The value of $\frac{2\sin \mathrm{x}}{\sin 3\mathrm{x}}+\frac{\tan \mathrm{x}}{\tan 3\mathrm{x}}$ is ______ .

$\begin{array}{l}\frac{2\sin \mathrm{x}}{\sin 3\mathrm{x}}+\frac{\tan \mathrm{x}}{\tan 3\mathrm{x}}\\ =\frac{2\sin \mathrm{x}}{\sin \mathrm{x}\cdot \left(3-4{\sin }^2\mathrm{x}\right)}+\frac{\tan \mathrm{x}}{\frac{\left(3\tan \mathrm{x}-{\tan }^3\mathrm{x}\right)}{\left(1-3{\tan }^2\mathrm{x}\right)}}\end{array}$
$\begin{array}{l}=\frac{2}{3-4{\sin }^2\mathrm{x}}+\frac{\left(1-3{\tan }^2\mathrm{x}\right)}{\left(3-{\tan }^2\mathrm{x}\right)}\\ =\frac{2}{3-4{\sin }^2\mathrm{x}}+\frac{1-\left(\frac{3{\sin }^2\mathrm{x}}{1-{\sin }^2\mathrm{x}}\right)}{3-\left(\frac{{\sin }^2\mathrm{x}}{1-{\sin }^2\mathrm{x}}\right)}\\ =\frac{2}{3-4{\sin }^2\mathrm{x}}+\frac{\left(1-4{\sin }^2\mathrm{x}\right)}{\left(3-4{\sin }^2\mathrm{x}\right)}\\ =\frac{\left(3-4{\sin }^2\mathrm{x}\right)}{\left(3-4{\sin }^2\mathrm{x}\right)}=1\end{array}$