12.15 $f(x)=2 \sin x+\sin 2 x$ on $\left[0, \frac{3 \pi}{2}\right]$.
12.15 First, let's find the values of $f(x)$ at the ends of the given interval: $f(0)=0, f\left(\frac{3 \pi}{2}\right)=-2$, and then the critical points belonging to this interval. We have $f^{\prime}(x)=2 \cos x+2 \cos 2 x ; f^{\prime}(x)=0$, if $\cos x+\cos 2 x=0$, from which $2 \cos \frac{3 x}{2} \cos \frac{x}{2}=0$. From the equation $\cos \frac{3 x}{2}=0$ we find $\frac{3 x}{2}=\frac{\pi}{2}+\pi n$, i.e., $x=\frac{\pi}{3}+\frac{2}{3} \pi n$, and from the equation $\cos \frac{x}{2}=0$ we get $\frac{x}{2}=\frac{\pi}{2}+\pi n$, i.e., $x=\pi+2 \pi n, n \in Z$. The second solution is part of the first, so the solution to the equation is $x=\frac{\pi}{3}+\frac{2}{3} \pi n, \quad n \in \boldsymbol{Z}$. The interval $\left[0, \frac{3 \pi}{2}\right]$ contains the points $x_{1}=\frac{\pi}{3}$ and $x_{2}=\pi$. We find the values of $f(x)$ at the critical points: $f\left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{2}, f(\pi)=0$. Comparing the numbers $f(0), \quad f\left(\frac{3 \pi}{2}\right), \quad f\left(\frac{\pi}{3}\right)$, and $f(\pi)$, we conclude that $y_{\text {min }}=-2, y_{\text {max }}=\frac{3 \sqrt{3}}{2}$.

Answer: $y_{\text {min }}=-2, y_{\text {max }}=1.5 \sqrt{3}$.