Question

Find the principal solutions of the following equation:

tan 3θ = - 1

Answer 1

tan 3θ = - 1

Since, θ ∈ (0, 2π), 3θ ∈ (0, 6π)

`tan 3θ = - 1 = - tan  π/4 = tan (π - π/4)` 

`= tan (2π - π/4) = tan(3π - π/4)` 

`= tan(4π - π/4) = tan(5π - π/4)`

`= tan(6π - π/4)` ......[∵ tan (π - θ) = tan(2π - θ) = tan(3π - θ) = tan(4π - θ) = tan (5π - θ) = tan (6π - θ) = - tan θ]

∴ `"tan" 3θ = "tan" (3π)/4 = "tan" (7π)/4 = "tan" (11π)/4 = "tan" (15π)/4 = "tan" (19π)/4 = "tan" (23π)/4`

∴ `3θ = (3π)/4 or 3θ = (7π)/4 or 3θ = (11π)/4 or 3θ = (15π)/4 or 3θ = (19π)/4 or 3θ = (23π)/4`

∴ `θ = π/4 or θ = (7π)/12 or θ = (11π)/12 or θ = (5π)/4 or θ = (19π)/12 or θ = (23π)/12`

Hence, the required principal solutions are,

`{π/4, (7π)/12, (11π)/12, (5π)/4, (19π)/12, (23π)/12}`.