[Note :[k] denotes the largest integer less than or equal to k ]
Let
$$f(\theta ) = {1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$$
Again let
$$g(\theta ) = {\sin ^2}\theta + 3\sin \theta \cos \theta + 5{\cos ^2}\theta $$
$$ = {{1 - \cos 2\theta } \over 2} + 5\left( {{{1 + \cos 2\theta } \over 2}} \right) + {3 \over 2}\sin 2\theta $$
$$ = 3 + 2\cos 2\theta + {3 \over 2}\sin 2\theta $$
$$\therefore$$ $$g{(\theta )_{\min }} = 3 - \sqrt {4 + {9 \over 4}} = 3 - {5 \over 2} = {1 \over 2}$$
$$\therefore$$ $$f(\theta ) = {1 \over {g{{(\theta )}_{\min }}}} = 2$$
Given, $$\tan \theta = \cot 5\theta $$
$$ \Rightarrow \tan \theta = \tan \left( {{\pi \over 2} - 5\theta } \right)$$
$$ \Rightarrow {\pi \over 2} - 5\theta = n\pi + \theta $$
$$ \Rightarrow 6\theta = {\pi \over 2} - {{n\pi } \over 6}$$
$$ \Rightarrow \theta = {\pi \over {12}} - {{n\pi } \over 6}$$
Also $$\cos 4\theta = \sin 2\theta = \cos \left( {{\pi \over 2} - 2\theta } \right)$$
$$ \Rightarrow 4\theta = 2n\pi \, \pm \,\left( {{\pi \over 2} - 2\theta } \right)$$
Taking positive
$$6\theta = 2n\pi + {\pi \over 2} \Rightarrow \theta = {{n\pi } \over 3} + {\pi \over {12}}$$
Taking negative
$$2\theta = 2n\pi - {\pi \over 2} \Rightarrow \theta = n\pi - {\pi \over 4}$$
Above values of $$\theta$$ suggests that there are only 3 common solutions.
have a solution $$\left( {{x_0},{y_0},{z_0}} \right)$$ with $${y_0}{z_0}{\mkern 1mu} \ne {\mkern 1mu} 0,$$ is
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.