Answer 1

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

`2sin4x.cos2x+sin4x=0`

`sin4x(2cos2x+1)=0`

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

`sin4x=0 or cos2x=-1/2=-cospi/3=cos(pi-pi/3)`

Using `sinx=0=>x=npi`

`sin4x=0`

`4x=npi`

The genral solution is x

`x=(npi)/4` 

using `cosx=cosalpha=>x=2mx+-alpha`

`cos2x=cos((2pi)/3)`

`2x=2mpi+-(2pi)/3`

The genral solution is x

`x=mpi+-pi/3` where `m,n in z`