## Calculus 8th Edition

$y= 2\sqrt 3 (x) - (\pi/\sqrt 3) +1$
$y= 4(sin^{2} x)$ $y' = 8 sinx cosx$ $At (\pi/6, 1)$ $y' = 8 sin (\pi/6) cos (\pi/6)$ $y' = 2\sqrt 3$ Slope is equal to $2\sqrt 3$ Substitue $(\pi/6, 1)$ $1 = 2\sqrt 3 (\pi/6) + b$ $b= 1-2\sqrt 3 (\pi/6)$ $b= 1- (\pi/\sqrt 3)$ Thus the equation is $y= 2\sqrt 3 (x) - (\pi/\sqrt 3) +1$