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Sin(2θ) = -sinθ

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Solve each equation on the interval 0 ≤ θ < 2π Sin(2θ) = -sinθ

trigonometric-functions

asked Dec 1, 2017 in PRECALCULUS by anonymous

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1 Answer

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sin(2θ) = -sinθ

2sinθcosθ=-sinθ

sinθ(2cosθ+1)=0

sinθ = 0

θ=0, π

2cosθ+1=0

2cosθ+1=0

cosθ =-1/2

θ=4π/3, 2π/3

θ=0, 2π/3, π, 4π/3

answered Dec 7, 2017 by homeworkhelp Mentor
reshown Dec 7, 2017 by bradely

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