1.

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

Answer» The given equation may be written as `(sin2x+sin4x+sin6x=0`.
`rArr 2" sin "((6x+2x))/(2)" cos "((6x-2x))/(2)+sin4x=0`
`[becausesinC+sinD=2"sin"((C+D))/(2)"cos"((C-D))/(2)]`
`rArr2sin4xcos2x+sin4x=0`
`rArrsin4x(2cos2x+1)=0`
`rArrsin4x=0or cos2x=-(1)/(2)=-"cos"(pi)/(3)=cos(pi-(pi)/(3))="cos"(2pi)/(3)`
`rArrsin4x=0orcos2x="cos"(2pi)/(3)`
`rArr4x=npior2x=(2mpi+-(2pi)/(3))`, where m , `ninI`
`rArrx=(npi)/(4)orx=(mpi+-(pi)/(3))` where m,`ninI`.
Hence , the general solution is given by `x=(npi)/(4)orx=(mpi+-(pi)/(3))`, where m, `ninI`.


Discussion

No Comment Found

Related InterviewSolutions