The number of integral values of $\mathrm{\alpha }$ such that  $\sin \mathrm{xcos}3\mathrm{x}-\mathrm{\alpha cos}\mathrm{xsin}3\mathrm{x}=0$  does not have any real root other then $(2\mathrm{n}+1)\frac{\mathrm{\pi }}{2},\mathrm{n}\in 1$ for any real value of x is _______.

$\sin \mathrm{xcos}3\mathrm{x}-\mathrm{\alpha cos}\mathrm{xsin}3\mathrm{x}=0$       ……..(i)

$\begin{array}{l}\Rightarrow \mathrm{\alpha }=\frac{\sin \mathrm{xcos}3\mathrm{x}}{\cos \mathrm{xsin}3\mathrm{x}} ..........\left(\mathrm{ii}\right)\\ \mathrm{\alpha }=\frac{\tan \mathrm{x}}{\tan 3\mathrm{x}}\\ \mathrm{\alpha }=\frac{\tan \mathrm{x}\left(1-3{\tan }^2\mathrm{x}\right)}{3\tan \mathrm{x}-{\tan }^2\mathrm{x}}\\ \mathrm{\alpha }=\frac{1-3{\tan }^2\mathrm{x}}{3-{\tan }^2\mathrm{x}}\end{array}$

For  real  value  of .x RHS  of 2nd never  lies  between $\left(\frac{1}{3},3\right)$

= Number of integral  points  [1, 2]
= 2 points