Given, $f(x)=\sin x+2 \cos ^{2} x, x \in\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$
$\therefore \quad f^{\prime}(x)=\cos x-4 \cos x \cdot \sin x$
and $f^{\prime}(x)=-\sin x-4 \cos 2 x$
For maximum or minimum of $f(x)$
Put $\quad f^{\prime}(x)=0$
$\Rightarrow \quad \cos x-4 \cos x \cdot \sin x=0$
$\Rightarrow \quad \cos x(1-4 \sin x)=0$
$\Rightarrow \quad \cos x=0=\cos \frac{\pi}{2}$ and $\sin x \neq \frac{1}{4}$
$\because x \in\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right] \Rightarrow x=\frac{\pi}{2}$
Now, $\quad f^{''}\left(\frac{\pi}{2}\right)=-\sin \frac{\pi}{2}-4 \cos \pi$
$$
=-1+4=3>0(\mathrm{min})
$$
So, $f(x)$ is minimum at $x=\frac{\pi}{2}$
and its minimum value is $f\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}+2 \cos ^{2} \frac{\pi}{2}=1-2 \times 0=1$