
Mark M. answered 04/21/21
Tutor
5.0
(278)
Mathematics Teacher - NCLB Highly Qualified
sin2 θ - 2 sin θ = 0
sin θ (sin θ - 2) = 0
Use Zero Product Rule (from Algebra 2) to solve for θ
Emma S.
asked 04/21/21What is the solution set of the equation sin^2θ = 2sinθ over the interval 0º ≤ θ < 360º?
Mark M. answered 04/21/21
Mathematics Teacher - NCLB Highly Qualified
sin2 θ - 2 sin θ = 0
sin θ (sin θ - 2) = 0
Use Zero Product Rule (from Algebra 2) to solve for θ
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