## Calculus 8th Edition

$\frac{\pi}{4}$
$sinx>cosx$ Let us simplify the above equation. $sinx=cosx$ Thus, $tanx=1$ Since, $tanx>0$ on first and third quadrant and $tanx<0$ on second and fourth quadrant. Thus, x must lie on $\frac{\pi}{4},\frac{5 \pi}{4}$ Hence, the solutions for the given equation is $\frac{\pi}{4}$