Odpowiedź:
[tex]\huge\boxed{~~sin^{2} \alpha -sin\alpha > 2~~\Rightarrow~~sin\alpha \in \varnothing~~}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy :
Rozwiązanie:
[tex]sin^{2} \alpha -sin\alpha > 2\\\\t=sin\alpha ~~\land~~sin\alpha \in < -1;1 > ~~\Rightarrow~~t\in < -1;1 > \\\\t^{2}-t > 2~~\mid -2\\\\t^{2}-t-2 > 0\\a=1,~~b=-1,~~c=-2\\\\\Delta=(-1)^{2}-4\cdot 1\cdot (-2)\\\\\Delta=1+8\\\\\Delta=9\\\\\sqrt{\Delta} =3\\\\t_{1} =\dfrac{1-3}{2}=-1~~\lor~~ t_{2} =\dfrac{1+3}{2}=2\\\\a=1~~\Rightarrow~~a > 0~~\Rightarrow~~ramiona~~paraboli~~skierowane~~do~~gory\\\\[/tex]
[tex]t\in (-\infty;-1)\cup(2;+\infty)~~\land~~t=sin\alpha ~~\land~~sin\alpha \in < -1;1 > \\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~\huge\boxed{~~sin\alpha \in \varnothing~~}[/tex]
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Odpowiedź:
[tex]\huge\boxed{~~sin^{2} \alpha -sin\alpha > 2~~\Rightarrow~~sin\alpha \in \varnothing~~}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy :
Rozwiązanie:
[tex]sin^{2} \alpha -sin\alpha > 2\\\\t=sin\alpha ~~\land~~sin\alpha \in < -1;1 > ~~\Rightarrow~~t\in < -1;1 > \\\\t^{2}-t > 2~~\mid -2\\\\t^{2}-t-2 > 0\\a=1,~~b=-1,~~c=-2\\\\\Delta=(-1)^{2}-4\cdot 1\cdot (-2)\\\\\Delta=1+8\\\\\Delta=9\\\\\sqrt{\Delta} =3\\\\t_{1} =\dfrac{1-3}{2}=-1~~\lor~~ t_{2} =\dfrac{1+3}{2}=2\\\\a=1~~\Rightarrow~~a > 0~~\Rightarrow~~ramiona~~paraboli~~skierowane~~do~~gory\\\\[/tex]
[tex]t\in (-\infty;-1)\cup(2;+\infty)~~\land~~t=sin\alpha ~~\land~~sin\alpha \in < -1;1 > \\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~\huge\boxed{~~sin\alpha \in \varnothing~~}[/tex]