Trigonometrische Gleichungen < Klassen 8-10 < Schule < Mathe < Vorhilfe
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| Status: |
(Frage) beantwortet  | | Datum: | 22:08 Mi 23.02.2011 | | Autor: | hase-hh |
| Aufgabe | Berechne alle Lösungen
a) [mm] 4*sin(x-3\pi) [/mm] = 3,5
b) cos(5x+7) = 0,8
c) sin(4x) = [mm] \bruch{1}{2} [/mm] |
Moin,
ist das so richtig?
a) [mm] 4*sin(x-3\pi) [/mm] = 3,5
[mm] sin(x-3\pi) [/mm] = 0,875
Substitution
z = x - [mm] 3\pi
[/mm]
sin(z) = 0,875 | arcsin
[mm] z_1 [/mm] = 1,0654 [mm] +2k\pi [/mm] ----- [mm] z_2 [/mm] = [mm] \pi [/mm] - [mm] z_1 +2k\pi
[/mm]
[mm] z_2 [/mm] = 2,0762 [mm] +2k\pi
[/mm]
Resubstitution
x [mm] -3\pi [/mm] = z
[mm] x_1 [/mm] - [mm] 3\pi [/mm] = 1,0654 [mm] +2k\pi [/mm] ----- [mm] x_2 [/mm] - [mm] 3\pi [/mm] = 2,0762 [mm] +2k\pi
[/mm]
[mm] x_1 [/mm] = 10,4902 [mm] +2k\pi [/mm] ----- [mm] x_2 [/mm] = 11,5010 [mm] +2k\pi
[/mm]
b) cos(5x+7) = 0,8
Substitution
z = 5x +7
cos(z) = 0,8 | arccos
[mm] z_1 [/mm] = 0,6435 [mm] +2k\pi [/mm] ----- [mm] z_2 [/mm] = [mm] \pi [/mm] - [mm] z_1 +2k\pi
[/mm]
[mm] z_2 [/mm] = 2,4981 [mm] +2k\pi
[/mm]
Resubstitution
5x + 7 = z
[mm] 5*x_1 [/mm] +7 = 0,6435 [mm] +2k\pi [/mm] ----- [mm] 5*x_2 [/mm] + 7 = 2,4981 [mm] +2k\pi
[/mm]
[mm] 5*x_1 [/mm] = -6,3565 [mm] +2k\pi [/mm] ----- [mm] 5*x_2 [/mm] = -4,5019 [mm] +2k\pi
[/mm]
[mm] x_1 [/mm] = -1,2713 [mm] +\bruch{2}{5}k\pi [/mm] ----- [mm] x_2 [/mm] = -0,90038 + [mm] \bruch{2}{5}k\pi
[/mm]
c) sin(4x) = [mm] \bruch{1}{2}
[/mm]
Substitution
z = 4x
sin(z) = [mm] \bruch{1}{2} [/mm] | arcsin
[mm] z_1 [/mm] = 0,5236 [mm] +2k\pi [/mm] ----- [mm] z_2 [/mm] = [mm] \pi [/mm] - [mm] z_1 +2k\pi
[/mm]
[mm] z_2 [/mm] = 2,618 [mm] +2k\pi
[/mm]
Resubstitution
4x = z
[mm] 4*x_1 [/mm] = 0,5236 [mm] +2k\pi 4*x_2 [/mm] = 2,618 [mm] +2k\pi
[/mm]
[mm] x_1 [/mm] = 0,1309 [mm] +\bruch{2}{4}k\pi [/mm] ----- [mm] x_2 [/mm] = 2,618 [mm] +\bruch{2}{4}k\pi
[/mm]
[mm] x_1 [/mm] = 0,1309 [mm] +\bruch{k\pi}{2} [/mm] ----- [mm] x_2 [/mm] = 2,618 + [mm] \bruch{k\pi}{2}
[/mm]
Danke & Gruß
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Hallo hase-hh,
> Berechne alle Lösungen
>
> a) [mm]4*sin(x-3\pi)[/mm] = 3,5
>
> b) cos(5x+7) = 0,8
>
> c) sin(4x) = [mm]\bruch{1}{2}[/mm]
> Moin,
>
> ist das so richtig?
