Write the equations of the tangent and normal

Write the equations of the tangent and normal to the curve $y=\ln(\cos x)$ at the point $x=2\pi$.

Step 1: Value at $x=2\pi$

\[ y(2\pi) = \ln(\cos2\pi) = \ln1 = 0. \]

The point of tangency is $(2\pi,0)$.

Step 2: Derivative and slope

\[ y' = \frac{-\sin x}{\cos x} = -\tan x, \qquad y'(2\pi) = -\tan2\pi = 0. \]

The tangent is horizontal; the normal is vertical.

Step 3: Verification

\[y(2\pi)=0,\ y'(2\pi)=0\ \checkmark;\qquad \text{tangent } y=0 \text{ and normal } x=2\pi \text{ meet at } (2\pi,0)\ \checkmark\]