Answer
$y=2x$
Work Step by Step
Given
$$y= 2x \sin x\ \ \ \ \ \ (\pi/2,\pi )$$
Since
\begin{aligned}
y'&= 2\left(\frac{d}{dx}\left(x\right)\sin \left(x\right)+\frac{d}{dx}\left(\sin \left(x\right)\right)x\right)\\
&= 2\left(\sin \left(x\right)+x\cos \left(x\right)\right)
\end{aligned}
Then
\begin{aligned}
m&= y'\bigg|_{(\pi/2,\pi)}\\
&= 2[\sin(\pi/2)+ \frac{\pi}{2} \cos(\pi/2)]\\
&= 2
\end{aligned}
Hence the tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-\pi}{x-\frac{\pi}{2}}&= 2\\
y-\pi &= 2x-\pi\\
y&= 2x
\end{aligned}
