19. $(\operatorname{tg} x)^{\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1$.

19. $(\tan x)^{\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1$.
19. If $\operatorname{tg} x \neq 0$ and $\operatorname{tg} x \neq 1$, then $\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}=0$. This equation is equivalent to two equations $\sin x=1$ and $\sin x=\frac{1}{2}$. When $\sin x=1$, the function $\operatorname{tg} x$ is undefined. From the equation $\sin x=\frac{1}{2}$, we find $x_{\mathrm{I}}=\pi k+(-1)^{k} \frac{\pi}{6}$. If $\operatorname{tg} x=1$, then $x_{2}=\frac{\pi}{4}+\pi k$