Question

answer with full explanation ​

Answer 1

Answer:

-24/25

Step-by-step explanation:

Given---->

-----------

25 cos² θ + 5 cos θ -12 =0 and α is a root

of this equation

π

-----< α < π

2

To find---->

-----------

sin 2α = ?

Solution---->

---------------

25 cos²θ +5 cosθ - 12= 0

25cos²θ +(20-15) cosθ -12=0

25cos²θ + 20cosθ - 15 cosθ -12=0

5cosθ(5cosθ+4)- 3(5cosθ + 4)=0

(5cosθ+4) (5cosθ-3) = 0

if 5cosθ -3 = 0

5cosθ =3

cosθ =3/5

cosθ is positive in first and fourth

quadrant

so α does not satisfy equation for

cosθ=3/5

if 5cosθ + 4 = 0

5 cosθ = -4

cos θ =-4/5

cosθ is negative in second and third quadrant ie

π 3π

-----< θ < -----

2 2

and α lies between

π/2 < α <π ( given)

it means θ = α for cosθ = -4/5

so cos α =-4/5

now sinα is positive becauae

π/2< α<π

and sinα is positive in second quadrant

sinα =√(1 -cos²α)

=√{(1)-(-4/5)²}

=√1-(16/25)

=√(9/25)

=3/5

sin2α=2sinα cosα

=2(3/5)(-4/5)

=-24/25

Hope it helps you

Thanks