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

Explanation

Given, `sin2x+sin4x+sin6x=0`
`:.(sin6xsin2x)+sin4x=0`
`implies2sin((6x+2x)/(2)).cos((6x-2x)/(2))+sin4x =0`
`[:'sinA+sinB=2sin((A+B)/(2)).cos((A-B)/(2))]`
`implies2sin4x.cos2x+sin4x=0`
`impliessin4x.(2cos2x+1)=0`
`impliessin4x=0orcos2x=-(1)/(2)`
`impliessin4x=0`
`impliescos2x=-cos((pi)/(3))`
`=cos(pi-(pi)/(3))`
`=cos""(2)/(3)pi`
`:'sintheta=0" inmplies "theta=npi`
and `costheta=cosalpha" implies "theta=2npi+-alpha,ninZ`
`:.4x=npior2x=2mpi+-(2pi)/(3)`
`:.` Required general solution is
`x=(npi)/(4)orx=mpi+-(pi)/(3)" where "n,m""inz`.