If 3 tan2θ - 9 = 0, find the value of 2cosθ + sinθ

3 tan²θ - 9 = 0

tan²θ = \(\frac{9}{3}\) =  3 

tanθ = \(\sqrt { 3}\)

θ = 60°

Now,

⇒ 2cosθ + sin(θ) 

= 2 × \(\frac{1}{2}\) + \(\frac{√3}{2}\) = \(\frac{2 + √3}{2}\)