1.

`sinx+sin3x+sin5x=0`

Answer» Given equation:
`sinx+sin3x+sin5x=0`,
or `sin3x+(sin5x+sinx) =0`,
or `sin3x+2sin(5x+x)/(2)cos(5x-x)/(2)=0`,
or `sin3x+2sin3xcos2x=0`,
or `sin3x(1+2cos2x)=0`,
`rArr sin3x=0` or `1+2cos2x=0`,
`sin3x=0`
`3x=npi`
`x=(npi)/(3)`
If `1+2cos2x=0`
`rArr cos2x=-1/2= cos(pi-pi/3)=cos(2pi)/(3)`
`rArr x=npi+-pi/3` where `n in Z`
Therefore, the general solution of given equation is
`x=(npi)/(3)` or `npi +- pi/3, n in Z`. Ans.


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