QJEEAdv10Ma628b: JEE Advanced 2010 Mathematics - Class 11 JEEAdv10Ma628b Solution: The number of all possible values of where for which the system …

View the equation in xyz, y and t.

We have,

$(xyz)\sin 3\theta - y\cos 3\theta - z\cos 3\theta = 0$

$(xyz)\sin 3\theta - 2y\sin 3\theta - 2z\cos 3\theta = 0$

$(xyz)\sin 3\theta - y(\cos 3\theta + \sin 3\theta ) - 2z\cos 3\theta = 0$

$xyz \ne 0$

Hence, the equation has non-trivial solution which gives

$$\left|\begin{array}{ccc}\sin 3\theta & -\cos 3\theta & -\cos 3\theta \\ \sin 3\theta & -2\sin 3\theta & -2\cos 3\theta \\ \sin 3\theta & -(\cos 3\theta +\sin 3\theta )& -2\cos 3\theta \end{array}\right|=0$$

$\Rightarrow \sin 3\theta \cos 3\theta (\sin 3\theta - \cos 3\theta ) = 0$

$\Rightarrow \sin 3\theta = 0$ then $xyz = 0$ (not possible)

$\cos 3\theta = 0$ not possible

$\sin 3\theta = \cos 3\theta \Rightarrow \tan 3\theta = 1$

$3\theta = n\pi + {\pi \over 4},n \in z$

$\theta = {{n\pi } \over 3} + {\pi \over {12}}$ ; $\theta = {\pi \over {12}},{{5\pi } \over {12}},{{9\pi } \over {12}}$

Thus there are 3 solutions.