Ответ:
cosx\\\\sinx-cosx>0\\\\sinx-sin(\dfrac{\pi}{2}-x)>0\\\\2sin(x-\dfrac{\pi}{4})\cdot cos\dfrac{\pi}{4}>0\\\\\sqrt2\cdot sin(x-\dfrac{\pi}{4})>0\\\\\sin(x-\dfrac{\pi}{4})>0\\\\2\pi n" alt="sinx>cosx\\\\sinx-cosx>0\\\\sinx-sin(\dfrac{\pi}{2}-x)>0\\\\2sin(x-\dfrac{\pi}{4})\cdot cos\dfrac{\pi}{4}>0\\\\\sqrt2\cdot sin(x-\dfrac{\pi}{4})>0\\\\\sin(x-\dfrac{\pi}{4})>0\\\\2\pi n" align="absmiddle" class="latex-formula">