\begin{aligned} &\text{Given curve: } y = 2x \sin x \\ &\text{Taking derivative on both sides, we get:} \\ &\frac{dy}{dx} = 2x \cos x + 2 \sin x \\ &\text{Substituting } x = \frac{\pi}{2} \text{ to find the slope at the point } \left( \frac{\pi}{2}, \pi \right): \\ &\frac{dy}{dx} = 2 \cdot \frac{\pi}{2} \cos \frac{\pi}{2} + 2 \sin \frac{\pi}{2} \\ &\frac{dy}{dx} = 2 \cdot 0 + 2 \cdot 1 = 2 \\ &\text{The equation of the tangent at } (x_1, y_1) \text{ with slope } m \text{ is given by:} \\ &(y - y_1) = m(x - x_1) \\ &\text{Substituting } (x_1, y_1) = \left( \frac{\pi}{2}, \pi \right) \text{ and } m = 2: \\ &(y - \pi) = 2 \left( x - \frac{\pi}{2} \right) \\ &\text{Expanding:} \\ &y - \pi = 2x - \pi \\ &y = 2x \\ &\text{Hence, the equation of the tangent line to the curve } y = 2x \sin x \text{ at the point } \left( \frac{\pi}{2}, \pi \right) \text{ is } 2x. \end{aligned}