Question

Find the principal solutions of the following equation:

sin 2θ = `-1/2`

Answer 1

sin 2θ = `- 1/2`

Since, θ ∈ (0, 2π), 2θ ∈ (0, 4π)

`sin 2θ = - 1/2 = - sin  π/6 = sin (π + π/6) = sin (2π - π/6)`

= `sin (3π + π/6) = sin(4π - π/6)`  .......[∵ sin (π + θ) = sin(2π – θ) = sin(3π + θ) = sin(4π – θ) = – sin θ]

∴ `sin 2θ = sin  (7π)/6 = sin  (11π)/6 = sin  (19π)/6 = sin  (23π)/6`

∴ `2θ = (7π)/6 or 2θ = (11π)/6  or 2θ = (19π)/6 or 2θ = (23π)/6`

∴ `θ = (7π)/12 or θ = (11π)/12 or θ = (19π)/12 or θ = (23π)/12`

Hence, the required principal solutions are `{(7π)/12, (11π)/12, (19π)/12, (23π)/12}`.