Đặt $\alpha = \frac{2x}{1+x^2},$ ta có: $y=\sin \alpha + \cos 2\alpha=-2 \sin ^{2} \alpha + \sin \alpha + 1$
Đặt $t=\sin \alpha,\Rightarrow t \in [-1;1],$ ta có:
$y=-2t^2+t+1=f(t)$
$f'(t)=-4t+1;f'(t)=0\Leftrightarrow t=\frac{1}{4}.$
$f(-1)=-2;f(\frac{1}{4})=\frac{9}{8};f(1)=0$
Suy ra:
$\min y=f(-1)=-2$ tại $\sin \alpha =-1\Leftrightarrow \sin \frac{2x}{1+x^2}=-1\Leftrightarrow \frac{2x}{1+x^2}=-\frac{\pi}{2}+k2\pi$
$\max y=f(\frac{1}{4})=\frac{9}{8}$ tại $\sin \alpha =\frac{1}{4}\Leftrightarrow \sin \frac{2x}{1+x^2}=\frac{1}{4}\Leftrightarrow \frac{2x}{1+x^2}= \dots $