Question

Find the principal solutions of the following equation:
tan 5θ = -1

Answer 1

tan 5θ = -1

tan 5θ = `-tan  pi/4` ...`(∵ tan  pi/4 = 1)`

tan 5θ = `tan(pi - pi/4)` ....`(∵ -tanθ = tan (pi - θ))`

∴ tan 5θ = `tan ((3pi)/4)`

tan θ  = tanα ⇒ θ  = nπ + α, n ∈ 2

∴ 5θ = `"n"pi + (3pi)/4`, n ∈ 2

∴ θ = `("n"pi)/5 + (3pi)/20`, n ∈ 2

Put n = 0, θ = `(3pi)/20` ∈ [0, 2π)

Put n = 1, θ = `(pi)/5 + (3pi)/20 = (4pi + 3pi)/20 = (7pi)/20` ∈ [0, 2π)

Put n = 2, θ = `(2pi)/5 + (3pi)/20 = (8pi + 3pi)/20 = (11pi)/20` ∈ [0, 2π)

Put n = 3, θ = `(3pi)/5 + (3pi)/20 = (12pi + 3pi)/20 = (15pi)/20` ∈ [0, 2π)

Put n = 4, θ = `(4pi)/5 + (3pi)/20 = (16pi + 3pi)/20 = (19pi)/20` ∈ [0, 2π)

Put n = 5, θ = `(5pi)/5 + (3pi)/20 = (20pi + 3pi)/20 = (23pi)/20` ∈ [0, 2π)

Put n = 6, θ = `(6pi)/5 + (3pi)/20 = (24pi + 3pi)/20 = (27pi)/20` ∈ [0, 2π)

Put n = 7, θ = `(7pi)/5 + (3pi)/20 = (28pi + 3pi)/20 = (31pi)/20` ∈ [0, 2π)

Put n = 8, θ = `(8pi)/5 + (3pi)/20 = (32pi + 3pi)/20 = (35pi)/20` ∈ [0, 2π)

Put n = 9, θ = `(9pi)/5 + (3pi)/20 = (36pi + 3pi)/20 = (39pi)/20` ∈ [0, 2π)

Put n = 10, θ = `(10pi)/5 + (3pi)/20 = (43pi)/20` ∉ [0, 2π)

∴ `{(3π)/20, (7π)/20, (11π)/20, (15π)/20, (19π)/20, (23π)/20, (27π)/20, (31π)/20, (35π)/20, (39π)/20}`