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

$$\left| {\matrix{ {\sin 3\theta } & { - \cos 3\theta } & { - \cos 3\theta } \cr {\sin 3\theta } & { - 2\sin 3\theta } & { - 2\cos 3\theta } \cr {\sin 3\theta } & { - (\cos 3\theta + \sin 3\theta )} & { - 2\cos 3\theta } \cr } } \right| = 0$$

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

$$ \Rightarrow \sin 3\theta = 0$$ 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.