Question

The number of all possible values of $\theta ,where0<\theta <\pi$, for which the system of equations $(y+z)cos3\theta =\left(xyz\right)sin3\theta$
$xsin3\theta =\frac{2cos3\theta }{y}+\frac{2sin3\theta }{z}\left(xyz\right)sin3\theta =(y+2z)cos3\theta +ysin3\theta$
have a solution $(x_0,y_0,z_0)with{y}_0{z}_0\ne 0$ is___

Answer 1

Let xyz =t
$tsin3\theta -ycos3\theta -zcos3\theta =0\left(1\right)$
$tsin3\theta -2ysin3\theta -2zcos3\theta =0\left(2\right)$
$tsin3\theta -y(cos3\theta +sin3\theta )-2zcos3\theta =0\left(3\right)$
$y_0.z_0\ne 0$ hence homegoeneous equation has non -trivial solution.
$\therefore D=\left|\begin{array}{ccc}sin3\theta & -cos3\theta & -cos3\theta \\ sin3\theta & -2sin3\theta & -2cos3\theta \\ sin3\theta & -(cos3\theta +sin3\theta )& -2cos3\theta \end{array}\right|=0$
$\Rightarrow sin3\theta cos3\theta (sin3\theta -cos3\theta )=0$
If $sin3\theta =0$, then from equation (2)
z =0, which is not possible
If $cos3\theta =0$ and $sin3\theta \ne 0$, then
$t.sin3\theta =0$
$\Rightarrow t=0$
$\Rightarrow x=0$
From equation (2), y=0 which is not possible
If $sin3\theta -cos3\theta =0$, then
$tan3\theta =1$
$\Rightarrow 3\theta =n\pi +\frac{\pi }{4},n\in I$
$\Rightarrow x.y.zsin3\theta \ne 0$
$\Rightarrow \theta =\frac{\pi }{12},\frac{5\pi }{12},\frac{9\pi }{12}$
Hence, three solutions.