If 2cosθ + 8sin2θ = 11, where

Solution

Given, 2cosθ + 8sin²θ = 11 Or, 2cosθ + 8(1 – cos²θ) = 11 [Since, sin²θ = 1 – cos²θ] Or, 2cosθ + 8 – 8cos²θ = 11 Or, 8cos²θ – 2cosθ + 3 = 0 Or, 8cos²θ – 6cosθ + 4cosθ – 3 = 0 Or, 2cosθ(4cos – 3) + 1(4cosθ – 3) = 0 Or, (2cosθ + 1)(4cosθ – 3) = 0 Or, cosθ = (-1/2), (3/4) [(-1/2) is rejected as 0^o < θ < 90^o] We know that = cosθ = (Base)/(Hypotenuse) = (3/4) (Perpendicular)² = (Hypotenuse)² – (Base)² (Perpendicular)² = 4² – 3² = 16 – 9 = 7 Perpendicular = √7 So, cotθ = base/perpendicular = (3/√7)