Answer
$y= 2\sqrt 3 (x) - (\pi/\sqrt 3) +1 $
Work Step by Step
$ 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 $