Answer
$$y=\left(1-\sqrt{3}\right)\left(x-\pi \right)+\pi+3$$
Work Step by Step
Given
$$y= 3x +6\cos x\ \ \ \ \ \ (\pi/3,\pi+3 )$$
Since
\begin{aligned}
y'&= 3-6\sin(x)
\end{aligned}
Then
\begin{aligned}
m&= y'\bigg|_{(\pi/3,\pi+3)}\\
&= 3-\sin(\pi/3)\\
&=3-3\sqrt{3}
\end{aligned}
Hence the tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-\pi-3}{x-\pi/3}&=3-3\sqrt{3}\\
y-\pi-3&=( 3-3\sqrt{3})(x-\pi/3)\\
y &=\left(1-\sqrt{3}\right)\left(x-\pi \right)+\pi+3
\end{aligned}
