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Find the gereral of the equation sinx-3...

Find the gereral of the equation `sinx-3sin2x+sin3x=cosx-3cos2x+cos3x`.

A

`npi+(pi)/(8)`

B

`(npi)/(2)+(pi)/(8)`

C

`(-1)^(n)(npi)/(2)+(pi)/(8)`

D

`2npi+"cos"^(-1)(3)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
We have , `sinx-3sin2x+sin3x=cosx-3cos2x+cos3x`
`rArrsinx+sin3x-3sin2x=cosx+cos3x-3cos2x2sin2xcosx-3sin2x-2cos=2xcosx+3cos2x=0`
`rArrsin2x(2cosx-3)-cos2x(2cosx-3)=0`
`rArr(sin2x-cos2x)(2cosx-3)=0`
`rArrcos2x=sin2x " " [becausecosxne(3)/(2)]`
`rArr2x=2npi+-((pi)/(2)-2x)`
Taking (+ve) sign `x=(npi)/(2)+(pi)/(8)`
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