Answer
$y=2\sqrt{3}x+\dfrac{3-\pi\sqrt{3}}{3}$
Work Step by Step
$y=4\sin^{2}x$ $,$ $(\pi/6,1)$
Use the chain rule to evaluate the derivative of the function:
$y'=8\sin x(\sin x)'=8\sin x\cos x$
Evaluate the derivative in $\dfrac{\pi}{6}$ and simplify to obtain the slope of the tangent line at the given point:
$m=8\sin\dfrac{\pi}{6}\cos\dfrac{\pi}{6}=8\Big(\dfrac{1}{2}\Big)\Big(\dfrac{\sqrt{3}}{2}\Big)=2\sqrt{3}$
Both the slope of the tangent line and a point through which it passes are known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line:
$y-1=2\sqrt{3}\Big(x-\dfrac{\pi}{6}\Big)$
$y-1=2\sqrt{3}x-\dfrac{\pi\sqrt{3}}{3}$
$y=2\sqrt{3}x-\dfrac{\pi\sqrt{3}}{3}+1$
$y=2\sqrt{3}x+\dfrac{3-\pi\sqrt{3}}{3}$
