Question:
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]
Solution:
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