Question

Помогите решить пожалуйста
SinX< cosX

Answer 1

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

$boxed{;;Bigg{sinalpha*cosbeta-cosalpha*sinbeta=sinBig(alpha-betaBig)};;}$

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

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