The number of θ ∈ (0, 4π) for which the system of linear equations

3(sin 3θ) x – y + z = 2

3(cos 2θ) x + 4y + 3z = 3

6x + 7y + 7z = 9

has no solution, is

Calculation:

Given, 

3(sin 3θ) x – y + z = 2

3(cos 2θ) x + 4y + 3z = 3

6x + 7y + 7z = 9

∴ 

= 3sin3θ(7) + 1(21cos2θ – 18) + 1(21cos2θ – 24)

⇒ Δ = 21sin3θ + 42cos2θ – 42

For no solution, Δ = 0

⇒ sin3θ + 2cos2θ = 2

⇒ sin3θ = 2 - 2cos2θ = 2(1 - cos2θ)

⇒ sin3θ = 2⋅2sin2θ

⇒ 3sinθ – 4sin3θ = 4sin2θ

⇒ sinθ(3 – 4sinθ – 4sin2θ) = 0

⇒ sinθ = 0 OR = sinθ = 

⇒ θ = 

∴ Number of θ's is 7.

The correct answer is Option 2.