The general solution of sin θ = 1/2 is:
A$\theta = \pi/6$ only (principal value)
B$\theta = n\pi/6$ for any n
C$\theta = 2n\pi + \pi/6$ only
D$\theta = n\pi + (-1)^n \pi/6$, n ∈ ℤ
Answer & Solution
Correct answer: D. $\theta = n\pi + (-1)^n \pi/6$, n ∈ ℤ
General solution for sin θ = sin α: θ = nπ + (-1)^n α, n ∈ ℤ. Here α = π/6, so θ = nπ + (-1)^n (π/6). At n=0: π/6 ✓. At n=1: π − π/6 = 5π/6 ✓.
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