Question

Let `θ∈(0, π/2)`. If the system of linear equations,

(1 + cos²θ)x + sin²θy + 4sin3θz = 0

cos²θx + (1 + sin²θ)y + 4sin3θz = 0

cos²θx + sin²θy + (1 + 4sin3θ)z = 0

has a non-trivial solution, then the value of θ is

 ______.

Options

• `(4π)/9`

• `π/18`

• `(5π)/18`

• `(7π)/18`

Answer 1

Let `θ∈(0, π/2)`. If the system of linear equations,

(1 + cos²θ)x + sin²θy + 4sin3θz = 0

cos²θx + (1 + sin²θ)y + 4sin3θz = 0

cos²θx + sin²θy + (1 + 4sin3θ)z = 0

has a non-trivial solution, then the value of θ is `underlinebb((7π)/18)`.

Explanation:

For non-trivial solution

`|(1 + cos2θ, sin^2θ, 4sin3θ),(cos^2θ, 1 + sin^2θ, 4sin3θ),(cos^2θ, sin^2θ, 1 + 4sin3θ)|` = 0

Apply R₃→R₃ – R₂, we get

`|(1 + cos^2θ, sin^2θ, 4sin3θ),(cos^2θ, 1 + sin^2θ, 4sin3θ),(0, -1, 1)|` = 0

Apply C₂→C₂ – C₃, we get

⇒ `|(1 + cos^2θ, sin^2θ + 4sin3θ, 4sin3θ),(cos^2θ, 1 + sin^2θ + 4sin3θ, 4sin3θ),(0, 0, 1)|` = 0

⇒ 2(1 + 2sin3θ) = 0

⇒ 2sin3θ = –1

⇒ sin3θ = `-1/2`

⇒ `3θ∈(0, (3π)/2)` as `θ∈(0, π/2)`

⇒ sin3θ = `sin(π + π/6) = sin((7π)/6)`  ...[∵ sin is –ve, lies in third quardant]

⇒ 3θ = `(7π)/6` 

⇒ θ = `(7π)/18`