the derivative gives the slope of the tangent line at any value of x, so let's start by finding that.

y=(e^2cos4x)+1 ..........i suppose 2cos4x is the power of e

=> y ' = -8sin4x*e^(2cos4x)

when x = pi/6

y' = -8sin(2pi/3)*e^(2cos(2pi/3)) = -4sqrt(3)*e^-1 = -4sqrt(3)/e ...this is the slope at x=pi/6

now when x = pi/6

y = e^(2cos(2pi/3)) + 1 = e^-1 + 1

using the point slope form:

y-y1 = m(x - x1)

=> y - e^-1 - 1 = [-4sqrt(3)/e]x + [-4sqrt(3)/e](pi/6)

=> y = [-4sqrt(3)/e]x - (2sqrt(3)pi/3e - e^-1 - 1) ...........equation of the tangent line

and you can simplify this if you wish

...wow, this is a weird one, you might want to wait till someone checks this