>
>
> a) [mm]4*sin(x-3\pi)[/mm] = 3,5
>
> [mm]sin(x-3\pi)[/mm] = 0,875
>
> Substitution
>
> z = x - [mm]3\pi[/mm]
>
> sin(z) = 0,875 | arcsin
>
> [mm]z_1[/mm] = 1,0654 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
>
> [mm]z_2[/mm] = 2,0762 [mm]+2k\pi[/mm]
>
> Resubstitution
>
> x [mm]-3\pi[/mm] = z
>
> [mm]x_1[/mm] - [mm]3\pi[/mm] = 1,0654 [mm]+2k\pi[/mm] ----- [mm]x_2[/mm] - [mm]3\pi[/mm] = 2,0762
> [mm]+2k\pi[/mm]
>
> [mm]x_1[/mm] = 10,4902 [mm]+2k\pi[/mm] ----- [mm]x_2[/mm] = 11,5010 [mm]+2k\pi[/mm]
>
![[ok]](/images/smileys/ok.gif "[ok]")
>
>
> b) cos(5x+7) = 0,8
>
> Substitution
>
> z = 5x +7
>
> cos(z) = 0,8 | arccos
>
> [mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
>
> [mm]z_2[/mm] = 2,4981 [mm]+2k\pi[/mm]
>
> Resubstitution
>
> 5x + 7 = z
>
> [mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 2,4981
> [mm]+2k\pi[/mm]
>
> [mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -4,5019 [mm]+2k\pi[/mm]
>
> [mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,90038 +
> [mm]\bruch{2}{5}k\pi[/mm]
>
[mm]x_{1}[/mm] stimmt, [mm]x_{2}[/mm] mußt nochmal nachrechnen.
>
>
> c) sin(4x) = [mm]\bruch{1}{2}[/mm]
>
> Substitution
>
> z = 4x
>
> sin(z) = [mm]\bruch{1}{2}[/mm] | arcsin
>
> [mm]z_1[/mm] = 0,5236 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
>
> [mm]z_2[/mm] = 2,618 [mm]+2k\pi[/mm]
>
> Resubstitution
>
> 4x = z
>
> [mm]4*x_1[/mm] = 0,5236 [mm]+2k\pi 4*x_2[/mm] = 2,618 [mm]+2k\pi[/mm]
>
> [mm]x_1[/mm] = 0,1309 [mm]+\bruch{2}{4}k\pi[/mm] ----- [mm]x_2[/mm] = 2,618
> [mm]+\bruch{2}{4}k\pi[/mm]
>
> [mm]x_1[/mm] = 0,1309 [mm]+\bruch{k\pi}{2}[/mm] ----- [mm]x_2[/mm] = 2,618 +
> [mm]\bruch{k\pi}{2}[/mm]
>
[mm]x_{1}[/mm] stimmt., bei [mm]x_{2}[/mm] ist vergessen worden,
die Zahl 2,618 durch 4 zu dividieren.
>
> Danke & Gruß
Gruss
MathePower
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| Status: |
(Frage) beantwortet | | Datum: | 20:03 Do 24.02.2011 | | Autor: | hase-hh |
Moin Moin,
b) cos(5x+7) = 0,8
Substitution
z = 5x +7
cos(z) = 0,8 | arccos
[mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
[mm]z_2[/mm] = 2,4981 [mm]+2k\pi[/mm]
Resubstitution
5x + 7 = z
[mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 2,4981 [mm]+2k\pi[/mm]
[mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -4,5019 [mm]+2k\pi[/mm]
[mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,90038 + [mm]\bruch{2}{5}k\pi[/mm]
[mm]x_{1}[/mm] stimmt, [mm]x_{2}[/mm] mußt nochmal nachrechnen.
Bei [mm] x_2 [/mm] finde ich keinen Fehler?! Oder ist der Ansatz falsch???
