If the points - 2 , 0 , - 1 , 1 3 and cos θ , sin θ are collinear, then the number of values of θ ∈ 0 , 2 π is

Step-by-step solution

Let A ≡ - 2 , 0 ,   B = - 1 , 1 3 ,   C ≡ cos θ , sin θ Equation to A B , y - 0 = 1 3 - 0 - 1 + 2 x + 2 ⇒ 3 y = x + 2 , Point C lies on this line. ⇒ 3 sin θ - cos θ = 2 ⇒ 3 2 sin θ - 1 2 cos θ = 1 ⇒ sin θ - π 6 = 1 , where θ ∈ 0 , 2 π θ - π 6 = π 2 ⇒ θ = π 2 + π 6 = 2 π 3 So, only one value of θ is possible.