Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 23

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