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If sinx +sin3x+sin5x=0, then...

If sinx +sin3x+sin5x=0, then

A

`x=(npi)/(3)`

B

`x=npi+(pi)/(3)`

C

`x=npi-(pi)/(3)`

D

All are correct

Text Solution

Verified by Experts

The correct Answer is:
D
`sinx+sin3x+sin5x=0`
`rArr(sin5x+sinx)+sin3x=0`
`rArr2" sin"(5x+x)/(2)"cos"(5x-x)/(2)+sin3x=0`
`rArr2sin3xcos2x+sin3x=0`
`rArrsin3x(2cos2x+1)=0`
Either sin3x=0
or `2cos2x+1=0`
when sin 3x=0,
Then , `3x=npirArrx=(npi)/(3)`
when `2cos2x+1=0`
`rArrcos2x=-(1)/(2)rArrcos2x=-"cos"(pi)/(3)`
`rArrcos2x=cos(pi-(pi)/(3))rArr"cos "2x="cos"(2pi)/(3)`
`rArr2x=2npi+-(2pi)/(3)rArrx=npi+-(pi)/(3),ninZ`
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