If 4 cos2θ - 3 sin2θ + 2 = 0, then the value of tanθ is (where 0 ≤ θ < 90°) A2\sqrt{2}2B6\sqrt{6}6C13\frac{1}{\sqrt{3}}31D1Answer: 6\sqrt{6}6 Read Explanation: Solution:Given4 cos2 θ - 3 sin2 θ + 2 = 0Formula:sin2θ + cos2θ = 1tan2θ = sin2θ/cos2θCalculation:4 cos2θ - 3 sin2θ + 2 = 0⇒ 4 cos2θ - 3 (1 - cos2θ) + 2 = 0⇒ 4 cos2θ - 3 + 3 cos2θ + 2 = 0⇒ 7 cos2θ - 1 = 0⇒ 7 cos2θ = 1⇒ cos2θ = 1/7sin2θ + cos2θ = 1⇒ sin2θ = 1 - 1/7⇒ sin2θ = 6/7Now,tan2θ = sin2θ/cos2θ⇒ tan2θ = (6/7)/(1/7)⇒ tan2θ = 6∴ tanθ = √6 Read more in App