JEE Advanced 2010 Paper 1Question 1
The number of all possible values of $\theta$, where $0<\theta<\pi$, for which the system of equations $$(y+z)\cos3\theta=(xyz)\sin3\theta$$ $$x\sin3\theta=\frac{2\cos3\theta}{y}+\frac{2\sin3\theta}{z}$$ $$(xyz)\sin3\theta=(y+2z)\cos3\theta+y\sin3\theta$$ have a solution $(x_0,y_0,z_0)$ with $y_0z_0\neq0$, is
Determinant
Answer$3$
Solution
Let $c=\cos3\theta$, $s=\sin3\theta$. Subtracting eq.1 from eq.3 gives $zc+ys=0$; eq.2 times $yz$ gives $xyzs=2cz+2sy=0$, so eq.1 gives $(y+z)c=0$. If $c=0$ then $ys=0$, forcing $y=0$, so $c\ne0$, $y=-z$ and then $s=c$, i.e. $\tan3\theta=1$ ($x=0$). In $(0,\pi)$: $3\theta=\frac\pi4,\frac{5\pi}4,\frac{9\pi}4$, so $3$ values.