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