Solve the following equation.
$θ : 2 \sqrt{3}sin^2 θ + cos θ - \sqrt{3} = 0 $ where θ is an cute angle
Answer & explanation
Correct answer: option 1
2√3 sin²θ + cosθ - √3 = 0
2√3( 1 - cos²θ) + cosθ - √3 = 0
2√3 - 2√3cos²θ + cosθ - √3 = 0
2√3cos²θ - cosθ - √3 = 0
2√3cos²θ -3 cosθ + 2 cosθ - √3 = 0
√3cosθ ( 2cosθ - √3)+ 1. ( 2cosθ - √3) = 0
( 2cosθ - √3). ( √3cosθ + 1) = 0
Either √3cosθ + 1 = 0 or 2cosθ - √3 = 0
√3cosθ + 1 = 0 is not possible because θ is an acute angle triangle.
So, 2cosθ - √3 = 0
cosθ = = \(\frac{√3}{2}\)
{ we know, cos30º = \(\frac{√3}{2}\) }
so, θ = 30º