The number of all possible values of θ, where 0 < θ < π, for which the system of equations (y+z)cos 3θ =(xyz)sin 3θ, xsin 3θ=2cos 3θy+2sin 3θz, (xyz)sin 3θ=(y+2z)cos 3θ+ysin 3θ have a solution (x, y, z) with yz ≠ 0, is

Correct option is D. 3
Two equations are given in the problem statement
(y+z)cos 3θ −(xyz)sin 3θ=0 (1)
xyzsin 3θ=(2cos 3θ)z+(2sin 3θ)y (2)
(xyz)sin 3θ=(y+2z)cos 3θ+ysin 3θ (3)
From equation (1) and (2), we get
∴(y+z)cos 3θ=(2cos 3θ)z+(2sin 3θ)y=(y+2z)cos 3θ+ysin 3θ
y(cos 3θ −sin 3θ)=zcos 3θ
and
From equation (2) and (3)
y (sin 3θ −cos 3θ) =0 ⇒sin 3θ −cos 3θ =0 ⇒sin 3θ =cos 3θ
∴3θ =nπ+π/4