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cosα is a root of the equation 25x^2+5x−...

`cosα` is a root of the equation `25x^2+5x−12=0,−1 lt x lt 0`, then find the value of `sin2α` is:

A

`24/25`

B

`-12/25`

C

`-24/25`

D

`20/25`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`:'x=(-5+-sqrt(25+1200))/50)=(-5+-35)/50=30/50,(-40)/50`
or `cos alpha =3/5,(-4)/5`
But `-1ltlxt0`
`:.cos alpha =-4/5` [lies in II and III quadrants]
`:.sin alpha=3/5` [lies in II quadrant]
`:.sin alpha=-3/5` [lies in III quadrant]
`:.sin 2 alpha=2.sinalpha.cos alpha=-24/25` [lies i II quadrant]
`:.sin 2 alpha =2.sin alpha . cos alpha =24/25` [lies in III quadrant]
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