If f(Pi/8) = 0 and f '(x) = 3/2 + (cos (2x))^2, find the equation of the tangent line to the graph of this function at the point (Pi/8, 0).
A tangent line has the following form for all functions
$\displaystyle y-y_0=m(x-x_0)$
Where $\displaystyle (x_0,y_0)$ is the point at which you want the line to be tangent and m= the derivative evalauted at the point $\displaystyle (x_0,y_0)$
So $\displaystyle f'(\frac{\pi}{8})=\frac{3}{2}+\cos^2(2\cdot\frac{\ pi}{8})$
$\displaystyle =\frac{3}{2}+\cos^2(\frac{\pi}{4})=\frac{3}{2}+\le ft ( \frac{\sqrt{2}}{2}\right )^2$
$\displaystyle =\frac{3}{2}+\frac{2}{4}=2$
I leave it to you to plug it into the formula I provided