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The equation 3sin^2x+10cosx-6=0 is sat...

The equation `3sin^2x+10cosx-6=0` is satisfied , if

A

`x=npi+-cos^(-1)((1)/(3))`

B

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

C

`x=npi+-cos^(-1)((1)/(6))`

D

`x=2npi+-cos^(-1)((1)/(6))`

Text Solution

Verified by Experts

The correct Answer is:
B
We have ,
`3sin^(2)x+10cosx-6=0`
`rArr3(1-cos^(2)x)+10cosx-6=0`
`rArr3cos^(2)x-10cosx+3=0`
`rArr(cosx-3)(3cosx-1)=0`
`rArrcosxne3orcosx=(1)/(3)`
`[thereforecosxne3`, as cos x cannot be greater than
`rArrx=2npi+-cos^(-1)((1)/(3))`
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