Question

If θ = tan^-1(2tan²θ) - 1/2sin^-1(3sin2θ/(5 + 4cos2θ)), then find the general value of θ

Answer 1

Given, θ = tan^-1(2tan²θ) - 1/2sin^-1(3sin2θ/(5 + 4cos2θ))

⇒ 3tanθ + 2tan⁴θ - 6tan²θ + tanθ = 0

⇒ 2tan⁴θ - 6tan²θ + 4tanθ = 0

⇒ tan⁴θ - 3tan²θ + 2tanθ = 0

⇒ tanθ(tan3θ - 3tanθ + 2) = 0

⇒ tanθ(tanθ - 1)² (tanθ + 2) = 0

⇒ tanθ = 0, 1, - 2

where tanθ = 0 ⇒ θ = nπ ∉ I

when tanθ = 1 ⇒ θ = mπ + π/4, m ∈ I

when tanθ = - 2 ⇒ θ = pπ + tan^-1(-2), p ∈ I