Let θ ∈ 0 , π 2 . If the system of linear equations 1 + cos 2 ⁡ θ x + sin 2 ⁡ θ y + 4 sin ⁡ 3 θ z = 0 cos 2 ⁡ θ x + 1 + sin 2 ⁡ θ y + 4 sin ⁡ 3 θ z = 0 cos 2 ⁡ θ x + sin 2 ⁡ θ y + ( 1 + 4 sin ⁡ 3 θ ) z = 0 has a non-trivial solution, then the value of θ is:

Step-by-step solution

1 + cos 2 θ sin 2 θ 4 sin 3 θ cos 2 θ 1 + sin 2 θ 4 sin 3 θ cos 2 θ sin 2 θ 1 + 4 sin 3 θ = 0 R 3 → R 3 - R 2 1 + cos 2 θ sin 2 θ 4 sin 3 θ cos 2 θ 1 + sin 2 θ 4 sin 3 θ 0 - 1 1 C 2 → C 2 + C 1 1 + cos 2 θ 2 4 sin 3 θ cos 2 θ 2 4 sin 3 θ 0 - 1 1 C 2 → C 2 + C 3 1 + cos 2 θ 2 + 4 sin 3 θ 4 sin 3 θ cos 2 θ 2 + 4 sin 3 θ 4 sin 3 θ 0 0 1 = 0 Expanding along R 3 ⇒ 1 + cos 2