Matematyka
gosia123
2017-06-25 02:10:59
Rozwiąż równanie: [latex]|1- sqrt{2} sinx|=2[/latex]
Odpowiedź
Katrix15
2017-06-25 04:50:13

[latex]|1-sqrt{2}sin x|=2[/latex] [latex]1-sqrt{2}sin x=2qquadlorqquad1-sqrt{2}sin x=-2[/latex] [latex]sin x=-cfrac{sqrt{2}}{2}qquadlorqquadsin x=cfrac{3sqrt{2}}{2}>1[/latex] [latex]x=-cfrac{pi}{4}+2kpiqquadlorqquad{x}=-cfrac{3pi}{4}+2kpi,qquad{k}inmathbb{C}[/latex]

rympał
2017-06-25 04:51:28

[latex]|1- sqrt{2}sinx| = 2[/latex] [latex]1- sqrt{2}sinx = 2[/latex] ∨ [latex]1- sqrt{2}sinx = -2[/latex] [latex] sqrt{2}sinx= -1 [/latex] ∨ [latex] sqrt{2}sinx = 3[/latex] [latex]sinx = -frac{ 1}{ sqrt{2} } = -frac{1}{ sqrt{2} }* frac{ sqrt{2} }{sqrt{2}} = -frac{ sqrt{2} }{2} [/latex] ∨ [latex]sinx = frac{3}{ sqrt{2}} extgreater 1[/latex] ⇒ x∈Ф x = [latex] -frac{3}{4} [/latex]π + 2kπ, k∈C ∨ x = [latex] -frac{1}{4} [/latex]π + 2k

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