[tex]\text{Given that,}\\\\\tan x = \dfrac 3{10}\\\\\text{Now,}\\\\\sin 2x = \dfrac{2\tan x}{1+ \tan^2 x}\\\\\\~~~~~~~~=\dfrac{2 \cdot \tfrac 3{10}}{1+ \left(\tfrac 3{10} \right)^2}\\\\\\~~~~~~~~=\dfrac{\tfrac 35}{1+ \tfrac 9{100}}\\\\\\~~~~~~~~=\dfrac{ \tfrac 35}{\tfrac{109}{100}}\\\\\\~~~~~~~~=\dfrac 35 \times \dfrac{100}{109}\\\\\\~~~~~~~~=\dfrac{60}{109}\\\\\\~~~~~~~~=0.5505[/tex]
