[tex]\displaystyle\bf\\Sinx > Cosx\\\\Sinx-Cosx > 0\\\\Sinx-Sin\Big(\frac{\pi }{2} -x\Big) > 0\\\\\\2Sin\frac{x-\frac{\pi }{2}+x }{2} \cdot Cos\frac{x+\frac{\pi }{2} -x}{2} > 0\\\\\\2Sin\Big(x-\frac{\pi }{4} \Big) Cos\frac{\pi }{4} > 0\\\\\\2Sin\Big(x-\frac{\pi }{4} \Big) \cdot\frac{\sqrt{2} }{2} > 0\\\\\\\sqrt{2} Sin\Big(x-\frac{\pi }{4} \Big) > 0\\\\\\Sin\Big(x-\frac{\pi }{4} \Big) > 0\\\\\\2\pi n < x-\frac{\pi }{4} < \pi +2\pi n ,n\in Z[/tex]

[tex]\displaystyle\bf\\\boxed{\frac{\pi }{4} +2\pi n < x < \frac{5\pi }{4} +2\pi n,n\in Z}[/tex]

Ответ:

.........................................................

Объяснение:

sinx - cosx > 0 | · [tex]\frac{\sqrt{2} }{2}[/tex]  ⇔ sinx · [tex]\frac{\sqrt{2} }{2}[/tex] - cosx · [tex]\frac{\sqrt{2} }{2}[/tex] >0 ⇔ sinx · cos[tex]\frac{\pi }{4}[/tex] - cosx · sin[tex]\frac{\pi }{4}[/tex] >0

⇔ sin(x - [tex]\frac{\pi }{4}[/tex]) >0  ⇔ 2πk + π > x - [tex]\frac{\pi }{4}[/tex] > 0 + 2πk ⇔ 2πk + [tex]\frac{5\pi }{4}[/tex] > x > [tex]\frac{\pi }{4}[/tex] + 2πk ; k ∈ Z