x=4cos^2theta , y=2sintheta Find all points (if any) of horizontal and vertical tangency to the curve.

Parametric curve (x(t),y(t)) has a horizontal tangent if its slope dy/dx is zero i.e when dy/dt=0 and dx/dt!=0
It has a vertical tangent if its slope approaches infinity i.e it is undefined, which implies that dx/dt=0 and dy/dt!=0  
Given parametric equations are:
x=4cos^2(theta) ,y=2sin(theta)
Here the parameter is theta
Let's take the derivative of x and y with respect to theta
dx/(d theta)=4(2cos(theta)d/(d theta)cos(theta))
dx/(d theta)=4(2cos(theta)(-sin(theta)))
dx/(d theta)=-4(2sin(theta)cos(theta))
Use trigonometric identity: sin(2theta)=2sin(theta)cos(theta)
dx/(d theta)=-4sin(2theta)
dy/(d theta)=2cos(theta)
For Horizontal tangents, set the derivative of y equal to zero
dy/(d theta)=2cos(theta)=0
=>cos(theta)=0
=>theta=pi/2,(3pi)/2
Let's check dx/(d theta) for the above angles,
For theta=pi/2
dx/(d theta)=-4sin(2*pi/2)=-4sin(pi)=0
For theta=(3pi)/2
dx/(d theta)=-4sin(2*(3pi)/2)=-4sin(3pi)=0
So, there are no horizontal tangents.
Now for vertical tangents, set the derivative of x equal to zero,
dx/(d theta)=-4sin(2theta)=0
=>sin(2theta)=0
=>2theta=0,pi,2pi,3pi
=>theta=0,pi/2,pi,(3pi)/2
Let's check for the above angles,
For theta=0
dy/(d theta)=2cos(0)=2 
For theta=pi/2
dy/(d theta)=2cos(pi/2)=0
For theta=pi
dy/(d theta)=2cos(pi)=-2
For theta=(3pi)/2
dy/(d theta)=2cos((3pi)/2)=0
So, the curve has vertical tangents at theta=0,pi
Now let's find the corresponding x and y coordinates by plugging theta in the parametric equation,
For theta=0
x=4cos^2(0)=4
y=2sin(0)=0
For theta=pi
x=4cos^2(pi)=4
y=2sin(pi)=0
So, the given parametric curve has vertical tangent at (4,0).