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The number of all possible values of $\theta$, where $0<\theta<\pi$, for which the system of equations
$ (y+z) \cos 3 \theta=(x y z) \sin 3 \theta $
$ x \sin 3 \theta=\frac{2 \cos 3 \theta}{y}+\frac{2 \sin 3 \theta}{z} $
$ (x y z) \sin 3 \theta=(y+2 z) \cos 3 \theta+y \sin 3 \theta$ have a solution $\left(\mathrm{x}_0, \mathrm{y}_0, \mathrm{z}_0\right)$ with $\mathrm{y}_0 \mathrm{z}_0 \neq 0$, is
$2$
$3$
$4$
$5$
Solution
$ (y+z) \cos 3 \theta-(x y z) \sin 3 \theta=0 $
$ x y z \sin 3 \theta=(2 \cos 3 \theta) z+(2 \sin 3 \theta) y $
$ \therefore(y+z) \cos 3 \theta=(2 \cos 3 \theta) z+(2 \sin 3 \theta) y=(y+2 z) \cos 3 \theta+y \sin 3 \theta $
$ y(\cos 3 \theta-2 \sin 3 \theta)=z \cos 3 \theta \text { and } $
$ y(\sin 3 \theta-\cos 3 \theta)=0 \Rightarrow \sin 3 \theta-\cos 3 \theta=0 \Rightarrow \sin 3 \theta=\cos 3 \theta $
$ \therefore 3 \theta=n \pi+\pi / 4$
