Question

Find the principal solutions of the following equation:

sin 2θ = `− 1/(sqrt2)`

Answer 1

sin 2θ = `− 1/(sqrt2)`

Since, θ ∈ (0, 2π), 2θ ∈ (0, 4π)

`∴ sin 2θ = − 1/(sqrt2) = − sin  π/4 = sin (π + π/4) = sin (2π − π/4) = sin (3π + π/4) = sin (4π − π/4)  ...[∵ sin (π + θ) = sin (2π − θ) = sin (3π + θ) = sin (4π − θ) = − sin θ]`

∴ `sin 2θ = sin  (5π)/4  = sin  (7π)/4 = sin  (13π)/4 = sin  (15π)/4`

∴ `2θ = (5π)/4  or 2θ = (7π)/4  or 2θ = (13π)/4 or 2θ = (15π)/4`

∴ `θ = (5π)/8  or θ = (7π)/8  or θ = (13π)/8 or θ = (15π)/8`

Hence, the required principal solutions are `{(5π)/8, (7π)/8, (13π)/8, (15π)/8}`