Find the general solution of the equation (√3 – 1) cos θ + (√3

1.	Find the general solution of the equation (√3 – 1) cos θ + (√3 + 1) sin θ = 2[Hint: Put √3 – 1 = r sin α, √3 + 1 = r cos α which gives tan α = tan((π/4) – (π/6)) α = π/12]
Answer» Let, r sinα = √3 – 1 and r cosα = √3 + 1 Therefore, r = √{(√3 – 1)² + (√3 + 1)²} = √8 = 2√2 And, tan α = (√3 – 1) / (√3 + 1) Therefore, r(sinα cos θ + cosα sin θ) = 2 ⇒ r sin (θ+α) = 2 ⇒ sin (θ+α) = 1/√2 ⇒ sin (θ+α) = sin (π/4) ⇒ θ+α = nπ + (– 1)ⁿ (π/4), n ∈ Z ⇒ θ = nπ + (– 1)ⁿ (π/4) – (π/12), n ∈ Z

Find the general solution of the equation (√3 – 1) cos θ + (√3 + 1) sin θ = 2[Hint: Put √3 – 1 = r sin α, √3 + 1 = r cos α which gives tan α = tan((π/4) – (π/6)) α = π/12]

Answer»

Let, r sinα = √3 – 1 and r cosα = √3 + 1

Therefore, r = √{(√3 – 1)² + (√3 + 1)²} = √8 = 2√2

And, tan α = (√3 – 1) / (√3 + 1)

Therefore, r(sinα cos θ + cosα sin θ) = 2

⇒ r sin (θ+α) = 2

⇒ sin (θ+α) = 1/√2

⇒ sin (θ+α) = sin (π/4)

⇒ θ+α = nπ + (– 1)ⁿ (π/4), n ∈ Z

⇒ θ = nπ + (– 1)ⁿ (π/4) – (π/12), n ∈ Z