The number of solutions of the pair of equations 2 sin2θ - cos2θ = 0 and 2 sin2θ - 3 sin θ = 0, in the interval [0, 2π] is
Two
We want to find the values of θ which will satisfy both the equation.
2 sin2θ - cos2θ = 0
⇒ 2 sin2θ - (1- 2 sin2θ) = 0
⇒ 4 sin2θ - 1= 0
⇒ sin2θ = 14
⇒ θ = nπ± π6 ..........(1)
[if sin2θ = sin2α, θ = nπ±α]
2 sin2θ - 3 sin θ = 0
⇒ we can form a quadratic in sin θ and solve this.
2(1- sin2θ) - 3 sinθ = 0
⇒ 2 - 2 sin2θ - 3 sin θ = 0
⇒ 2 sin2θ + 3 sin θ - 2 = 0
⇒ 2 sin2θ - sin θ + 4 sin θ - 2 = 0
⇒ 2 sin θ (sin θ - 12) + 4 (sinθ - 12) = 0
⇒ (2 sin θ + 4) (sin θ – 12) = 0
⇒ sin θ = 12 [sin θ ≠ –2]
⇒ θ = nπ + (–1)n 12 . . . . . (2)
We can substitute different values of n (n = 0, 1, 2, 3 . . . . .) in (1) and (2), then pick the common values. If we do that we will get π6 and 5π6 as common values
The first equation gave us sin2 θ = 14 or sin θ = ∓ 12 and second equation is sin θ = 12. The common part is sin θ = 12. This has two solutions in [0, 2π]. They are π6 and 5π6.
⇒ two values