KVPY2019MathematicsTrigonometric Equations
Let A = θ ∈ R : 1 3 sin θ + 2 3 cos θ 2 = 1 3 sin 2 θ + 2 3 cos 2 θ . Then
Options
- AA ∩ 0 , π is an empty set
- BA ∩ 0 , π has exactly one point
- CA ∩ 0 , π has exactly two points
- DA ∩ 0 , π has more than two points
Correct answer
B. A ∩ 0 , π has exactly one point
Step-by-step solution
Given trigonometric relation is 1 3 sin θ + 2 3 cos θ 2 = 1 3 sin 2 θ + 2 3 cos 2 θ ⇒   1 9 sin 2 θ + 4 9 cos 2 θ + 4 9 sin θ cos θ = 1 3 sin 2 θ + 2 3 cos 2 θ ⇒   2 9 sin 2 θ + 2 9 cos 2 θ - 4 9 sin θ cos θ = 0 ⇒   sin 2 θ + cos 2 θ - 2 sin θ cos θ = 0 ⇒   sin 2 θ = 1 ⇒   2 θ = 2 n π + π 2 ,   n ∈ I ⇒   θ = n π + π