Solve the following equation.
$θ : 2 cos^2 θ + ( 4 + \sqrt{3}) sin θ - 2( 1 + \sqrt{3}) = 0 $ where θ is an acute angle.
Answer & explanation
Correct answer: option 4
$ 2 cos^2 θ + ( 4 + \sqrt{3}) sin θ - 2( 1 + \sqrt{3}) = 0 $
{ sin²θ + cos²θ = 1 }
2 ( 1 - sin²θ ) + 4 sinθ + √3 sinθ - 2 - √3 = 0
2 - 2 sin²θ + 4 sinθ + √3 sinθ - 2 - √3 = 0
2 sin²θ- 4 sinθ - √3 sinθ - 2√3 = 0
2 sinθ ( sinθ - 2 ) - √3 ( sinθ - 2 ) = 0
( 2sinθ - √3 ) . ( sinθ - 2 ) = 0
Either ( 2sinθ - √3 ) = 0 or ( sinθ - 2 ) = 0
( sinθ - 2 ) = 0 is not possible .
So, 2sinθ - √3 = 0
sinθ = \(\frac{ √3}{2}\)
{ we know, sin60º = \(\frac{ √3}{2}\) }
So, θ = 60º