We need the Cartesian coordinates of the point of tangency:
#r((2pi)/3) = (2pi)/3 -sin(-(2pi)/3 + pi/3) = (2pi)/3 - sqrt(3)/2#
#x = rcos(theta) and y = rsin(theta)#
#x = ((2pi)/3 - sqrt(3)/2)cos((2pi)/3) and y = ((2pi)/3 - sqrt(3)/2)sin((2pi)/3)#
#x = sqrt(3)/4 - (2pi)/6 and y = (sqrt(3)pi)/3 - 3/4#
For the slope of the line we need #dy/dx# in polar form evaluated at #theta = (2pi)/3#. This reference Tangent Lines to polar equations gives us the equation:
#dy/dx = ((dr)/(d""theta)sin(theta) + rcos(theta))/((dr)/(d""theta)cos(theta) - rsin(theta)#
#(dr)/(d""theta) = 1 + cos( -theta + pi/3)#
Evaluated at #theta = (2pi)/3#
#(dr)/(d""theta) = 3/2 #
The slope, m, of the tangent line is:
#m = ((3/2)sin((2pi)/3) + ((2pi)/3 - sqrt(3)/2)cos((2pi)/3))/((3/2)cos((2pi)/3) - ((2pi)/3 - sqrt(3)/2)sin((2pi)/3)#
#m = ((3/2)(sqrt3/2) + ((2pi)/3 - sqrt(3)/2)(-1/2))/((3/2)(-1/2) - ((2pi)/3 - sqrt(3)/2)(sqrt3/2)#
#m = -(9sqrt(3) + sqrt(3) - 2pi)/(9 + (2pi - sqrt(3))(sqrt3)#
#m ~~ - 0.48#
Use the point-slope form of the equation of a line:
#y = m(x - x_0) + y_0#
#y = -0.48(x - sqrt(3)/4 + (2pi)/6) + (sqrt(3)pi)/3 - 3/4#
