Answer:
Θ = 0, π, [tex]\frac{7\pi }{6}[/tex], [tex]\frac{11\pi }{6}[/tex]
Step-by-step explanation:
Using the identity
cos2Θ = 1 - 2sin²Θ, then
sinΘ + 1 = 1 - 2sin²Θ ( subtract 1 - 2sin²Θ from both sides )
2sin²Θ + sinΘ = 0 ← factor out sinΘ from each term
sinΘ(2sinΘ + 1) = 0
Equate each factor to zero and solve for Θ
sinΘ = 0, hence
Θ = [tex]sin^{-1}[/tex] (0) = 0, π, 2π
2sinΘ + 1 = 0 → sinΘ = - [tex]\frac{1}{2}[/tex] , thus
Θ = [tex]sin^{-1}[/tex] ( - [tex]\frac{1}{2}[/tex] ) = - [tex]\frac{\pi }{6}[/tex]
Since sinΘ < 0 then Θ is in third/ fourth quadrants, thus
Θ = π + [tex]\frac{\pi }{6}[/tex] = [tex]\frac{7\pi }{6}[/tex] or Θ = 2π - [tex]\frac{\pi }{6}[/tex] = [tex]\frac{11\pi }{6}[/tex]
Solution is
Θ = 0, π, [tex]\frac{7\pi }{6}[/tex], [tex]\frac{11\pi }{6}[/tex] for 0 ≤ Θ < 2π
