alleryGeneral solution of sin θ = sin α:
Aθ = nπ + (-1)ⁿα where n ∈ ℤ
Bθ = π - α
Cθ = 2π - α
Dθ = α
Answer & Solution
Correct answer: A. θ = nπ + (-1)ⁿα where n ∈ ℤ
General solution of sin θ = sin α: θ = nπ + (-1)ⁿα.
For even n, θ = α + 2π·(n/2); for odd n, θ = π - α + 2π·((n-1)/2).
Captures both branches of sin.
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