The solution:
The given
#r=2theta-3 sin((13theta)/8-(5 pi)/3)# at #theta=(7pi)/6#
#dy/dx=(r cos theta+r' sin theta)/(-r sin theta+r' cos theta)#
#dy/dx=([2theta-3 sin((13theta)/8-(5 pi)/3)] cos theta+[2-3(13/8)cos ((13theta)/8-(5 pi)/3)]*sin theta)/(-[2theta-3 sin((13theta)/8-(5 pi)/3)] sin theta+[2-3(13/8)cos ((13theta)/8-(5 pi)/3)] cos theta)#
Evaluating #dy/dx# at #theta=(7pi)/6#
#dy/dx=([2((7pi)/6)-3 sin((13((7pi)/6))/8-(5 pi)/3)] cos ((7pi)/6)+[2-3(13/8)cos ((13((7pi)/6))/8-(5 pi)/3)]*sin ((7pi)/6))/(-[2((7pi)/6)-3 sin((13((7pi)/6))/8-(5 pi)/3)] sin ((7pi)/6)+[2-3(13/8)cos ((13((7pi)/6))/8-(5 pi)/3)] cos ((7pi)/6))#
#dy/dx=([(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] cos ((7pi)/6)+[2-(39/8)cos ((91pi)/48-(5 pi)/3)]*sin ((7pi)/6))/(-[(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] sin ((7pi)/6)+[2-(39/8)cos ((91pi)/48-(5 pi)/3)] cos ((7pi)/6))#
#color(blue)(dy/dx=([(7pi)/3-3 sin((11pi)/48)] cos ((7pi)/6)+[2-(39/8)cos ((11pi)/48)]*sin ((7pi)/6))/(-[(7pi)/3-3 sin((11pi)/48)] sin ((7pi)/6)+[2-(39/8)cos ((11pi)/48)] cos ((7pi)/6)))#
#color(blue)(dy/dx=-0.92335731861741)#
#x=r cos theta=(2theta-3 sin((13theta)/8-(5 pi)/3))*cos theta#
#x=[(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] cos ((7pi)/6)#
#x=[(7pi)/3-3 sin((11pi)/48)] cos ((7pi)/6)#
#x=-4.6352670975528#
#y=r sin theta=(2theta-3 sin((13theta)/8-(5 pi)/3))*sin theta#
#y=[(7pi)/3-3 sin((91pi)/48-(5 pi)/3)] sin ((7pi)/6)#
#y=[(7pi)/3-3 sin((11pi)/48)] sin ((7pi)/6)#
#y=-2.6761727065385#
Using Point-Slope Form:
Equation of the tangent line is
#y-y_1=m(x-x_1)#
#y--2.6761727065385=-0.92335731861741(x--4.6352670975528)#
Check the graph:
God bless....I hope the explanation is useful.
