The interval in which the $f(x)=\sin x-\cos x, 0 \leq x \leq 2 \pi$ is strictly decreasing is:

$\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$

The correct answer is Option (3) - $\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$

$f(x)=\sin x-\cos x$

$f'(x)=\cos x+\sin x=0$

so $x=\frac{3π}{4},\frac{7π}{4}$

$f'(x)<0$ for $x∈(\frac{3π}{4},\frac{7π}{4})$

so f is decreasing in $(\frac{3π}{4},\frac{7π}{4})$