Posted by sh on .
Find an equation of the tangent line to the given curve at the given point
y=(1/cosx)2cosx at((pi/3),1)
y'=0+2sinx
slope = √3
y=mx+b
1=(√3)(pi/3)+b
b=(3√3pi)/3
3√3x3y+3√3pi=0
What did I do wrong? Thanks in advance.

Calculus 
sh,
Got it.