Inactive Tutor answered 04/13/22
Tutor
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2 - 2sin2θ = 3sinθ + 3; 2sin2θ + 3sinθ + 1 = 0; (2sinθ + 1)(sinθ + 1) = 0;
sinθ = - 1; θ = 3π/2;
sinθ = - 1/2; θ = 7π/6 or θ = 11π/6
Abby J.
asked 04/13/22Fine exactly all solutions to the given equation:
2cos^2θ = 3sinθ + 3, 0 ≤ θ ≤ 2pi
Inactive Tutor answered 04/13/22
2 - 2sin2θ = 3sinθ + 3; 2sin2θ + 3sinθ + 1 = 0; (2sinθ + 1)(sinθ + 1) = 0;
sinθ = - 1; θ = 3π/2;
sinθ = - 1/2; θ = 7π/6 or θ = 11π/6
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