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Hallo hase-hh,
> Moin Moin,
>
> b) cos(5x+7) = 0,8
>
> Substitution
>
> z = 5x +7
>
> cos(z) = 0,8 | arccos
>
> [mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
>
>
> [mm]z_2[/mm] = 2,4981 [mm]+2k\pi[/mm]
>
> Resubstitution
>
> 5x + 7 = z
>
> [mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 2,4981
> [mm]+2k\pi[/mm]
>
> [mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -4,5019 [mm]+2k\pi[/mm]
>
> [mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,90038 +
> [mm]\bruch{2}{5}k\pi[/mm]
>
>
>
> [mm]x_{1}[/mm] stimmt, [mm]x_{2}[/mm] mußt nochmal nachrechnen.
>
> Bei [mm]x_2[/mm] finde ich keinen Fehler?! Oder ist der Ansatz
> falsch???
>
Dann prüfe, ob [mm]x_{2}[/mm] die Gleichung
[mm]\cos\left(5*x_{2}+7\right)=0,8[/mm]
erfüllt.
Gruss
MathePower
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| Status: |
(Frage) beantwortet | | Datum: | 20:21 Do 24.02.2011 | | Autor: | hase-hh |
Moin Mathe-P.,
> Hallo hase-hh,
>
> > Moin Moin,
> >
> > b) cos(5x+7) = 0,8
> >
> > Substitution
> >
> > z = 5x +7
> >
> > cos(z) = 0,8 | arccos
> >
> > [mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
> >
>
> >
> > [mm]z_2[/mm] = 2,4981 [mm]+2k\pi[/mm]
> >
> > Resubstitution
> >
> > 5x + 7 = z
> >
> > [mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 2,4981
> > [mm]+2k\pi[/mm]
> >
> > [mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -4,5019 [mm]+2k\pi[/mm]
> >
> > [mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,90038 +
> > [mm]\bruch{2}{5}k\pi[/mm]
> >
> >
> >
> > [mm]x_{1}[/mm] stimmt, [mm]x_{2}[/mm] mußt nochmal nachrechnen.
> >
> > Bei [mm]x_2[/mm] finde ich keinen Fehler?! Oder ist der Ansatz
> > falsch???
> >
>
>
> Dann prüfe, ob [mm]x_{2}[/mm] die Gleichung
>
> [mm]\cos\left(5*x_{2}+7\right)=0,8[/mm]
>
> erfüllt.
cos(2,4981) = -0,8
und nun?
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Hallo hase_hh,
> Moin Mathe-P.,
>
> > Hallo hase-hh,
> >
> > > Moin Moin,
> > >
> > > b) cos(5x+7) = 0,8
> > >
> > > Substitution
> > >
> > > z = 5x +7
> > >
> > > cos(z) = 0,8 | arccos
> > >
> > > [mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
>
> > >
> >
> > >
> > > [mm]z_2[/mm] = 2,4981 [mm]+2k\pi[/mm]
> > >
> > > Resubstitution
> > >
> > > 5x + 7 = z
> > >
> > > [mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 2,4981
> > > [mm]+2k\pi[/mm]
> > >
> > > [mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -4,5019 [mm]+2k\pi[/mm]
> > >
> > > [mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,90038 +
> > > [mm]\bruch{2}{5}k\pi[/mm]
> > >
> > >
> > >
> > > [mm]x_{1}[/mm] stimmt, [mm]x_{2}[/mm] mußt nochmal nachrechnen.
> > >
> > > Bei [mm]x_2[/mm] finde ich keinen Fehler?! Oder ist der Ansatz
> > > falsch???
> > >
> >
> >
> > Dann prüfe, ob [mm]x_{2}[/mm] die Gleichung
> >
> > [mm]\cos\left(5*x_{2}+7\right)=0,8[/mm]
> >
> > erfüllt.
>
>
> cos(2,4981) = -0,8
>
> und nun?
[mm]z_{2}[/mm] ist falsch berechnet worden.
Vielmehr muß
[mm]z_{2}=\red{2}\pi-z_{1}+2*k*\pi, \ k \in \IZ[/mm]
sein.
