Solve the following equation.

$θ : 2 \sqrt{3}sin^2 θ + cos θ - \sqrt{3} = 0 $ where θ is an cute angle

30°

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