Question

помогите решить тригонометрию

Answer 1

$-4sinbigg(dfrac {3x}4+dfrac {pi}4bigg)>-2sqrt2~~~~Big|~:(-4)\\sinbigg(dfrac {3x}4+dfrac {pi}4bigg)<dfrac{sqrt2}2~~~~~~~~~~~~~~sinbeta=dfrac{sqrt2}2\\~~~~~Downarrow~Downarrow~Downarrow~~~~~~~~~~~~beta_1=dfrac{pi}4+2pi n;~~beta_2=dfrac{3pi}4+2pi k;~~n,kinmathbb Z\\\dfrac{3pi}4-2pi+2pi n<dfrac {3x}4+dfrac{pi}4<dfrac{pi}4+2pi n~~~~~Big|-dfrac{pi}4\\dfrac{3pi}4-dfrac{pi}4-2pi+2pi n<dfrac {3x}4<dfrac{pi}4-dfrac{pi}4+2pi n$

$dfrac{pi}2-2pi+2pi n<dfrac {3x}4<2pi n~~~~Big|cdot 4\\2pi-8pi+8pi n<3x<8pi n\~~~~-6pi+8pi n<3x<8pi n~~~~big|:3\\~~~~-2pi+dfrac{8pi n}3<x<dfrac{8pi n}3\\boxed{boldsymbol{xinbigg(-2pi+dfrac{8pi }3n;~~dfrac{8pi}3nbigg);~~ninmathbb Z}}$

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Графическое решение неравенства  $sinbeta<dfrac{sqrt2}2$ в приложении.