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| Status: |
(Frage) beantwortet | | Datum: | 22:45 Do 24.02.2011 | | Autor: | hase-hh |
Jetzt habe ich:
b) cos(5x+7) = 0,8
Substitution
z = 5x +7
cos(z) = 0,8 | arccos
[mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] -----
[mm]z_2[/mm] = [mm]2*\pi[/mm] - [mm]z_1 +2k\pi[/mm]
[mm]z_2[/mm] = 5,6397 [mm]+2k\pi[/mm]
Resubstitution
5x + 7 = z
[mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 5,6397 [mm]+2k\pi[/mm]
[mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -1,3603 [mm]+2k\pi[/mm]
[mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,2721 + [mm]\bruch{2}{5}k\pi[/mm]
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Hallo hase-hh,
> Jetzt habe ich:
>
> b) cos(5x+7) = 0,8
>
> Substitution
> z = 5x +7
>
> cos(z) = 0,8 | arccos
>
> [mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] -----
>
> [mm]z_2[/mm] = [mm]2*\pi[/mm] - [mm]z_1 +2k\pi[/mm]
>
> [mm]z_2[/mm] = 5,6397 [mm]+2k\pi[/mm]
>
> Resubstitution
> 5x + 7 = z
>
> [mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 5,6397
> [mm]+2k\pi[/mm]
>
> [mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -1,3603 [mm]+2k\pi[/mm]
>
> [mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,2721 +
> [mm]\bruch{2}{5}k\pi[/mm]
>
Jetzt stimmt's.
Gruss
MathePower
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| Status: |
(Mitteilung) Reaktion unnötig  | | Datum: | 20:34 Do 24.02.2011 | | Autor: | abakus |
> Hallo hase-hh,
>
> > Moin Moin,
> >
> > b) cos(5x+7) = 0,8
> >
> > Substitution
> >
> > z = 5x +7
> >
> > cos(z) = 0,8 | arccos
> >
> > [mm]z_1[/mm] = 0,6435 [mm]+2k\pi[/mm] ----- [mm]z_2[/mm] = [mm]\pi[/mm] - [mm]z_1 +2k\pi[/mm]
Hallo,
hier hast du eine Quadrantenbeziehung angewendet, die NICHT für den Kosinus, sondern für den Sinus gilt.
Richtig ist z.B. [mm] z_2=-z_1 [/mm] (oder [mm] 2\pi-z_1).
[/mm]
Gruß Abakus
> >
>
> >
> > [mm]z_2[/mm] = 2,4981 [mm]+2k\pi[/mm]
> >
> > Resubstitution
> >
> > 5x + 7 = z
> >
> > [mm]5*x_1[/mm] +7 = 0,6435 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] + 7 = 2,4981
> > [mm]+2k\pi[/mm]
> >
> > [mm]5*x_1[/mm] = -6,3565 [mm]+2k\pi[/mm] ----- [mm]5*x_2[/mm] = -4,5019 [mm]+2k\pi[/mm]
> >
> > [mm]x_1[/mm] = -1,2713 [mm]+\bruch{2}{5}k\pi[/mm] ----- [mm]x_2[/mm] = -0,90038 +
> > [mm]\bruch{2}{5}k\pi[/mm]
> >
> >
> >
> > [mm]x_{1}[/mm] stimmt, [mm]x_{2}[/mm] mußt nochmal nachrechnen.
> >
> > Bei [mm]x_2[/mm] finde ich keinen Fehler?! Oder ist der Ansatz
> > falsch???
> >
>
>
> Dann prüfe, ob [mm]x_{2}[/mm] die Gleichung
>
> [mm]\cos\left(5*x_{2}+7\right)=0,8[/mm]
>
> erfüllt.
>
>
> Gruss
> MathePower
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| Status: |
(Mitteilung) Reaktion unnötig | | Datum: | 22:40 Do 24.02.2011 | | Autor: | hase-hh |
Moin abakus,
vielen Dank!!
Endlich eine Information, die mir hilft, zu verstehen, was an meinem Ansatz und vor allem warum mein Ansatz falsch ist.
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