Let's begin by finding the y coordinate corresponding to the given x coordinate:
#y = ((pi)/6)^2sin^2(2(pi)/6)#
#y = (pi^2/36)(3/4) = pi^2/48#
The slope, n, of the normal line is:
#n = -1/m# where m is the slope of the tangent line.
To obtain the slope of the tangent line, we must compute the first derivative:
#f'(x) = 2xSin(2 x) (2xCos(2 x) + Sin(2 x))#
Evaluate at #x = pi/6#:
#m = f'(pi/6) = 2(pi/6)Sin(2 pi/6) (2(pi/6)Cos(2 pi/6) + Sin(2 pi/6))#
#m = 2(pi/6)(sqrt3/2) ((pi/6)+ sqrt3/2)#
#m = sqrt3(pi/6) ((pi/6)+ sqrt3/2)#
#m = sqrt3(pi/6)^2+ pi/4#
#m = (sqrt3pi^2 + 9pi)/36#
#n = -36/(sqrt3pi^2 + 9pi)#
Using the point-slope form of the equation of a line:
#y - y_1 = n(x - x_1)#
#y - pi^2/48 = -36/(sqrt3pi^2 + 9pi)(x - pi/6)#
