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

1 0 \(\frac{2 + √3}{2}\) \(\frac{3 + √3}{2}\)

\(\frac{2 + √3}{2}\)

3 tan2θ - 9 = 0 tan2θ = \(\frac{9}{3}\) =  3  tanθ = \(\sqrt { 3}\) θ = 60° Now, ⇒ 2cosθ + sin(θ)  = 2 × \(\frac{1}{2}\) + \(\frac{√3}{2}\) = \(\frac{2 + √3}{2}\)