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If 3 cos x!=2 sin x, then the general so...

If `3 cos x!=2 sin x,` then the general solution of `sin^2x-cos 2x=2-sin 2x` is `x`

A

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

B

`(npi)/(2),ninZ`

C

`(4n+1)(pi)/(2),ninZ`

D

`(2n-1)pi,inZ`

Text Solution

Verified by Experts

The correct Answer is:
C

`sin^(2)x-cos2x=2-sin2x`
`rArr1-cos^(2)x-(2cos^(2)x-1)=2-2sinxcosx`
`rArr-3cos^(2)x+2sinxcosx=0`
`rArr cosx(2sinx-3cosx)=0`
`rArrcosx=0 " " [because2sinx-3cosxne0]`
`rArrx=2npi+(pi)/(2)rArrx=(4n+1)(pi)/(2),ninz`
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