Question 1179606

you found {{{x[1]=-pi/3}}} and got {{{y[1]=(e^(sqrt(3)(pi/3)))/2}}}

 which is a point of tangency


Secondly, find the slope of the tangent line, which is the derivative of the function, evaluated at the point: {{{m}}}={{{f}}}′{{{x}}}


{{{m= -e^(-sqrt(3)(-pi/3)) (sin(-pi/3) + sqrt(3) cos(-pi/3))}}}


{{{m=0}}} => if so, we have horizontal line {{{y=a}}}


in your case {{{y=y[1]}}} or 


{{{y=(e^(sqrt(3)(pi/3)))/2}}}-> tangent


check if that is tangent line

{{{ graph( 600, 600, -10, 10, -10, 10,e^(-sqrt(3)x)*cos(x),(e^(sqrt(3)pi/3))/2) }}}