Given parametric equations are:

`x=2sin(2t)`

`y=3sin(t)`

Let's make a table of x and y values for different values of t. (Refer the attached image).The point where the curve crosses itself will have same x and y values for different values of t.

So from the table, the curve crosses itself at the point (0,0) for t=0 and t=`pi`

The derivative `dy/dx` is the slope of the line tangent to the parametric graph `(x(t),y(t))`

`dy/dx=(dy/dt)/(dx/dt)`

`x=2sin(2t)`

`dx/dt=2cos(2t)*2=4cos(2t)`

`y=3sin(t)`

`dy/dt=3cos(t)`

`dy/dx=(3cos(t))/(4cos(2t))`

At t=0, `dy/dx=(3cos(0))/(4cos(2*0))=3/4`

Equation of the tangent line can be found by the point slope form of the line,

`y-0=3/4(x-0)`

`y=3/4x`

At t=`pi` , `dy/dx=(3cos(pi))/(4cos(2pi))=-3/4`

`y-0=-3/4(x-0)`

`y=-3/4x`

Equations of the tangent lines at the point where the curve crosses itself are :

`y=3/4x , y=-3/4x`