Given

#f(x)=tan x-sin 2x# at #x_1=(4pi)/3#

Solve for the point #(x_1, y_1)# first

#y_1=tan x_1-sin 2*x_1#

#y_1=tan ((4pi)/3)-sin (2*(4pi)/3)#

#y_1=sqrt3-sqrt3/2#

#y_1=sqrt3/2#

Our point #(x_1, y_1)=((4pi)/3, sqrt3/2)#

Solve for the slope #m#

#f(x)=tan x-sin 2x#

find the first derivative #f' (x)=m#

#f' (x)=sec^2 x- 2*cos 2x#

#m=f' ((4pi)/3)=sec^2 ((4pi)/3)- 2*cos (2((4pi)/3))#

#m=(-2)^2-2(-1/2)#

#m=4+1=5#

For the normal line

#m_n=-1/m#

#m_n=-1/5#

Solve for the normal line:

#y-y_1=m_n(x-x_1)#

#y-sqrt3/2=-1/5(x-(4pi)/3)#

#y-sqrt3/2=-1/5x+(4pi)/15#

#color(blue)(y=-1/5x+(4pi)/15+sqrt3/2)#

Kindly see the graph of #f(x)=tan x-sin 2x# and the normal line #y=-1/5x+(4pi)/15+sqrt3/2# at the point #((4pi)/3, sqrt3/2)#

God bless....I hope the explanation is useful.