The number of all possible values of $ heta$, where $0

Question

The number of all possible values of $	heta$, where $0<	heta<\pi$, for which the system of equations$$(y+z) \cos 3 	heta=(x y z) \sin 3 	heta$$$$x \sin 3 	heta=\frac{2 \cos 3 	heta}{y}+\frac{2 \sin 3 	heta}{z}$$$$( x y z) \sin 3 	heta=(y+2 z) \cos 3 	heta+y \sin 3 	heta$$ have a solution $\left(x_{0}, y_{0}, z_{0}ight)$ with $y_{0} z_{0} 
eq 0$, is

Professor PC Thomas Classes · Answer

(3) : View the equation in $x y z, y$ and $t$. We have, $(x y z) \sin 3 	heta-y \cos 3 	heta-z \cos 3 	heta=0$ $(x y z) \sin 3 	heta-2 y \sin 3 	heta-2 z \cos 3 	heta=0$\ $(x y z) \sin 3 	heta-y(\cos 3 	heta+\sin 3 	heta)-2 z \cos 3 	heta=0$ $x y z 
eq 0$\ Hence the equation has non-trivial solution\ which gives\ $\left|\begin{array}{lll}\sin 3 	heta & -\cos 3 	heta & -\cos 3 	heta \ \sin 3 	heta & -2 \sin 3 	heta & -2 \cos 3 	heta \ \sin 3 	heta & -(\cos 3 	heta+\sin 3 	heta) & -2 \cos 3 	heta\end{array}ight|=0$\ $\Rightarrow \sin 3 	heta \cos 3 	heta(\sin 3 	heta-\cos 3 	heta)=0$\ $\Rightarrow \quad \sin 3 	heta=0$ then $x y z=0 \quad$ (not possible) $\cos 3 	heta=0$ not possible\ $\sin 3 	heta=\cos 3 	heta \Rightarrow 	an 3 	heta=1$\ $3 	heta=n \pi+\frac{\pi}{4}, n \in Z \Rightarrow 	heta=\frac{n \pi}{3}+\frac{\pi}{12} ;$\ $	heta=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{9 \pi}{12}$\ Thus there are 3 solutions.\