Find the general solution of the equation `sin2x+sin4x+sin6x=0`.

Explanation

`sin2x+sin4x+sin6x=0`.
`therefore (sin6x+sin2x)+sin4x=0`
`therefore 2sin((6x+2x)/(2))cos((6x-2x)/(2))+sin4x=0`
`therefore 2 sin 4xcos 2x+sin4x=0`
`therefore sin 4x(2cos2x+1)=0`
`therefore` either `sin4x=0 or 2cos 2x+1=0`
`therefore` either `sin4x=0 or 2 cos 2x=-1`
`therefore` either `sin4x=0 or cos 2x=-(1)/(2)`
`therefore` either ` sin 4x=0 or cos 2x =-cos.(pi)/(3)`
`therefore` either `sin4x=0 or cos 2x=cos(pi-(pi)/(3))......[because cos(pi-theta)=-costheta]`
`therefore `either ` sin4x=0 or cos 2x=cos.(2pi)/(3)`
The general solution of `sintheta=0 is theta = n pi , n in Z` and `cos theta = cos alpha` is `theta =2 n pi +-alpha, n in Z`.
`therefore` the required general solution is given by
`4x= n pi , n in Z or 2x=2 n pi +-alpha(2pi)/(3), n in Z`
i.e., `x=(n pi)/(4), n in Z or x=n pi +-(pi)/(3), n in Z`.