sin4x
sin4x(2cos2x-1)=0
Therefore,
sin4x=0 or 2cos2x-1=0
sin4x=0 or 2cos2x=1
sin4x=0 or cos2x=1/2
Now
Let sinx=siny
sin4x=sin4y
now sin4x=0
Therfore sin4y=0
sin4y=sin(0)
4y=0
y=0
Now cos2x=1/2
Let cosx=cosy
Cos2x=cos2y
cos2y=1/2 ;cos(2y)=π/3;2y=π/3;
Now,
for sin4x=sin4y the general solution will be
4x=nπ±(-1)ⁿ4y
Putting y=0
4x=nπ
x=nπ/4
For cos2x=cos2y is
2x=2nπ±2y
now 2y=π/3
2x=2nπ±π/3
on solving we get
x=nπ±π/6
Therfore for sin4x=0 ,x=nπ/4
and
for cos2x=1/2,x=nπ±π/6

Hope it helps