Question

Решить неравенство: sinx>cosx​

Answer 1

Ответ:

$sinx>cosx\\\\sinx-cosx>0\\\\sinx-sin(\dfrac{\pi}{2}-x)>0\\\\2sin(x-\dfrac{\pi}{4})\cdot cos\dfrac{\pi}{4}>0\\\\\sqrt2\cdot sin(x-\dfrac{\pi}{4})>0\\\\\sin(x-\dfrac{\pi}{4})>0\\\\2\pi n<x-\dfrac{\pi}{4}<\pi +2\pi n\ ,\ n\in Z\\\\\dfrac{\pi}{4}+2\pi n<x<\dfrac{5\pi}{4}+2\pi n\ ,\ n\in Z$