Answer
The equation of the tangent line is $y=\sqrt{3}x+\dfrac{12-\pi\sqrt{3}}{6}$
Work Step by Step
$y=1+2\sin x$ $;$ $x=\dfrac{\pi}{6}$
Evaluate the derivative of the function given:
$y'=2\cos x$
Substitute $x=\dfrac{\pi}{6}$ in the derivative found and simplify to obtain the slope of the tangent line at the given point:
$m=2\cos\dfrac{\pi}{6}=2\Big(\dfrac{\sqrt{3}}{2}\Big)=\sqrt{3}$
Substitute $x=\dfrac{\pi}{6}$ in the original expression to obtain the $y$-coordinate of the point given:
$y$-coordinate $=1+2\sin\dfrac{\pi}{6}=1+2\Big(\dfrac{1}{2}\Big)=1+1=2$
The point is $\Big(\dfrac{\pi}{6},2\Big)$
The slope of the tangent line and a point through which it passes are now 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 at the given point:
$y-2=\sqrt{3}\Big(x-\dfrac{\pi}{6}\Big)$
$y-2=\sqrt{3}x-\dfrac{\pi\sqrt{3}}{6}$
$y=\sqrt{3}x-\dfrac{\pi\sqrt{3}}{6}+2$
$y=\sqrt{3}x+\dfrac{12-\pi\sqrt{3}}{6}$
The graph of the function and the tangent line is shown in the answer section.
