Question

Помогите решить пожалуйста
SinX< cosX

Answer 1

$\boldsymbol{sinx<cosx}\\\\\boldsymbol{sinx-cosx<0}\;\;\;\;\;\;\;|\;\;*\dfrac{\sqrt{2}}{2}\\\\\\\boldsymbol{\dfrac{\sqrt{2}}{2}*sinx-\dfrac{\sqrt{2}}{2}*cosx<\dfrac{\sqrt{2}}{2}*0}\\\\\\\boldsymbol{cos\dfrac{\pi}{4}*sinx-sin\dfrac{\pi}{4}*cosx<0}$

$\boxed{\;\;\Bigg{sin\alpha*cos\beta-cos\alpha*sin\beta=sin\Big(\alpha-\beta\Big)}\;\;}$

$\boldsymbol{sin\bigg(x-\dfrac{\pi}{4}\bigg)<0}\\\\\\\boldsymbol{-\pi+2\pi{n}<x-\dfrac{\pi}{4}<2\pi{n}}\\\\\\\boldsymbol{-\dfrac{3\pi}{4}+2\pi{n}<x<\dfrac{\pi}{4}+2\pi{n}}$

$\boldsymbol{\Bigg{\boxed{\;\;x\;\in\;\Bigg(-\dfrac{3\pi}{4}+2\pi{n}\;;\;\dfrac{\pi}{4}+2\pi{n}\Bigg)\;,\;\;n\;\in\;\mathbb{\boldsymbol{Z}}\;\;}}}$