Answer
$x \in(\frac{\pi}{4} , \frac{\pi}{2}) \cup (\frac{5\pi}{4},\frac{3\pi}{2})$
Work Step by Step
$sinx>cosx$
Let us simplify the above equation.
$sinx=cosx$
Thus, $tanx=1$
This is true for x = $\frac{\pi}{4},\frac{5 \pi}{4}$
Therefore $tan(x) > 1$, for
$x \in(\frac{\pi}{4} , \frac{\pi}{2}) \cup (\frac{5\pi}{4},\frac{3\pi}{2})$
