If 4 - 2sin2θ - 5cosθ = 0, 0° < θ < 90°, then find sinθ+ tanθ.

4 - 2sin²θ - 5cosθ = 0

Put θ = 60°

⇒ 4 - 2 × (\(\frac{\sqrt{3}}{2}\))² - \(\frac{5}{2}\)

⇒ 4 - \(\frac{6}{4}\) - \(\frac{5}{2}\) = 0 satisfied

So,

⇒ tanθ+ sinθ = \(\sqrt {3}\) + \(\frac{\sqrt {3}}{2}\) = \(\frac{3\sqrt {3}}{2}\)