Let
Δ(θ)=1sinθ1−sinθ1sinθ−1−sinθ1,0≤θ≤2π
Solution of ∆(θ) = 3 is
π2,3π2
π4,3π4,5π4,7π4
π4,3π4
π4,π2,3π4,π
Using C1→C1+C3 we get
Δ(θ)=2sinθ101sinθ0−sinθ1=21+sin2θ
∴ Δ(θ)=3⇒2sin2θ=1⇒ sinθ=±12⇒θ=π4,3π4,5π4,7π4