The number of all possible values of theta,wh | vedclass

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(B) Given the system:$1$) $(y+z) \cos 3	heta = (xyz) \sin 3	heta$$2$) $x \sin 3	heta = \frac{2 \cos 3	heta}{y} + \frac{2 \sin 3	heta}{z}$ $\Right

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$3$

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(B) Given the system:$1$) $(y+z) \cos 3	heta = (xyz) \sin 3	heta$$2$) $x \sin 3	heta = \frac{2 \cos 3	heta}{y} + \frac{2 \sin 3	heta}{z}$ $\Rightarrow xyz \sin 3	heta = 2z \cos 3	heta + 2y \sin 3	heta$$3$) $(xyz) \sin 3	heta = (y+2z) \cos 3	heta + y \sin 3	heta$Equating $(1)$ and $(3)$:$(y+z) \cos 3	heta = (y+2z) \cos 3	heta + y \sin 3	heta$$y \cos 3	heta + z \cos 3	heta = y \cos 3	heta + 2z \cos 3	heta + y \sin 3	heta$$z \cos 3	heta + y \sin 3	heta = 0 \Rightarrow y \sin 3	heta = -z \cos 3	heta$From $(1)$ and $(2)$:$(y+z) \cos 3	heta = 2z \cos 3	heta + 2y \sin 3	heta$$y \cos 3	heta + z \cos 3	heta = 2z \cos 3	heta + 2y \sin 3	heta$$y \cos 3	heta - 2y \sin 3	heta = z \cos 3	heta$Substituting $z \cos 3	heta = -y \sin 3	heta$:$y \cos 3	heta - 2y \sin 3	heta = -y \sin 3	heta$$y \cos 3	heta - y \sin 3	heta = 0$Since $y_0 
eq 0$,$\cos 3	heta = \sin 3	heta \Rightarrow 	an 3	heta = 1$$3	heta = n\pi + \frac{\pi}{4} \Rightarrow 	heta = \frac{n\pi}{3} + \frac{\pi}{12}$For $0 $n=0 \Rightarrow 	heta = \frac{\pi}{12}$$n=1 \Rightarrow 	heta = \frac{5\pi}{12}$$n=2 \Rightarrow 	heta = \frac{9\pi}{12} = \frac{3\pi}{4}$There are $3$ such values.

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